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
values |
|---|---|---|---|---|---|---|---|
Best approach for this -----easy to understand------- | combinations | 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$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
One Liner 99.91% Beats | combinations | 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:9ms\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 two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Backtracking cheatsheet + simple solution | combinations | 0 | 1 | ### Backtracking\nBacktracking is a general algorithm for finding all (or some) solutions to some computational problems which incrementally builds candidates to the solution and abandons a candidate ("backtracks") as soon as it determines that the candidate cannot lead to a valid solution. \n\nIt is due to this backtr... | 235 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Recursive Python3 Solution | combinations | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI am new to recursion so this combinations i solved by making it a subpart of *All possible sequences of given numbers*\nThis solution is not the best/optimal solution dont go by this just my attempt to make anyone also new like me to und... | 1 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Python short and clean. Generic solution. Functional programming. | combinations | 0 | 1 | # Approach\n\n# Complexity\n- Time complexity: $$O(n!)$$\n\n- Space complexity: $$O(n!)$$\n\n# Code\n```python\nclass Solution:\n def combine(self, n: int, k: int) -> list[list[int]]:\n T = TypeVar(\'T\')\n def combinations(pool: Iterable[T], r: int = None) -> Iterator[Iterator[T]]:\n pool =... | 1 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
🔥 Easy Backtracking for Generating Combinations | combinations | 1 | 1 | # Intuition\nIn this problem, we\'re asked to generate all possible combinations of \nk numbers out of n. My initial instinct was to use a recursive strategy, particularly the backtracking approach. Backtracking is a powerful method that systematically explores all potential combinations, distinguishing the ones that m... | 14 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
✅ [C++ , Java , Python] Simple solution with explanation (Backtracking)✅ | combinations | 1 | 1 | # Intuition\nIt is pretty clear from the constraints that the solution of 2<sup>n</sup> could work. We could use a backtracking approch in which if we see that the going path doesn\'t leads towards answer we could backtrack.\n<table>\n<tr>\n<th>Input Parameters</th>\n<th>Required time Complexity</th>\n</tr>\n<tr>\n<td>... | 12 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Python3 one-liner Solution | combinations | 0 | 1 | \n```\nclass Solution:\n def combine(self, n: int, k: int) -> List[List[int]]:\n return combinations(range(1,n+1),k)\n``` | 5 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Easy to understand [C++/Java/python3] | combinations | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding all possible combinations of k numbers chosen from the range [1, n]. To solve this, we can use a backtracking approach. The idea is to start with an empty combination and keep adding numbers to it one by one u... | 5 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Beats 98.68% | combinations | 0 | 1 | # Code\n```\nclass Solution:\n def combine(self, n: int, k: int) -> List[List[int]]:\n import itertools\n\n # Create an iterable (e.g., a list)\n iterable = [i+1 for i in range(n)]\n\n\n # Define the size of the combinations you want\n combination_size = k\n\n # Use itertool... | 1 | Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
You may return the answer in **any order**.
**Example 1:**
**Input:** n = 4, k = 2
**Output:** \[\[1,2\],\[1,3\],\[1,4\],\[2,3\],\[2,4\],\[3,4\]\]
**Explanation:** There are 4 choose 2 = 6 total combin... | null |
Python Intution using itertools module (combinations) | subsets | 0 | 1 | #Intuition and Approach:\nThe code generates all possible combinations of the elements in the nums list. It uses the itertools.combinations function, which returns all possible combinations of a given length.\n\nThe outer loop (for i in range(len(nums) + 1)) iterates over all possible lengths of combinations, starting ... | 1 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Python For beginners || Easy to understand || Recursion way for beginners | subsets | 0 | 1 | # Intuition\nTo solve the problem of generating all possible subsets of a given list of integers, one common approach is to use backtracking. The general idea is to explore all possible combinations by making choices at each step (either including or excluding an element) and backtrack when necessary. The recursive nat... | 4 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Recursive Solution in Python! | subsets | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nExplore all the possibilities, which is only possible using recursion. \n# Approach\n<!-- Describe your approach to solving the problem. -->\nFor each element in the array, you have two options either you will pick to include in subset or... | 0 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Simple backtracking solution in python3 (Beats 98% at runtime) | subsets | 0 | 1 | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to generate all possible subsets of a given set of distinct integers. The int... | 1 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
General Backtracking questions solutions in Python for reference : | subsets | 0 | 1 | I have taken solutions of @caikehe from frequently asked backtracking questions which I found really helpful and had copied for my reference. I thought this post will be helpful for everybody as in an interview I think these basic solutions can come in handy. Please add any more questions in comments that you think mig... | 210 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
🔥Easy Python 1-Line 🗣️🗣️ Math and Bitwise Operations🔥 | subsets | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCommonly the powerset is referred to as $2^S$ which convienently relates to the cardinality of the powerset $|P(S)| = 2^{|S|}$. This leads to a natural way of thinking of the powerset as a bit string where the bits determine if an element... | 3 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
One line, efficient, easy Python(with explanation) || Beats 95.50% 🔥🔥🔥 | subsets | 0 | 1 | \n```\nclass Solution:\n def subsets(self, s: List[int]) -> List[List[int]]:\n return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))\n\n```\nThis solution uses the combinations function from the itertools module to generate all possible combinations of elements in the set s, and then uses ch... | 2 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Simple For Loop, no DFS, easy to understand | subsets | 0 | 1 | # Intuition\nFor each level of iteration, we just need to firstly copy the powerset of the preivous iteration, secondly add the current number for the powerset that we are keep tracking on, and lastly extend the powersetCopy to the current powerset.\n\nNote: we need to add [] to the powerset in each iteration to cater ... | 3 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Recursive Backtracking for Generating Subsets in Python (Beats %99.39 at runtime) | subsets | 0 | 1 | # Intuition\nThe idea is to use a helper method that recursively constructs subsets\n\n# Approach\nWe\'ll define a helper_method that takes a \'start\' index as an argument.\nWe use a \'result\' list to store the generated subsets and a \'twos\' list to\nbuild each subset. We start by appending a copy of \'twos\' to \'... | 2 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Bit Manipulation same as Power set problem | subsets | 0 | 1 | # Power set Approach-->Bit Manipulations\n```\nclass Solution:\n def subsets(self, nums: List[int]) -> List[List[int]]:\n list1 = [[]]\n n=len(nums)\n for i in range(1,2**n):\n list2=[]\n for j in range(n):\n if (i&1<<j):\n list2.append(num... | 2 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
✔️ [Python3] BIT MANIPULATION (99.55% faster) (´▽`ʃƪ), Explained | subsets | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe idea is to use a bitmask where every bit represents an element in the `nums` list. If a bit is set to one, that means the corresponding element is active and goes to a subset. By subtracting the mask by 1 until i... | 33 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Beats 92.72% 78. Subsets with step by step explanation | subsets | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- The function backtrack takes two arguments: start and subset.\n- start is the start index of the array nums to consider in the current iteration.\n- subset is a list that contains the elements of the current subset.\n- In ... | 3 | Given an integer array `nums` of **unique** elements, return _all possible_ _subsets_ _(the power set)_.
The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
**Example 1:**
**Input:** nums = \[1,2,3\]
**Output:** \[\[\],\[1\],\[2\],\[1,2\],\[3\],\[1,3\],\[2,3\],\[1,2,3\]\]
... | null |
Solution | word-search | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n bool isExist = false;\n void backtrack(string &word, string &solution, int row, int col, int const rowSize, int const colSize, vector<vector<char>> &board,vector<vector<int>> &visited){\n if(solution.back() != word.at(solution.size()-1) || visited.at(row).at(col) > 0)... | 447 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Python | Faster than 98% w/ Proof | Easy to Understand | word-search | 0 | 1 | ```\ndef exist(self, board: List[List[str]], word: str) -> bool:\n\t# Count number of letters in board and store it in a dictionary\n\tboardDic = defaultdict(int)\n\tfor i in range(len(board)):\n\t\tfor j in range(len(board[0])):\n\t\t\tboardDic[board[i][j]] += 1\n\n\t# Count number of letters in word\n\t# Check if boa... | 205 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Python3, Backtracking, mega easy to understand | word-search | 0 | 1 | # Code\n```\nclass Solution:\n def exist(self, board: List[List[str]], word: str) -> bool:\n ans = False\n m, n = len(board[0]), len(board)\n dir = [[-1, 0], [0, 1], [1, 0], [0, -1]]\n\n def backtracking(c, r, cnt, visited):\n if cnt == len(word):\n nonlocal ans\... | 3 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Python DFS with word reverse to prevent TLE | word-search | 0 | 1 | Python DFS with word reverse based on last and first char frequency to reduce the unnecessary searches\n```\nclass Solution:\n def exist(self, board: List[List[str]], word: str) -> bool:\n ROWS, COLS = len(board), len(board[0])\n path = set()\n\n def dfs(r, c, i):\n if i == len(word):... | 3 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Exploring Words: Solving the Grid Word Search Problem with DFS Algorithm | word-search | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to find whether a given word can be constructed from the characters of the given grid such that adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once. The pr... | 32 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Easy Python Solution : 90.29 beats% || With Comments | word-search | 0 | 1 | # Code\n```\nclass Solution:\n def exist(self, board: List[List[str]], word: str) -> bool:\n ROWS, COLS = len(board), len(board[0])\n visited = set()\n\n def dfs(r,c,idx):\n # if idx == len(word), then word has been found\n if idx == len(word):\n return True\n\n ... | 10 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Click this if you're confused | word-search | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe must explore all possible word paths in the grid and prune this combinatorial search using backtracking.\n\nAt each cell we can explore the next adjacent cell by going Up, Down, Left, and Right. If we find a cell that matches the next ... | 1 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
With Pruning, Runtime reduced from 6000ms to 50ms | word-search | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt\'s better to check the edge case where it will return False.\n1. total characters in board is less than length of word\n2. minimum character of each word required to get answer are not in board\n\nThe answer will be same if one travers... | 2 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Python Elegant & Short | No TLE | word-search | 0 | 1 | ```\nclass Solution:\n """\n Time: O(n*m*k)\n Memory: O(k)\n\n where k - word length\n """\n\n VISITED = \'#\'\n\n def exist(self, board: List[List[str]], word: str) -> bool:\n n, m = len(board), len(board[0])\n\n beginning = end = 0\n for r in range(n):\n for c in... | 6 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Readable Python dfs solution | word-search | 0 | 1 | \n# Code\n```\nclass Solution:\n def exist(self, board: List[List[str]], word: str) -> bool:\n rows = len(board)\n cols = len(board[0])\n vis = set()\n\n def dfs(r, c, cur):\n if cur == len(word):\n return True\n if r < 0 or c < 0 or r >= rows or c >= ... | 2 | Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
**Example 1:**... | null |
Very Easy 100% (Fully Explained)(Java, C++, Python, JavaScript, C, Python3) | remove-duplicates-from-sorted-array-ii | 1 | 1 | Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.\n\t\t\tSince it is impossible to change the length of the array in some languages, you must instead have the result... | 80 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
EASIEST TO UNDERSTAND METHOD PYTHON3 BEATS 90% MEMORY | remove-duplicates-from-sorted-array-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n d = 0\n for i in range(0, len(nums) - 2):\n if nums[i + 1] < nums[i]:\n break\n elif nums[i + 1] == nums[i]:\n count = 1\n x = i + 2\n ... | 1 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Easy & Clear Solution Python3 | remove-duplicates-from-sorted-array-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n for i in range(len(nums)-2,0,-1):\n if(nums[i]==nums[i-1] and nums[i]==nums[i+1]):\n nums.pop(i+1)\n return len(nums)\n\n``` | 20 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Simple Python 3 Solution BEATS 98.7% 💻🤖🧑💻💻🤖 | remove-duplicates-from-sorted-array-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n if len(nums) <= 2:\n return len(nums)\n\n currentIndex = 2\n for i in range(2, len(nums)):\n if nums[i] != nums[currentIndex - 2]:\n nums[currentIndex] = nums[i]\n ... | 10 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Easiest Solution using Counter | remove-duplicates-from-sorted-array-ii | 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 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
simple 2-pointers solution. speed 94.67%, memory 86.44% | remove-duplicates-from-sorted-array-ii | 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. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 4 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Python3 || O(n) || Simple solution | remove-duplicates-from-sorted-array-ii | 0 | 1 | # Code\n```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n counter = Counter(nums)\n l = []\n for i,j in counter.items():\n if j>2:\n l+=[i]*2\n else:\n l+=[i]*j\n nums[:] = l \n```\n# **Please upvote guys if y... | 5 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
[Python] Intuitive optimal 2 pointers solution | remove-duplicates-from-sorted-array-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n2 pointers, one goes first to find the next unique element, the slower one waits until the first one find a new element and fill its current position.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe iterate over... | 2 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
O(n) | Simple Python Solution | remove-duplicates-from-sorted-array-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCompared to "Remove Elementes From A Sorted Array I": The key differences are the ">" symbol and l- 2.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nPlaceholder \'l\' stops at the 3rd repeated value while \'i\' con... | 6 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Beating 92.86% Memory Python Easiest Solution | remove-duplicates-from-sorted-array-ii | 0 | 1 | \n\n\n# Code\n```\nclass Solution(object):\n def removeDuplicates(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n c=0\n s=set(nums)\n for... | 1 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Simple Python solution with two pointers | remove-duplicates-from-sorted-array-ii | 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. -->\nThis function iterates through the sorted array using two pointers, slow and fast. The slow pointer keeps track of the first position to store a unique element, while ... | 3 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
Two Pointers Approach | remove-duplicates-from-sorted-array-ii | 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:O(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... | 4 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
best ever solution on leetcode i can bet on it | remove-duplicates-from-sorted-array-ii | 1 | 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: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)... | 2 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
3 line easy to understand code beats 86% in runtime & 99.31% in memory Very easy to understand | remove-duplicates-from-sorted-array-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 5 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
80. Remove Duplicates from Sorted Array II with step by step explanation | remove-duplicates-from-sorted-array-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- Initialize a variable i to 0.\n- Loop through each n in nums.\n- Check if i is less than 2 or n is greater than the nums[i-2].\n - If the condition is true, assign n to nums[i] and increment i by 1.\n- Return i.\n\n# Comp... | 9 | Given an integer array `nums` sorted in **non-decreasing order**, remove some duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears **at most twice**. The **relative order** of the elements should be kept the **same**.
Since it is impossible to change the len... | null |
✅ 100% Binary Search [VIDEO] with Rotation Handling - Optimal | search-in-rotated-sorted-array-ii | 1 | 1 | # Problem Understanding\n\nThe problem presents us with a modified search challenge where we\'re provided an array that has been rotated at an undetermined pivot. Unlike a standard sorted array, this comes with the added complexity of having been altered. Imagine a scenario where you take a sorted array, make a cut at ... | 66 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Easiest Solution Using Binary Search | search-in-rotated-sorted-array-ii | 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)$$ -->\nO(logn)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(... | 1 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Ex-Amazon explains a solution with a video, Python, JavaScript, Java and C++ | search-in-rotated-sorted-array-ii | 1 | 1 | # Intuition\nUsing modified binary search and add a few conditions before narrow the searching range.\n\n---\n\n# Solution Video\n## *** Please upvote for this article. *** \n# Subscribe to my channel from here. I have 243 videos as of August 10th\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmatio... | 8 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Basic solution | search-in-rotated-sorted-array-ii | 0 | 1 | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def search(self, nums: List[int], target: int) -> bool:\n for num in range(len(nums)):\n if nums[num] == target:\n return True\n return False\n``` | 1 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Python Easy One-Line Solution || 100% | search-in-rotated-sorted-array-ii | 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. -->\nWe have check weather element in list or not.\n\n# Complexity\n- Time complexity: \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: \n<!--... | 2 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Python short and clean. Binary-Search. | search-in-rotated-sorted-array-ii | 0 | 1 | # Approach\nSimilar to [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/) problem and its [Solution](https://leetcode.com/problems/search-in-rotated-sorted-array/solutions/3879752/python-short-and-clean-single-binary-search-2-solutions/).\nAn additional check... | 1 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Python easy solution | search-in-rotated-sorted-array-ii | 0 | 1 | # Code\n```\nclass Solution:\n def search(self, nums: List[int], target: int) -> bool:\n return True if target in nums else False\n``` | 4 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Python O(n) Binary Search Discontinuity ✅🔥 | search-in-rotated-sorted-array-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSuppose some sorted nondecreasing array `original` gets rotated.\n\nSince `original` is nondecreasing, we know that either the entire array is flat, or the last element must be greater than the first element.\n\nNo matter how much the arr... | 2 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Python3 one liner Solution | search-in-rotated-sorted-array-ii | 0 | 1 | \n```\nclass Solution:\n def search(self,nums:List[int],target:int)->bool:\n return True if target in nums else False \n``` | 4 | There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).
Before being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nu... | null |
Very Easy 100% (Fully Explained)(Java, C++, Python, JS, C, Python3) | remove-duplicates-from-sorted-list-ii | 1 | 1 | Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.\n# **Java Solution:**\nRuntime: 1 ms, faster than 93.57% of Java online submissions for Remove Duplicates from Sorted List II.\n```\nclass So... | 63 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
straight forward python3 faster than 99% | remove-duplicates-from-sorted-list-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse previous node to track which part to cut off\n# Approach\n<!-- Describe your approach to solving the problem. -->\nAdd dummy node incase the first number repeats.\nIterate through and cut off repeating nodes.\n# Complexity\n- Time com... | 2 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
Beats 99.93% ||🔥🔥 Python Linked List || 🔥🔥 | remove-duplicates-from-sorted-list-ii | 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 the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
Use simple Brute force | remove-duplicates-from-sorted-list-ii | 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- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\... | 2 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
82. Remove Duplicates from Sorted List II with step by step explanation | remove-duplicates-from-sorted-list-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Create a dummy node dummy and set its next node as the head of the linked list.\n2. Create a prev variable and initialize it with dummy.\n3. Use a while loop to traverse the linked list while head and head.next are not No... | 8 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
[Python 3] ✅ O(1) Space Solution 🔥 | remove-duplicates-from-sorted-list-ii | 0 | 1 | ##### UPVOTE IF YOU LIKE ;)\n\n\n# Code\n```py\nclass Solution:\n def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n # Base Case\n if not head:... | 3 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
Linked List and HashMap (Python) | easy to understand | remove-duplicates-from-sorted-list-ii | 0 | 1 | # Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n), for HashMap\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\nNote - This code will also work for unsorted lists\n\nLink - https://leetcode.com/problems/remove-duplicates-from-an-un... | 2 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
Python || 93.83% Faster || Easy || O(N) Solution | remove-duplicates-from-sorted-list-ii | 0 | 1 | ```\nclass Solution:\n def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n if head==None or head.next==None:\n return head\n d={}\n temp=head\n while temp!=None:\n if temp.val in d:\n d[temp.val]+=1\n else:\n ... | 4 | Given the `head` of a sorted linked list, _delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,2,3,3,4,4,5\]
**Output:** \[1,2,5\]
**Example 2:**
**Input:** head = \[1,1,1,2,3\]
**Outpu... | null |
Easiest Solution | remove-duplicates-from-sorted-list | 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 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Delete Duplicates | remove-duplicates-from-sorted-list | 0 | 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 deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n if not head or not head.next:\n ... | 3 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Easy Python3 Solution with Explanation | remove-duplicates-from-sorted-list | 0 | 1 | # Explanation \n\n- Set curr to the head and iterate till `curr` and `curr.next` is not None\n- if `curr` node and `curr.next` have same value then just skip that node and set `curr.next = curr.next.next` else `curr = curr.next` and just return the `head`\n\n# Code\n```\nclass Solution:\n def deleteDuplicates(self, ... | 3 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
6 lines code in python3 | remove-duplicates-from-sorted-list | 0 | 1 | \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 deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n curr=head\n while curr:\n ... | 9 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Beats 94.95% 83. Remove Duplicates from Sorted List with step by step explanation | remove-duplicates-from-sorted-list | 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:\nBeats\n94.95%\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Definition for singl... | 5 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Python3 | O(n) | Beats 95.80% (41 ms) | Simple with Explanation! | remove-duplicates-from-sorted-list | 0 | 1 | # Approach\n- We first deal with edge case of head being `None` rather than a `ListNode`\n- Next we create a new variable `curr` to point at our current node, starting with the `head` node\n- If `curr.next` is another node, we compare `curr.val` and `curr.next.val`\n - If the values are the same, we must remove one ... | 35 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Awesome Trick With 7 Lines of Code | remove-duplicates-from-sorted-list | 0 | 1 | \n# One Pointers Approach--->Time:O(N)\n```\nclass Solution:\n def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n current=head\n while current and current.next:\n if current.val==current.next.val:\n current.next=current.next.next\n else:\n... | 9 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Very Easy 100% (Fully Explained) (Java, C++, Python, JS, C, Python3) | remove-duplicates-from-sorted-list | 1 | 1 | # **Java Solution:**\nRuntime: 1 ms, faster than 92.17% of Java online submissions for Remove Duplicates from Sorted List.\n```\nclass Solution {\n public ListNode deleteDuplicates(ListNode head) {\n // Special case...\n if(head == null || head.next == null)\n return head;\n // Initia... | 33 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Remove Duplicates from the linked list (Runtime - Beats 85.11%) || Brute-force solution | remove-duplicates-from-sorted-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs the values are ``sorted``, we can just compare the current value with the next value to check for ``duplicates``. Traversing through the linked list and checking for this condition in each node will solve the problem.\n\n# Approach\n<!... | 5 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Python3 | Beats 90% + | Efficient Removal of Duplicates from a Sorted Linked List | remove-duplicates-from-sorted-list | 0 | 1 | # Intuition\nOur goal is to remove duplicate nodes while maintaining the order. Since the list is sorted, duplicate nodes will be adjacent to each other. We just need to skip them.\n\n# Approach\n1. We start with a pointer called `curr` that initially points to the `head` of the linked list.\n1. We use a while loop to ... | 3 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Python || 98.89% Faster || 6 Lines || O(n) Solution | remove-duplicates-from-sorted-list | 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 deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:\n temp=head\n while temp:\n while tem... | 6 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
✅|| Python|| Beats 94%|| Explained with Diagrams | remove-duplicates-from-sorted-list | 0 | 1 | **Runtime: 42 ms, faster than 94.70% of Python3 online submissions for Remove Duplicates from Sorted List.**\n**Memory Usage: 13.8 MB, less than 70.70% of Python3 online submissions for Remove Duplicates from Sorted List.**\n\nThe theme to solve this question is just like how we solve regular linked list problems. \n\n... | 8 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Python3 || Easy and Simple solution. | remove-duplicates-from-sorted-list | 0 | 1 | # I request you guys to please upvote if you find the solution helpful.\uD83E\uDEF6\uD83E\uDEF6\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 deleteDuplicates(self, head: ... | 1 | Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.
**Example 1:**
**Input:** head = \[1,1,2\]
**Output:** \[1,2\]
**Example 2:**
**Input:** head = \[1,1,2,3,3\]
**Output:** \[1,2,3\]
**Constraints:**
* The numb... | null |
Segment Tree & Divide and Conquer in Python | largest-rectangle-in-histogram | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nImplement Segment Tree recursively (build & query, update is not needed in this problem).\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nBuild segment tree: O(n) \nSegment tree range query: O(log n)\n\nDivi... | 1 | Given an array of integers `heights` representing the histogram's bar height where the width of each bar is `1`, return _the area of the largest rectangle in the histogram_.
**Example 1:**
**Input:** heights = \[2,1,5,6,2,3\]
**Output:** 10
**Explanation:** The above is a histogram where width of each bar is 1.
The l... | null |
Beats 99%✅ | O( n )✅ | Python (Step by step explanation) | largest-rectangle-in-histogram | 0 | 1 | # Intuition\nThe problem involves finding the largest rectangle that can be formed in a histogram of different heights. The approach is to iteratively process the histogram\'s bars while keeping track of the maximum area found so far.\n\n# Approach\n1. Initialize a variable `maxArea` to store the maximum area found, an... | 7 | Given an array of integers `heights` representing the histogram's bar height where the width of each bar is `1`, return _the area of the largest rectangle in the histogram_.
**Example 1:**
**Input:** heights = \[2,1,5,6,2,3\]
**Output:** 10
**Explanation:** The above is a histogram where width of each bar is 1.
The l... | null |
✔️ [Python3] MONOTONIC STACK t(-_-t), Explained | largest-rectangle-in-histogram | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe idea is to use a monotonic stack. We iterate over bars and add them to the stack as long as the last element in the stack is less than the current bar. When the condition doesn\'t hold, we start to calculate area... | 64 | Given an array of integers `heights` representing the histogram's bar height where the width of each bar is `1`, return _the area of the largest rectangle in the histogram_.
**Example 1:**
**Input:** heights = \[2,1,5,6,2,3\]
**Output:** 10
**Explanation:** The above is a histogram where width of each bar is 1.
The l... | null |
Superb Logic with Stack | largest-rectangle-in-histogram | 0 | 1 | [https://leetcode.com/problems/largest-rectangle-in-histogram/discuss/3544276/Superb-Logic-with-Stack](http://)# Monotonic Stack\n```\nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n stack=[]\n maxarea=0\n for i,v in enumerate(heights):\n start=i\n ... | 3 | Given an array of integers `heights` representing the histogram's bar height where the width of each bar is `1`, return _the area of the largest rectangle in the histogram_.
**Example 1:**
**Input:** heights = \[2,1,5,6,2,3\]
**Output:** 10
**Explanation:** The above is a histogram where width of each bar is 1.
The l... | null |
Largest Rectangle in Histogram with step by step explanation | largest-rectangle-in-histogram | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe above solution uses the monotonic stack algorithm to calculate the largest rectangle in histogram. The idea is to maintain a stack of indices, where the heights of the elements in the stack are in increasing order.\n\n- ... | 3 | Given an array of integers `heights` representing the histogram's bar height where the width of each bar is `1`, return _the area of the largest rectangle in the histogram_.
**Example 1:**
**Input:** heights = \[2,1,5,6,2,3\]
**Output:** 10
**Explanation:** The above is a histogram where width of each bar is 1.
The l... | null |
📢Stack and DP: Unveiling the Secrets to Count Maximal Rectangles || Full Explanation || Mr. Robot | maximal-rectangle | 1 | 1 | # Unlocking the Power of Algorithms: Solving the Maximal Rectangle Problem\n\nIf you\'re a coding enthusiast or a budding programmer, you\'ve probably heard about the Maximal Rectangle Problem. This fascinating problem is a classic in the world of computer science and is often used to demonstrate the power of dynamic p... | 6 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Maximal rectangle with recursion | maximal-rectangle | 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 `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Ex-Amazon explains a solution with a video, Python, JavaScript, Java and C++ | maximal-rectangle | 1 | 1 | # My youtube channel - KeetCode(Ex-Amazon)\nI create 142 videos for leetcode questions as of April 10, 2023. I believe my channel helps you prepare for the coming technical interviews. Please subscribe my channel!\n\n### Please subscribe my channel - KeetCode(Ex-Amazon) from here.\n\n**I created a video for this questi... | 16 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
DP / python solution O(M * N ^ 2) | maximal-rectangle | 0 | 1 | # Intuition\nDP\n\n# Approach\nReduce overlapping counting\n\n# Complexity\n- Time complexity:\nO(M * N ^ 2)\n\n- Space complexity:\nO(M * N)\n\n# Code\n```\nclass Solution:\n def maximalRectangle(self, matrix: List[List[str]]) -> int:\n # 1 0 1 2 3 \n # 0 1 0 1 0 \n # 1 2 0 1 2\n ... | 2 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Identical to question number 85 | maximal-rectangle | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIdentical with [85. Maximal Square](https://leetcode.com/problems/maximal-rectangle/description/), and my [solution](https://leetcode.com/problems/maximal-square/solutions/2814174/o-n-space-detailed-explanation-with-intuition/) to that qu... | 1 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Super Similar problem like Histogram | maximal-rectangle | 0 | 1 | https://leetcode.com/problems/maximal-rectangle/discuss/3544364/Super-Similar-problem-like-Histogram-------------------------->Histogram problem\n# Similar to Histogram Problem\n```\nclass Solution:\n def maximalRectangle(self, matrix: List[List[str]]) -> int:\n if not matrix:\n return 0\n m... | 3 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python Easy Solution || 100% || Kadane's Algorithm || | maximal-rectangle | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing **Kadane\'s Algorithm**\nHere we need to find continuos 1.\nAs per the Kadane\'s Algorithm we can find maximum subarray in an Array, but Here Array is consist of +ve number, We will manuplate the array \nif "0"--> -40000, becouse as... | 2 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python Solution | maximal-rectangle | 0 | 1 | \n\n\n# Explanation\nThe solution first initializes the "left" and "right" arrays to the leftmost and rightmost boundaries possible, respectively, and the "height" array to all z... | 1 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python || 96.75% Faster || DP || Histogram | maximal-rectangle | 0 | 1 | ```\nclass Solution:\n def maximalRectangle(self, matrix: List[List[str]]) -> int:\n def mah(heights: List[int]) -> int:\n st=[]\n maxArea=0\n for bar in heights+[-1]:\n step=0\n while st and st[-1][1]>=bar:\n w,h=st.pop()\n ... | 2 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python My Soln | maximal-rectangle | 0 | 1 | class Solution:\n\n def maximalRectangle(self, matrix: List[List[str]]) -> int:\n def maxHistogramRectangleArea(heights):\n area = 0\n stack = []\n for i, h in enumerate(heights):\n start = i\n while stack and stack[-1][1] > h:\n ... | 1 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Maximal Rectangle with step by step explanation | maximal-rectangle | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis solution uses the concept of histogram to solve the problem. For each row, it converts the binary values into heights of the bars and then calculates the largest rectangle in the histogram using the stack data structure... | 5 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python Easy Understandable Solution | maximal-rectangle | 0 | 1 | \n# Code\n```\nclass Solution:\n def maximalRectangle(self, matrix: List[List[str]]) -> int:\n if not matrix or not matrix[0]:\n return 0\n n, m = len(matrix), len(matrix[0])\n height = [0] * (m + 1)\n res = 0\n for row in matrix:\n for a in range(m):\n ... | 3 | Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return _its area_.
**Example 1:**
**Input:** matrix = \[\[ "1 ", "0 ", "1 ", "0 ", "0 "\],\[ "1 ", "0 ", "1 ", "1 ", "1 "\],\[ "1 ", "1 ", "1 ", "1 ", "1 "\],\[ "1 ", "0 ", "0 ", "1 ", "0 "\]\]
**Ou... | null |
Python easy solution | partition-list | 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 ListNode:\n def __init__(self, val=0, next=None):\n self.val = val\n self.next = next\n\nclass Solution:\n def partition(self... | 2 | Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.
You should **preserve** the original relative order of the nodes in each of the two partitions.
**Example 1:**
**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:**... | null |
✅ 100% Two Pointer [VIDEO] & Linked List + Visualization - Optimal | partition-list | 1 | 1 | # Problem Understanding\n\nIn the "Partition a Linked List Around a Value" problem, we are provided with the head of a linked list and a value `x`. The task is to rearrange the linked list such that all nodes with values less than `x` come before nodes with values greater than or equal to `x`. A key point is that the o... | 126 | Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.
You should **preserve** the original relative order of the nodes in each of the two partitions.
**Example 1:**
**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:**... | null |
python easy | partition-list | 0 | 1 | # Complexity:\nTime complexity: O(n), where n is the number of nodes in the linked list. We traverse the linked list once, performing constant-time operations.\nSpace complexity: O(1), as we use a constant amount of extra space for creating the two dummy nodes and a few pointers regardless of the input size.\n# Code\n`... | 1 | Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.
You should **preserve** the original relative order of the nodes in each of the two partitions.
**Example 1:**
**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:**... | null |
Python3 Solution | partition-list | 0 | 1 | \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 partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:\n d1=c1=ListNode(0)\n d2=c2=ListNode(0)\n ... | 1 | Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.
You should **preserve** the original relative order of the nodes in each of the two partitions.
**Example 1:**
**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:**... | null |
⬆️ Ultra C++ / Python with Explenation Two Pointers | partition-list | 0 | 1 | # Intuition\nThe problem essentially requires us to partition a linked list around a value `x` such that nodes with values less than `x` appear before nodes with values greater than or equal to `x`, while maintaining the relative order. \n\nAt first glance, one might consider rearranging the nodes in the original list.... | 1 | Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.
You should **preserve** the original relative order of the nodes in each of the two partitions.
**Example 1:**
**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:**... | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.