title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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Roman to Integer - Jesse | roman-to-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 5 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
test | roman-to-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
Very Simple Python Solution | roman-to-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe essentially start scanning adding all of the corresponding values for each character regardless of order. (e.g. "IX" is 11 not 9) Then, we check the order of the elements, and if we find that the order is reversed (i.e. "IX"), we make ... | 51 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
Easiest Beginner Friendly Sol || HashMap || C ++, Java, Python | roman-to-integer | 1 | 1 | # Intuition of this Problem:\nThe given code is implementing the conversion of a Roman numeral string into an integer. It uses an unordered map to store the mapping between Roman numerals and their corresponding integer values. The algorithm takes advantage of the fact that in a valid Roman numeral string, the larger n... | 49 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
python3 | fast | space-O(1) | time - O(n) | roman-to-integer | 0 | 1 | ```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n a={"I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000}\n sum=0 \n x=0\n while x<len(s):\n if x<len(s)-1 and a[s[x]]<a[s[x+1]]:\n sum += (a[s[x+1]] - a[s[x]])\n x += 2\n ... | 8 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
3 Solutions for Roman to Integer (Python3 🐍) | roman-to-integer | 0 | 1 | \n**1.First Solution (list comprehantion)**\n* Runtime: *55ms -> 121ms*\n* Memory runtime: 13.7MB -> 14MB\n\n**2. Secound Solution (for loop)**\n* Runtime: 48ms -> 109ms\n* Memory runtime: 13.8MB -> 14MB\n\n\n**3. Third Solution (enum, for fun)**\n* Runtime: 69ms -> 179ms\n* Memory runtime: 13.9MB -> 14.1MB\n\n\n\n\n ... | 3 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
⭐C#, Java, Python3,JavaScript Solution (Easy) | roman-to-integer | 1 | 1 | ### C#,Java,Python3,JavaScript different solution with explanation\n**\u2B50[https://zyrastory.com/en/coding-en/leetcode-en/leetcode-13-roman-to-integer-solution-and-explanation-en/](https://zyrastory.com/en/coding-en/leetcode-en/leetcode-13-roman-to-integer-solution-and-explanation-en/)\u2B50**\n\n**\uD83E\uDDE1See ne... | 16 | Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
**Symbol** **Value**
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, `2` is written as `II` in Roman numeral, just two ones added... | Problem is simpler to solve by working the string from back to front and using a map. |
✅Python3 || C++|| Java✅ 19 ms || Beats 99.91% | longest-common-prefix | 1 | 1 | # Please UPVOTE\uD83D\uDE0A\n\n\nThis code implements the longestCommonPrefix function that takes a list of strings v as input and returns the longest common prefix of all the strings. Here is an explanatio... | 1,587 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
[VIDEO] Visualization of Vertical Scanning Solution | longest-common-prefix | 0 | 1 | https://youtube.com/watch?v=SiNDN2M4dtQ\n\nThis solution is the <b>vertical scanning</b> approach that is discussed in the official solution, slightly modified for Python. The idea is to scan the the first character of every word, then the second character, etc. until a mismatch is found. At that point, we return a s... | 9 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
Longest Common Prefix!!! | longest-common-prefix | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of this function is to find the longest common prefix among a list of strings. the "prefix" refers to the initial sequence of characters that all strings in the given list share in common. More formally, it is the sequence of cha... | 4 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
Longest Common Prefix in Python | longest-common-prefix | 0 | 1 | # Intuition\nCompare each letter of each word to check if they match, and add them to an empty string until you hit a character that doesn\'t match. Return the string obtained so far.\n\n# Approach\nInitialize an empty string. Zip the list, so you get the first characters of each word together in a tuple, the second le... | 105 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
Python & startswith | longest-common-prefix | 0 | 1 | ```\nclass Solution:\n def longestCommonPrefix(self, strs: List[str]) -> str:\n \n pre = strs[0]\n \n for i in strs:\n while not i.startswith(pre):\n pre = pre[:-1]\n \n return pre \n``` | 150 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
[Python3] list(zip(*str)) | longest-common-prefix | 0 | 1 | * list(zip(*strs))\nstrs = ["flower","flow","flight"]\n```\nstrs = ["flower","flow","flight"]\nl = list(zip(*strs))\n>>> l = [(\'f\', \'f\', \'f\'), (\'l\', \'l\', \'l\'), (\'o\', \'o\', \'i\'), (\'w\', \'w\', \'g\')]\n```\n```\nclass Solution:\n def longestCommonPrefix(self, strs: List[str]) -> str:\n l = l... | 215 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
⭐C#,Java,Python3,JavaScript Solution ( Explanation ) | longest-common-prefix | 1 | 1 | **Which have included C#, Java, Python3,JavaScript solutions**\n**\u2B50[https://zyrastory.com/en/coding-en/leetcode-en/leetcode-14-longest-common-prefix-solution-and-explanation-en/](https://zyrastory.com/en/coding-en/leetcode-en/leetcode-14-longest-common-prefix-solution-and-explanation-en/)\u2B50**\n\n\n#### Example... | 11 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
Python Solution!! | longest-common-prefix | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def longestCommonPrefix(self, strs: List[str]) -> str:\n\n minimun_word: int = len(min(strs, key = len))\n preffix: str = ""\n\n for index in range(minimun_word):\n \n pivot: str = strs[0][index]\n window: list = [word[index] fo... | 2 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
Python | Brilliant Approach taught by my bf xD | longest-common-prefix | 0 | 1 | # Intuition\n**EDIT** - Earlier I used sorting, which took O(M * NLOGN) complexity.Instead we can use min() and max() , which takes O(N*M) time.complexity. (N is no.of elements in the array and M is size of the string)\n\nIf you sort the given array (lexicographically), the **first** and **last** word will be the least... | 23 | Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string `" "`.
**Example 1:**
**Input:** strs = \[ "flower ", "flow ", "flight "\]
**Output:** "fl "
**Example 2:**
**Input:** strs = \[ "dog ", "racecar ", "car "\]
**Output:** " "... | null |
[Python] 5 Easy Steps - Beats 97.4% - Annotated | 3sum | 0 | 1 | ```python\ndef threeSum(self, nums: List[int]) -> List[List[int]]:\n\n\tres = set()\n\n\t#1. Split nums into three lists: negative numbers, positive numbers, and zeros\n\tn, p, z = [], [], []\n\tfor num in nums:\n\t\tif num > 0:\n\t\t\tp.append(num)\n\t\telif num < 0: \n\t\t\tn.append(num)\n\t\telse:\n\t\t\tz.append(nu... | 888 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Beats : 99.48% [44/145 Top Interview Question] | 3sum | 0 | 1 | # Intuition\n*3 solutions, Each latter is more optimized!*\n\n# Complexity\n- Time complexity:\nO(n^2)\n ***Note, these are worst case complexity, optimization improves the runtime.***\n\n- Space complexity:\nO(n)\n ***Note, This is the dominant or the higher order space complexity, while optimizing the space incurred ... | 114 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
[VIDEO] Visualization of O(n^2) Solution (Two Pointers) | 3sum | 0 | 1 | https://youtu.be/IIxoo93bmPQ\n\nThis problem is similar to Two Sum, but the main differentiator is that there can now be multiple solutions and we must not return duplicate ones. This requires us to use a different approach, since Two Sum assumes that there is only one solution.\n\nThe key is to sort the array first. ... | 1 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python & Java Solution - 100% EXPLAINED ✔ | 3sum | 0 | 1 | # PYTHON CODE\n\n```\nclass Solution:\n def threeSum(self, nums: List[int]) -> List[List[int]]: \n nums.sort() # sorting cause we need to avoid duplicates, with this duplicates will be near to each other\n l=[]\n for i in range(len(nums)): #this loop will help to fix the one number i.e, i\n ... | 123 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python 3 pointer solution | 3sum | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def threeSum(self, nums: List[int]) -> List[List[int]]:\n # sort the list in ascending order\n nums.sort()\n\n # initialize an empty list to store the triplets\n ans = []\n\n # get the length of the list\n n = len(nums)\n\n # set ptr... | 1 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python Detailed Explanation | 3sum | 0 | 1 | # Explanation\nleft : leftmost index (iterate)\nmid : left + 1\nright: rightmost index (len(nums) - 1)\n\nThe main idea is that we fix `left` index. Then we find the sum of three values (left, mid, right) where `mid` is **initially `left + 1`** and `right` is **initially `len(nums) - 1`**.\n\n**If** the current sum is ... | 3 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python3 | 3sum | 0 | 1 | # Complexity\n- Time complexity: o(n^2)\n\n# Code\n```\nclass Solution:\n def threeSum(self, nums: List[int]) -> List[List[int]]:\n out = [] # Initialize an empty list to store the output triplets.\n nums.sort() # Sort the input list in ascending order.\n\n for i in range(0, len(nums) - 2): ... | 1 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Beginners Brute Force to Optimised (Two Pointer) Solution | 3sum | 0 | 1 | ## Brute force approach\n\n### Code\n```python\n# Brute Force\n# TC: O(n*n*n)\nclass Solution:\n def threeSum(self, nums: List[int]) -> List[List[int]]:\n arrLength = len(nums)\n\n ans = []\n\n\n for i_idx in range(0, arrLength - 2):\n for j_idx in range(i_idx + 1, arrLength - 1):\n ... | 5 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Easy python solution || Using two Pointers | 3sum | 0 | 1 | # Code\n```\nclass Solution:\n def threeSum(self, nums: List[int]) -> List[List[int]]:\n nums.sort()\n dic={}\n lst=[]\n for i in range(len(nums)):\n low=i+1\n high=len(nums)-1\n while low<high:\n sm=nums[i]+nums[low]+nums[high]\n ... | 2 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python code for N-Sum and bonus simplistic way of solving this question | 3sum | 0 | 1 | \n# Code\n```\nclass Solution:\n def threeSum(self, nums: List[int], target = 0) -> List[List[int]]:\n def findNsum(l, r, target, N, tmp, res):\n if r-l+1<N or N<2 or target < nums[l]*N or target > nums[r]*N:\n return\n if N == 2:\n while l < r:\n ... | 3 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
3Sum threesome | 3sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nit uses the 2sum but it calculates the target based on what the first number is. Then you use pointers to search for numbers that sum to the target. These 2 numbers in the list will be the other 2 numbers in the triplet. Keep searching th... | 0 | Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
Notice that the solution set must not contain duplicate triplets.
**Example 1:**
**Input:** nums = \[-1,0,1,2,-1,-4\]
**Output:** \[\[-1,-1,2\],\[-1,0... | So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x... |
Python easy solution | letter-combinations-of-a-phone-number | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def letterCombinations(self, digits: str) -> List[str]:\n digits_d = {\'2\':[\'a\',\'b\',\'c\'], \n \'3\':[\'d\',\'e\',\'f\'],\n \'4\':[\'g\',\'h\',\'i\'], \n \'5\':[\'j\',\'k\',\'l\'], \n \'6\':[\'m\',\'n\',\'o\'],... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
📞 100% Backtracking & Iterative [VIDEO] Letter Combinations of a Phone Number | letter-combinations-of-a-phone-number | 1 | 1 | # Intuition\nGiven a string containing digits from 2-9 inclusive, we need to return all possible letter combinations that the number could represent, just like on a telephone\'s buttons. To accomplish this, we present two different approaches:\n\n1. **Backtracking Approach**: This approach leverages recursion to explor... | 92 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Simple Backtracking Algorithm|| Beats 90% 📶 || Python ✅ | letter-combinations-of-a-phone-number | 0 | 1 | # Intuition\nBacktracking Using Recursion.Iterative deepening search with depth as the length of given string ```s``` length.\n\n# Approach\nThe solution involves mainly **3 steps**:\n\n1. Starting the process with an empty string and index as 0 indicating the pointer digits string.\n2. Iterating it until we reach the ... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Simple solution | letter-combinations-of-a-phone-number | 0 | 1 | # Code\n```\nfrom typing import Dict, List\n\nclass Solution:\n \n def get_mapping(self) -> Dict[str, List[str]]:\n return {\n "2": ["a", "b", "c"],\n "3": ["d", "e", "f"],\n "4": ["g", "h", "i"],\n "5": ["j", "k", "l"],\n "6": ["m", "n", "o"],\n ... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Go/Python O(n * 4^n) time | O(n * 4^n) space Recursion/Iterative | letter-combinations-of-a-phone-number | 0 | 1 | # Recursion\n- Time complexity: $$O(n * 4^n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n * 4^n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```golang []\nfunc letterCombinations(digits string) []string {\n answer := []string{}\n if len(digits) == ... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
✅✅Python Simple Solution 🔥🔥 Easiest🔥 | letter-combinations-of-a-phone-number | 0 | 1 | # Please UPVOTE \uD83D\uDC4D\n\n**!! BIG ANNOUNCEMENT !!**\nI am Giving away my premium content videos related to computer science and data science and also will be sharing well-structured assignments and study materials to clear interviews at top companies to my first 1000 Subscribers. So, **DON\'T FORGET** to Subscri... | 122 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Simple backtracking : ✅ Java 🔥 | | Python 🔥 | letter-combinations-of-a-phone-number | 1 | 1 | > # Simple backtracking solution : \n\n```java []\nclass Solution {\n public void buildMap(HashMap<Character,String> hmap)\n {\n hmap.put(\'2\',"abc");\n hmap.put(\'3\',"def");\n hmap.put(\'4\',"ghi");\n hmap.put(\'5\',"jkl");\n hmap.put(\'6\',"mno");\n hmap.put(\'7\',"pq... | 3 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
🔥💯Beats 100% 💪 Very Easy C++ || Java || Pyhton3 || Backtracking Detailed Solution 🔥💯 | letter-combinations-of-a-phone-number | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Mapping of Digits to Letters:** The problem is similar to generating all possible combinations of characters you can get from pressing the digits on a telephone keypad. We are given a mapping of each digit to a set of letters.\n\n**Back... | 33 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Python 1-liner. Functional programming. | letter-combinations-of-a-phone-number | 0 | 1 | # Complexity\n- Time complexity: $$O(4^n)$$\n\n- Space complexity: $$O(4^n)$$, including output space.\n\nwhere, `n is length of digits`.\n\n# Code\n```python\nclass Solution:\n def letterCombinations(self, digits: str) -> list[str]:\n phone = {\n \'1\': \'\' , \'2\': \'abc\', \'3\': \'def\',\n ... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Hey Knight! You need to see this! | letter-combinations-of-a-phone-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Ex-Amazon explains a solution with a video, Python, JavaScript, Java and C++ | letter-combinations-of-a-phone-number | 1 | 1 | # Intuition\nUsing backtracking to create all possible combinations\n\n---\n\n# Solution Video\n## *** Please upvote for this article. *** \n\nhttps://youtu.be/q0RJHpE92Zk\n\n# Subscribe to my channel from here. I have 239 videos as of August 3rd\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation... | 19 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
🚀 One-Liner Solution to Letter Combinations of a Phone Number | letter-combinations-of-a-phone-number | 0 | 1 | # Intuition\nGiven a string containing digits from 2-9 inclusive, we need to return all possible letter combinations that the number could represent. This can be achieved using a compact one-liner code that leverages Python\'s list comprehension and the itertools module.\n\n# Approach\n1. **Create Mapping**: Use list c... | 4 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
3 Approaches || Faster than 99% | letter-combinations-of-a-phone-number | 0 | 1 | \n\n# Complexity\n- Time complexity:\nO(n* 4^N)\n\n- Space complexity:\nO(4^N)\n\n\n```python1st []\nclass Solution:\n def letterCombinations(self, digits: str) -> List[str]:\n res = []\n digitTochar= {"2": "abc", \n "3": "def",\n "4": "ghi",\n ... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Simple python solution using Dqueue and dictionary which beats 88% | letter-combinations-of-a-phone-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent. Return the answer in **any order**.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
**Example 1:**
**Input:** di... | null |
Sum MegaPost - Python3 Solution with a detailed explanation | 4sum | 0 | 1 | If you\'re a newbie and sometimes have a hard time understanding the logic. Don\'t worry, you\'ll catch up after a month of doing Leetcode on a daily basis. Try to do it, even one example per day. It\'d help. I\'ve compiled a bunch on `sum` problems here, go ahead and check it out. Also, I think focusing on a subject a... | 146 | Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Examp... | null |
Python3 | Brute-force > Better > Optimal | Full Explanation | 4sum | 0 | 1 | - Approach\n - Brute-force\n - We keep four-pointers `i`, `j`, `k` and `l`. For every quadruplet, we find the sum of `A[i]+A[j]+A[k]+A[l]`\n - If this sum equals the target, we\u2019ve found one of the quadruplets and add it to our data structure and continue with the rest\n - Time Complexity: $... | 7 | Given an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:
* `0 <= a, b, c, d < n`
* `a`, `b`, `c`, and `d` are **distinct**.
* `nums[a] + nums[b] + nums[c] + nums[d] == target`
You may return the answer in **any order**.
**Examp... | null |
Python3 if and while case solution | remove-nth-node-from-end-of-list | 0 | 1 | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def removeNthFromEnd(self, head: Optional[ListNode], n: int... | 1 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Python Two Pointer solution with comments Easy to Understand | remove-nth-node-from-end-of-list | 0 | 1 | Please upvote once you get this :)\n```\nclass Solution:\n def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:\n fast = head\n slow = head\n # advance fast to nth position\n for i in range(n):\n fast = fast.next\n \n if not fast:\n retur... | 187 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Beats 100%✅ | O( n )✅ | Easy-To-Understand✅ | Python (Step by step explanation)✅ | remove-nth-node-from-end-of-list | 0 | 1 | # Intuition\nTo remove the nth node from the end of a singly-linked list, we can calculate the length of the list, and then identify the target node to be removed. It\'s important to handle edge cases where `n` is equal to the length of the list or when the list contains only one node.\n\n# Approach\n1. Initialize `tem... | 1 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Beats 100%✅ | O( n )✅ | Easy-To-Understand✅ | Python (Step by step explanation)✅ | remove-nth-node-from-end-of-list | 0 | 1 | # Intuition\nThe intuition for solving this problem is to use two pointers, `l` and `r`, to create a "window" of size `n` between them. This "window" will help us identify the nth node from the end of the linked list. \n\n# Approach\n1. Initialize `temp` as a new ListNode with a value of 0 and `head` as its next node. ... | 3 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Please Upvote ( If you find it helpful ) | remove-nth-node-from-end-of-list | 0 | 1 | # Intuition\nSimple intution delay the slow_pointer by n+1 after fast pointer and then iterate till fast_pointer.next does not becomes None. Delete the required node by slow_pointer.next = slow_pointer.next.next \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\... | 2 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Python3 | Beats 95% | Efficient Removal of Nth Node from the End of a Linked List | remove-nth-node-from-end-of-list | 0 | 1 | # Solution no. 01\nThe first solution helps you understand the basics with simple steps. \n\n# Intuition\n\nTo determine the node to remove, which is n positions from the end, we need to figure out how many positions we should move from the front to reach the desired node. By counting the total number of nodes in the l... | 9 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
19. Two pointer technique using | Python | remove-nth-node-from-end-of-list | 0 | 1 | # Code\n```\nclass Solution:\n def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:\n first, second = head, head\n for x in range(n):\n second = second.next\n while second:\n nex = first.next\n second = second.next\n if s... | 1 | Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
**Example 1:**
**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]
**Example 2:**
**Input:** head = \[1\], n = 1
**Output:** \[\]
**Example 3:**
**Input:** head = \[1,2\], n = 1
**Output:** \[1\]
**C... | Maintain two pointers and update one with a delay of n steps. |
Very Easy || 100% || Fully Explained (C++, Java, Python, JS, Python3) | valid-parentheses | 1 | 1 | An input string is valid if:\n1. Open brackets must be closed by the same type of brackets.\n2. Open brackets must be closed in the correct order.\n\n# **C++ Solution:**\nRuntime: 0 ms, faster than 100.00% of C++ online submissions for Valid Parentheses.\n```\nclass Solution {\npublic:\n bool isValid(string s) {\n ... | 385 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
Very Easy || 100% || Fully Explained (C++, Java, Python, JS, Python3) | valid-parentheses | 1 | 1 | An input string is valid if:\n1. Open brackets must be closed by the same type of brackets.\n2. Open brackets must be closed in the correct order.\n\n# **C++ Solution:**\nRuntime: 0 ms, faster than 100.00% of C++ online submissions for Valid Parentheses.\n```\nclass Solution {\npublic:\n bool isValid(string s) {\n ... | 385 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
[VIDEO] Step-by-Step Visualization Using a Stack | valid-parentheses | 0 | 1 | https://youtube.com/watch?v=YwvHeouhy6s\n\nSince the **last** bracket that is opened must also be the **first** one to be closed, it makes sense to use a data structure that uses the **Last In, First Out** (LIFO) principle. Therefore, a **stack** is a good choice here.\n\nIf a bracket is an *opening* bracet, we\'ll ju... | 28 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
[VIDEO] Step-by-Step Visualization Using a Stack | valid-parentheses | 0 | 1 | https://youtube.com/watch?v=YwvHeouhy6s\n\nSince the **last** bracket that is opened must also be the **first** one to be closed, it makes sense to use a data structure that uses the **Last In, First Out** (LIFO) principle. Therefore, a **stack** is a good choice here.\n\nIf a bracket is an *opening* bracet, we\'ll ju... | 28 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
Is this a Valid Parentheses??? | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code checks whether a given string containing only parentheses, brackets, and curly braces is valid in terms of their arrangement. A valid string should have each opening bracket matched with its corresponding closing bracket in the c... | 7 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
Is this a Valid Parentheses??? | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code checks whether a given string containing only parentheses, brackets, and curly braces is valid in terms of their arrangement. A valid string should have each opening bracket matched with its corresponding closing bracket in the c... | 7 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
PythonEasy | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 114 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
PythonEasy | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 114 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
Python code for determining if the string has valid parentheses (TC&SC: O(n)) | valid-parentheses | 0 | 1 | # Intuition\nThe code utilizes a stack data structure to keep track of open brackets (e.g., \'(\', \'[\', \'{\'). It iterates through the input string, and for each bracket character, it performs the following checks:\n\nIf it\'s an open bracket, it pushes it onto the stack.\nIf it\'s a closing bracket, it checks if th... | 2 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
Python code for determining if the string has valid parentheses (TC&SC: O(n)) | valid-parentheses | 0 | 1 | # Intuition\nThe code utilizes a stack data structure to keep track of open brackets (e.g., \'(\', \'[\', \'{\'). It iterates through the input string, and for each bracket character, it performs the following checks:\n\nIf it\'s an open bracket, it pushes it onto the stack.\nIf it\'s a closing bracket, it checks if th... | 2 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
👹Simple Stack Checker, readable, python, java 🐍☕ | valid-parentheses | 1 | 1 | # Intuition\nthink of it with simple logic, if (, then i+1 aka next element needs to be ), if not return false\n\n# Approach\ncreate stack\ncreate mapping of \'(\'to\')\' and so on\norder of if this is keys or this is values doesn\'t matter\nitterate through string, if i in values (in my code), push to stack aka append... | 2 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
👹Simple Stack Checker, readable, python, java 🐍☕ | valid-parentheses | 1 | 1 | # Intuition\nthink of it with simple logic, if (, then i+1 aka next element needs to be ), if not return false\n\n# Approach\ncreate stack\ncreate mapping of \'(\'to\')\' and so on\norder of if this is keys or this is values doesn\'t matter\nitterate through string, if i in values (in my code), push to stack aka append... | 2 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
Python 4ms Faster then 100% with explanation | valid-parentheses | 0 | 1 | ```\nclass Solution(object):\n\tdef isValid(self, s):\n """\n :type s: str\n :rtype: bool\n """\n d = {\'(\':\')\', \'{\':\'}\',\'[\':\']\'}\n stack = []\n for i in s:\n if i in d: # 1\n stack.append(i)\n elif len(stack) == 0 or d[st... | 434 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
Python 4ms Faster then 100% with explanation | valid-parentheses | 0 | 1 | ```\nclass Solution(object):\n\tdef isValid(self, s):\n """\n :type s: str\n :rtype: bool\n """\n d = {\'(\':\')\', \'{\':\'}\',\'[\':\']\'}\n stack = []\n for i in s:\n if i in d: # 1\n stack.append(i)\n elif len(stack) == 0 or d[st... | 434 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
solutuion for u :) | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
solutuion for u :) | valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
✅Beats 100%|💡O(n) solution with 100% Acceptance Rate with Easy and Detailed Explanation 💡😄 | valid-parentheses | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe use a stack to maintain the order of open brackets encountered.\nWhen a closing bracket is encountered, we check if it matches the last encountered open bracket.\nIf the brackets are balanced and correctly ordered, the stack should be ... | 7 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
✅Beats 100%|💡O(n) solution with 100% Acceptance Rate with Easy and Detailed Explanation 💡😄 | valid-parentheses | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe use a stack to maintain the order of open brackets encountered.\nWhen a closing bracket is encountered, we check if it matches the last encountered open bracket.\nIf the brackets are balanced and correctly ordered, the stack should be ... | 7 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
Simple way for python | valid-parentheses | 0 | 1 | \n# Code\n```\nclass Solution:\n def isValid(self, s: str) -> bool:\n ack = []\n lookfor = {\')\':\'(\', \'}\':\'{\', \']\':\'[\'}\n\n for p in s:\n if p in lookfor.values():\n ack.append(p)\n elif ack and lookfor[p] == ack[-1]:\n ack.pop()\n ... | 12 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-express... |
Simple way for python | valid-parentheses | 0 | 1 | \n# Code\n```\nclass Solution:\n def isValid(self, s: str) -> bool:\n ack = []\n lookfor = {\')\':\'(\', \'}\':\'{\', \']\':\'[\'}\n\n for p in s:\n if p in lookfor.values():\n ack.append(p)\n elif ack and lookfor[p] == ack[-1]:\n ack.pop()\n ... | 12 | Given a string `s` containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corres... | An interesting property about a valid parenthesis expression is that a sub-expression of a valid expression should also be a valid expression. (Not every sub-expression) e.g.
{ { } [ ] [ [ [ ] ] ] } is VALID expression
[ [ [ ] ] ] is VALID sub-expression
{ } [ ] is VALID sub-expression
... |
✔️ [Python3] MERGING, Explained | merge-two-sorted-lists | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nFor simplicity, we create a dummy node to which we attach nodes from lists. We iterate over lists using two-pointers and build up a resulting list so that values are monotonically increased.\n\nTime: **O(n)**\nSpace:... | 1,078 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
(VIDEO) Step-by-Step Visualization and Explanation | merge-two-sorted-lists | 0 | 1 | https://www.youtube.com/watch?v=E5XXiY6QnAs\n\nWe\'ll create a starting node called `head` and use a pointer called `current` to traverse the two lists. At each iteration of the loop, we compare the values of the nodes at list1 and list2, and point `current.next` to the <i>smaller</i> node. Then we advance the pointe... | 125 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Binbin's knight needs some rest | merge-two-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python3 Solution with a Detailed Explanation - dummy explained | merge-two-sorted-lists | 0 | 1 | [Linkedlists](https://realpython.com/linked-lists-python/) can be confusing especially if you\'ve recently started to code but (I think) once you understand it fully, it should not be that difficult. \n\nFor this problem, I\'m going to explain several ways of solving it **BUT** I want to make something clear. Something... | 484 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
O( n )✅ | Python (Step by step explanation)✅ | merge-two-sorted-lists | 0 | 1 | # Intuition\n<!-- Your intuition or thoughts about solving the problem -->\nMy intuition is to merge two sorted linked lists into a single sorted linked list while keeping track of the current nodes in both input lists.\n\n# Approach\n<!-- Describe your approach to solving the problem -->\n1. Initialize a new linked li... | 7 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Awesome Trick For Linked List----->Python3 | merge-two-sorted-lists | 0 | 1 | \n\n# Single Linked List----->Time : O(N)\n```\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n out=dummy=ListNode()\n while list1 and list2:\n if list1.val<list2.val:\n out.next=list1\n li... | 14 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Iterative approach | merge-two-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince the lists are already sorted iterate through both lists and add minNode to new list until we reach end of any list\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Prepare head and tail of new result list, in... | 0 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python3 easy solution | merge-two-sorted-lists | 0 | 1 | # Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n dummy=ListNode()\n ... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
✔️ Python Easy Solution | 99% Faster | Merge Two Sorted Lists | merge-two-sorted-lists | 0 | 1 | **IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE**\n\nVisit this blog to learn Python tips and techniques and to find a Leetcode solution with an explanation: https://www.python-techs.com/\n\n**Solution:**\n```\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[... | 97 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Simplest and READABLE Python code | merge-two-sorted-lists | 0 | 1 | \n# Code\n```\nclass Solution:\n def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:\n # Create a dummy node \n dummy = ListNode(0)\n cur = dummy\n \n while l1 and l2:\n if l1.val <= l2.val:\n cur.next = l1\n l1 = l1.next\n ... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
python iterative method | merge-two-sorted-lists | 0 | 1 | # Intuition\nThis question is asking you to merge 2 linked lists and return the head.\n\nThink of this question like you are inserting values between the head and the tail of a new empty list.\n\nWe start off by initalizing a head and movingtail to ListNode(). This makes head and tail both equal to an empty node. the h... | 6 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python3 Iterative and Recursive | merge-two-sorted-lists | 0 | 1 | \n# Complexity\n- Time complexity: O(m + n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n\n- Iterative \n\n```\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> O... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Simplest python solution beats 99% | merge-two-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python Simplest Iterative Solution with Explanation | Beg to Adv | Linked List | merge-two-sorted-lists | 0 | 1 | ```python\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n dummy = ListNode(0) # ... | 13 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python | Easy | merge-two-sorted-lists | 0 | 1 | ```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n \n ans= ListNode()\n ... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
21. Merge Two Sorted Lists | merge-two-sorted-lists | 0 | 1 | # Intuition\nPretty simple to understand\n\n# Approach\n1.Get elements from both Linked lists into a normal list\n2. Sort the list (List.sort(reverse=True))\n3. Then convert the new list into a linked list\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python Easy Solution in O(n) Complexity | merge-two-sorted-lists | 0 | 1 | ```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n list3=ListNode()\n te... | 3 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
In Place Merging in python : O(N) time O(1) space | merge-two-sorted-lists | 0 | 1 | # Intuition\nSo there are two approaches to solve the problem,\nfirst by creating a wholesome new list which is pretty easy, \nsecond by modifying the list itself \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nthis question can be solved by recursive or iterative approach\nI\'ll di... | 2 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Merging Two Sorted Linked List Using a Dummy Node - Algorithm Explained with Time and Space Analysis | merge-two-sorted-lists | 0 | 1 | # Intuition\nThe intuition of the solution is to iteratively compare the values of the nodes in `list1` and `list2`, and add the node with the smaller value to the resulting linked list. A dummy node is used as the starting node of the resulting linked list to make the implementation easier.\n\n# Approach\n<!-- Describ... | 1 | You are given the heads of two sorted linked lists `list1` and `list2`.
Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.
Return _the head of the merged linked list_.
**Example 1:**
**Input:** list1 = \[1,2,4\], list2 = \[1,3,4\]
**Output:**... | null |
Python, Java w/ Explanation | Faster than 96% w/ Proof | Easy to Understand | generate-parentheses | 1 | 1 | 1. The idea is to add `\')\'` only after valid `\'(\'`\n2. We use two integer variables `left` & `right` to see how many `\'(\'` & `\')\'` are in the current string\n3. If `left < n` then we can add `\'(\'` to the current string\n4. If `right < left` then we can add `\')\'` to the current string\n\n**Python Code:**\n``... | 1,292 | Given `n` pairs of parentheses, write a function to _generate all combinations of well-formed parentheses_.
**Example 1:**
**Input:** n = 3
**Output:** \["((()))","(()())","(())()","()(())","()()()"\]
**Example 2:**
**Input:** n = 1
**Output:** \["()"\]
**Constraints:**
* `1 <= n <= 8` | null |
Beats 92% using iteration. Detailed journey of stack explained | generate-parentheses | 0 | 1 | # Code\n```\nclass Solution:\n def generateParenthesis(self, n: int) -> List[str]:\n\n ##########################################################################\n ##########################################################################\n\n \'\'\'for n=2 amazingly explained recursion tree\n ... | 1 | Given `n` pairs of parentheses, write a function to _generate all combinations of well-formed parentheses_.
**Example 1:**
**Input:** n = 3
**Output:** \["((()))","(()())","(())()","()(())","()()()"\]
**Example 2:**
**Input:** n = 1
**Output:** \["()"\]
**Constraints:**
* `1 <= n <= 8` | null |
Python Iterative Solution ( Non Recursive) | generate-parentheses | 0 | 1 | \n# Code\n```\nclass Solution:\n def generateParenthesis(self, n: int) -> List[str]:\n res = [\'\']\n n1 = [0]\n n2 = [0]\n\n j = 0;\n while j < 2*n :\n rest = []\n n1t = []\n n2t = []\n for i in range(0,len(res)):\n if( n1... | 2 | Given `n` pairs of parentheses, write a function to _generate all combinations of well-formed parentheses_.
**Example 1:**
**Input:** n = 3
**Output:** \["((()))","(()())","(())()","()(())","()()()"\]
**Example 2:**
**Input:** n = 1
**Output:** \["()"\]
**Constraints:**
* `1 <= n <= 8` | null |
Python3 | Recursive approach | Step by step explanation ✅✅ | generate-parentheses | 0 | 1 | # Intuition\nThe problem aims to generate valid combinations of parentheses given an integer `n`. We need to find all possible combinations of `n` pairs of open and close parentheses.\n\n# Approach\n1. Initialize an empty list `ans` to store the valid combinations of parentheses and an empty stack `stack` to help in ge... | 3 | Given `n` pairs of parentheses, write a function to _generate all combinations of well-formed parentheses_.
**Example 1:**
**Input:** n = 3
**Output:** \["((()))","(()())","(())()","()(())","()()()"\]
**Example 2:**
**Input:** n = 1
**Output:** \["()"\]
**Constraints:**
* `1 <= n <= 8` | null |
Python3 | Recursive approach | Simple Logic | Step by step explanation | generate-parentheses | 0 | 1 | One thing we need to understand is, we need a way to add \u201C(\u201D and \u201C)\u201D to all possible cases and \nthen find a way to validate so that we don\u2019t generate the unnecessary ones.\n\nThe first condition is if there are more than 0 open / left brackets, we recurse with the right\nones. And if we have m... | 61 | Given `n` pairs of parentheses, write a function to _generate all combinations of well-formed parentheses_.
**Example 1:**
**Input:** n = 3
**Output:** \["((()))","(()())","(())()","()(())","()()()"\]
**Example 2:**
**Input:** n = 1
**Output:** \["()"\]
**Constraints:**
* `1 <= n <= 8` | null |
Beats 98% Easy to Understand | merge-k-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Python Simple solution | merge-k-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nNot very space efficient, but pretty fast. Move everything into an array, sort the array, and move everything back into a linked list.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexi... | 2 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Day 71 || Divide and Conquer || Easiest Beginner Friendly Sol | merge-k-sorted-lists | 1 | 1 | **NOTE 1 - PLEASE READ INTUITION AND APPROACH FIRST THEN SEE THE CODE. YOU WILL DEFINITELY UNDERSTAND THE CODE LINE BY LINE AFTER SEEING THE APPROACH.**\n\n**NOTE 2 - BEFORE SOLVING THIS PROBELM, I WILL HIGHLY RECOMMEND YOU TO SOLVE BELOW PROBLEM FOR BETTER UNDERSTANDING.**\n**21. Merge Two Sorted Lists :** https://lee... | 65 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Easy python solution | merge-k-sorted-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n extracting nodevalues in nodes list and sorting them back then after putting them in linked lists.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity... | 2 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
[Python] Easy Intuitive approach | merge-k-sorted-lists | 0 | 1 | # Intuition\nSince the constraints were:\n* length of lists, $$k = 10^4$$\n* length of each linked lists say $$n = 500$$\n \nIt was not a difficult choice to go for time complexity of $$O(kn)$$ \n\nAs we will get a TLE in python (generally) if we try to exceed $$O(10^8)$$.\nBut our solution takes $$O(10^4 * 500) < O(10... | 7 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Python3 and C++ || 95 ms || Beats 95.60% and EASY | merge-k-sorted-lists | 0 | 1 | # Please UPVOTE\uD83D\uDE0A\n\n\n\n# Python3\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\n... | 33 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
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