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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💯Faster✅💯 Lesser✅4 Methods🔥Binary Search🔥Two Pointers🔥Modified Binary Search🔥Linear Search | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 4 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nConsider a long list of numbers that are listed from smallest to largest, in that order. ... | 88 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python3 Solution | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | \n```\nclass Solution:\n def searchRange(self, nums: List[int], target: int) -> List[int]:\n n=len(nums)\n left=0\n right=n-1\n if target in nums:\n for i in range(0,n):\n if nums[left]==nums[right]:\n return [left,right] \n elif... | 2 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python solution easy to understand with optimise space and time complexity | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | If you like the solution please give \u2B50 to me\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code aims to find the starting and ending positions of a target element in a sorted array using a linear search approach.\n# Approach\n<!-- Describe your approach to solving the probl... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
100% BEATS - BEGINNER FRIENDLY CODE IN BINARY SEARCH (TIME - O(LOG N) ,SPACE - O(1)) | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis code defines a Solution with three member functions: search_first, search_last, and searchRange. The searchRange function aims to find the first and last occurrence of a target value in a sorted array of integers.\n\nHere\'s a breakd... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
✅ 3 ms || Intuitive solution || C++ || 2 methods || Binary Search | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # Intuition\n\nWe have to find elements in $$O(logn)$$ time complexity, So the first thing that comes into mind is Binary Search!\n\n# Approach\n\n1. The `search` function is used to find the insertion point of the target value in the sorted vector. It returns the index where the target value should be inserted or the ... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
✅✅✅Easy, understandable Python solution | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe have to apply binary search twice to find the 2 ends of the subarray filled with target\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIt first searches for the left index where the target is found and then for... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
python3 code | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\nThis code defines a function called searchRange that takes in a list of integers (nums) and a target integer (target) as inputs. The function returns a list containing the starting and ending index of the target integer in the input list. If the target integer is not present in the list, the function retur... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python code for finding the starting and ending position of a given target value | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\nIt first checks if the target exists in the nums list using the in operator. If not, it returns [-1, -1] to indicate that the target is not present.\n\nIf the target is found in nums, it calculates the starting index by using the index() method to find the first occurrence of the target. It calculates the ... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
🤯 CRAZY ONE-LINE SOLUTION IN PYTHON | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Approach\nUse `bisect_left` and `bisect_right` to find leftmost and rightmost indexes.\n\nIf the leftmost index gets out of range or there is no target in array, then return `[-1, -1]`.\n\nElse return `[leftmost, rightmost - 1]`\n\n# Complexity\n- Time complexity: $$O(\\log n)$$\n- Space complexity: $$O(\\log n)$$\n\... | 3 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
【Video】Give me 10 minutes - How we think about a solution - Binary Search | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | Welcome to my post! This post starts with "How we think about a solution". In other words, that is my thought process to solve the question. This post explains how I get to my solution instead of just posting solution codes or out of blue algorithms. I hope it is helpful for someone.\n\n# Intuition\nUsing Binary Search... | 49 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python Solution for Finding First and Last Occurrence | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to find the first and last occurrence of the given target value in a sorted array of integers.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n1. Check if the target value exists in the input array.\n\n2.... | 2 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
✅ Python || Binary Search || Time Complexity O(log(n)) || Beats 98.66% in Python3 || python3 | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\nAs the array is sorted and time complexity is in terms of log(N) , so we can use binary search.\n\n# Approach\n1. Brute Force : we simply Iterate from 0 to n-1 and use flag= True for starting position and storing it, after storing we do flag = False for ending position \nIn this approach Time Complexity is... | 2 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python3 Solution: Using Bisect Module | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | ```\nclass Solution:\n def searchRange(self, nums: List[int], target: int) -> List[int]:\n import bisect\n start=bisect.bisect_left(nums,target) \n end=bisect.bisect_right(nums,target)\n if start>end-1:\n return [-1,-1]\n return [start,end-1]\n```\n**Steps:**\n1. Im... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python Solution: Using Bisect Module | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | ```\nclass Solution:\n def searchRange(self, nums: List[int], target: int) -> List[int]:\n import bisect\n start=bisect.bisect_left(nums,target) \n end=bisect.bisect_right(nums,target)\n if start>end-1:\n return [-1,-1]\n return [start,end-1]\n```\n**Steps:**\n1. Im... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Easy to Understand || c++ || java || python | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe use a two-pointer approach, where low starts at the beginning of the array, and high starts at the end of the array.\nWe compare the elements at low and high with the target value.\nIf the element at low is less than the ... | 7 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Beats : 95.32% [49/145 Top Interview Question] | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n*O(log n) was hinting, `binary search`.*\n\n# Approach\nThis code block contains two methods: `searchRange` and `binarySearch`. The purpose of this code is to find the range of indices in a sorted list (`nums`) where a given `target` integer appears.\n\nThe `searchRange` method takes two parameters: `nums... | 2 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
😎Brute Force to Efficient Method 100% beats🤩 | Java | C++ | Python | JavaScript | C# | PHP | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # 1st Method :- Brute force\nThe brute force approach involves iterating through the entire array and checking each element to see if it matches the target element. Here are the steps:\n\n1. Initialize two variables, `first` and `last`, to -1. These variables will be used to store the `first` and last occurrence of the... | 15 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
🚩Beats 100% 🔥with 📈easy and optimised approach | 🚀Using Binary Search with Expanding Range | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this question is to perform two binary searches separately to find the leftmost and rightmost occurrences of the target value.\n\n---\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Implemen... | 5 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
C++/Python lower_bound & upper_bound(bisect) vs self made binary search | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse C++ lower_bound & upper_bound to solve this question. Python version uses bisect.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n2nd approach is self made binarch search. Use it twice and solve this question. Th... | 6 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Simple Solution. Binary search. Python. JavaScript | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n# Binary search for left and right bounds.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use binary search to find the target\'s index from left side of an array and from the right side of an array.\n# Complexit... | 0 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Solution | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing binary search algorithm twice to get the leftmost and rightmost index of the target value.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWriting two binary search in one, one for left and other one for righ... | 0 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Easy understandable approach Python3 | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\nThe problem asks as to find the range of a given number in a sorted list which is in ascending order like [1,2,4,6,6,6,7] target=6 that means we have return the first occurrence index here it\'s 3 and last occurrence that is 5 something like [3,5] and have to return [-1,-1] if we dont have the given value ... | 0 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Using Binay search in python.. | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | \nclass Solution:\n def searchRange(self, nums: List[int], target: int) -> List[int]:\n l=0\n l1=[]\n h=len(nums)-1\n res1=-1\n res2=-1\n while(l<=h):\n mid=(l+h)//2\n if target==nums[mid]:\n res1=mid\n h=mid-1\n ... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Best solution with tricky method with 99.59% beats || Biggners friendly||line by line explanation | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n```\ndef binarySearch(nums,target):\n```\n\n\ndef binarySearch(nums, target):: This line defines an inner function named binarySearch. It takes two arguments: nums (a list of integers) and target (the value to search for). This function w... | 1 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
🚩O(n) 📈Simple Solution with detailed explanation | 🚀Using Simple Approach | Beats 100% | find-first-and-last-position-of-element-in-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this question is to count the number of occurrences of the target value in the sorted array and then calculate the starting and ending positions based on the count.\n\n---\n\n# Approach\n<!-- Describe your approach to... | 6 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
Python O(logn) | find-first-and-last-position-of-element-in-sorted-array | 0 | 1 | ```python\nclass Solution:\n def searchRange(self, nums: List[int], target: int) -> List[int]:\n \n def search(x):\n lo, hi = 0, len(nums) \n while lo < hi:\n mid = (lo + hi) // 2\n if nums[mid] < x:\n lo = mid+1\n ... | 222 | Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
If `target` is not found in the array, return `[-1, -1]`.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[5,7,7,8,8,10\], target = 8
*... | null |
[VIDEO] Visualization of Binary Search Solution | search-insert-position | 0 | 1 | https://youtu.be/v4J_AWp-6EQ\n\nIf we simply iterated over the entire array to find the target, the algorithm would run in O(n) time. We can improve this by using binary search instead to find the element, which runs in O(log n) time (see the video for a review of binary search).\n\nNow, why does binary search run in ... | 7 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
💯 % || Easy Full Explanation || Beginner Friendly🔥🔥 | search-insert-position | 1 | 1 | # Intuition\nSince the array is sorted and we need to achieve O(log n) runtime complexity, we can use binary search. Binary search is an efficient algorithm for finding a target element in a sorted array.\n\n# Approach\n**Initialize Pointers:**\n- Initialize two pointers, low and high, to represent the range of element... | 13 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Python Easy Solution || 100% || Binary Search || | search-insert-position | 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: $$O(logn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g.... | 1 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Easy solution | search-insert-position | 0 | 1 | if the first one doesn\'t work, the second one will work\n\n# Code\n```\nclass Solution:\n def searchInsert(self, nums: List[int], target: int) -> int:\n for i in range(len(nums)):\n if nums[i]==target:\n return i\n nums.append(target)\n nums = sorted(nums) \n ... | 0 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Index Finder in a list | search-insert-position | 0 | 1 | # Intuition\nFirst understand the problem if ur a beginner\n\n# Approach\nI divide these problem into subproblems i spilt it by 3 parts first part is target value is greater than max depends on the i code inside the first part,Second part as target value is present or not if presented then depended program is executed.... | 0 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Easy python solution | search-insert-position | 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 sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Solution of search insert position problem | search-insert-position | 0 | 1 | # Approach\nSolved using binary search\n\n# Complexity\n- Time complexity:\n$$O(logn)$$ - as we are using binary search, which runs in O(logn) time.\n\n- Space complexity:\n$$O(1)$$ - as, no extra space is required.\n\n# Code\n```\nclass Solution:\n def searchInsert(self, nums: List[int], target: int) -> int:\n ... | 1 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
Beginner Friendly Solution(96.78% || 73.66%) | search-insert-position | 0 | 1 | **Binary Search Approach**\nRuntime: 43ms beats 96.78%\nMemory: 14.7MB beats 73.66%\n\n# Code\n```\nclass Solution:\n def searchInsert(self, nums: List[int], target: int) -> int:\n high = len(nums) - 1\n low = 0\n while high>=low:\n mid = (low+high)//2\n if nums[mid] == ta... | 1 | Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with `O(log n)` runtime complexity.
**Example 1:**
**Input:** nums = \[1,3,5,6\], target = 5
**Output:** 2
**Exa... | null |
C++ || Java || Python Solution (7 ms Beats 98.94% of users with C++) | valid-sudoku | 1 | 1 | # Intuition\nThe intuition is to use three matrices (representing rows, columns, and 3x3 subgrids) to keep track of the presence of digits in the Sudoku board. We iterate through each cell in the board and check whether the placement of a digit is valid based on the conditions of Sudoku.\n\n# Approach\n1. Initialize th... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Valid Sudoku Explanation | valid-sudoku | 0 | 1 | # 2023-11-21 || 3rd\n\n# Code\n```\nclass Solution:\n def isValidSudoku(self, board: List[List[str]]) -> bool:\n for i in range(9):\n for j in range(9):\n if board[i][j] != ".":\n target = board[i][j]\n target_i, target_j = i, j\n ... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
3rd | valid-sudoku | 0 | 1 | 3\n\n# Code\n```\nclass Solution:\n def isValidSudoku(self, board: List[List[str]]) -> bool:\n digits = ["1","2","3","4","5","6","7","8","9"]\n a = []\n b = []\n c = []\n for i in range(9):\n for j in range(9):\n num = board[i][j]\n if num i... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Beats 96.78% || Short 7-line Python solution (with detailed explanation) | valid-sudoku | 0 | 1 | \n1)It initializes an empty list called "res", which will be used to store all the valid elements in the board.\n\n2)It loops through each cell in the board using two nested "for" loops.\nFor each cell, it retrieves the value of the element in that cell and stores it in a variable called "element".\n\n3)If the element ... | 196 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Detailed solution runs in 100ms | valid-sudoku | 0 | 1 | \n# Code\n```py\nclass Solution:\n def isValidSudoku(self, board: List[List[str]]) -> bool:\n def CheckBox(box,seen):\n if box ==".":\n return True\n if box not in seen:\n seen.append(box)\n return True\n else:\n retu... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Beats : 96.06% [50/145 Top Interview Question] | valid-sudoku | 0 | 1 | # Intuition\n*A simple straight-forward solution!*\n\n# Approach\nThis code block is an implementation of the solution to `isValidSudoku`. Let\'s break down the code step by step:\n\n1. The code starts by importing the `collections` module, which is used to create defaultdicts (Why? Will explain!).\n \n2. Within the ... | 15 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Python | 2 WAYS | Sets / Strings | O(1) Time / Space | 90% T 88% S | valid-sudoku | 0 | 1 | # Solved 2 ways - Utilizing 3 Sets and Building Strings\n\n**Approach 1: Utilizing 3 Sets with HashMaps**\n\nWe want to traverse the sudoku board whilst populating and checking against a suite of 3 sets that may invalidate our board. We know that duplicate elements invalidate the board if they fall within the same row... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Sudoku Validator: The Art of Efficient Board Inspection in Python | valid-sudoku | 0 | 1 | # Intuition\nThe intuition behind the solution is to iterate through each row, column, and 3x3 box of the Sudoku board, keeping track of the numbers encountered. If a number is repeated in any row, column, or box, the board is not valid.\n# Approach\nThe approach involves using three lists (`row`, `column`, and `box`) ... | 1 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
[Python] - Clean & Simple - Faster than 96% in runtime 🔥🔥🔥 | valid-sudoku | 0 | 1 | To solve this problem we have to check if each value in sudoku does not repeat in its:\n\n1. Row\n1. Column\n1. Block\n\n# Complexity\n- Time complexity: $$O(n*n)$$ \n where n is fiexed here.\n so, Time Complexity will be O(81) ==> O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:... | 23 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
[Python / C++] a simple-minded brute-force solution in Python & CPP | valid-sudoku | 0 | 1 | ```cpp\nclass Solution {\npublic:\n bool isValidSudoku(vector<vector<char>>& board)\n {\n int x, y;\n\n x = -1;\n while (++x < 9)\n {\n y = -1;\n while (++y < 9)\n {\n if (board[x][y] != \'.\' && (!check_row(board, x, y) \n ... | 3 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Python solutions | Single Traversal | Single Dictionary | valid-sudoku | 0 | 1 | Just store the indexs of the numbers in a dictionary in `(x, y)` format. Then for every number check for same row, same col and same box condition.\nThis will require a single traversal. The same box condition can be checked using `pos[0]//3 == x//3 and pos[1]//3 == y//3` since `i//3` and `j//3` give the box position.\... | 78 | Determine if a `9 x 9` Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
1. Each row must contain the digits `1-9` without repetition.
2. Each column must contain the digits `1-9` without repetition.
3. Each of the nine `3 x 3` sub-boxes of the grid must contain... | null |
Easy Solution | sudoku-solver | 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. $$O(n)$$ --... | 6 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Simple Backtracking Approach | sudoku-solver | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe Sudoku puzzle can be solved using a backtracking algorithm. We\'ll start by iterating through the entire board, looking for empty cells. For each empty cell, we\'ll try to fill it with numbers from \'1\' to \'9\' while ensuring that t... | 4 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Python Easiest Recursive Solution. | sudoku-solver | 0 | 1 | ```\nclass Solution:\n def solveSudoku(self, board: List[List[str]]) -> None:\n n = 9\n \n \n def isValid(row, col, ch):\n row, col = int(row), int(col)\n \n for i in range(9):\n \n if board[i][col] == ch:\n ... | 73 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Readable and intuitive solution in python3. | sudoku-solver | 0 | 1 | # Intuition\nMaintain a row array and a column array to store theh values that are entered in a row. Also, maintain an array for the numbers entered in the current block.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nStart looping from 1 to 9 for each empty cells and check if the n... | 2 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Superb Logic Backtracking | sudoku-solver | 0 | 1 | # Backtracking\n```\nclass Solution:\n def solveSudoku(self, grid: List[List[str]]) -> None:\n return self.solve(grid)\n def solve(self,grid):\n for i in range(len(grid)):\n for j in range(len(grid[0])):\n if grid[i][j]==".":\n for c in range(1,len(grid)+... | 8 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Python || Striver's Solution || Backtracking || Easy | sudoku-solver | 0 | 1 | ```\nclass Solution:\n def solveSudoku(self, board: List[List[str]]) -> None:\n def safe(row,col,k,board): #function to check whether it is safe to insert a no. in board or not\n for i in range(9):\n if board[row][i]==k: # for checking element in same row\n return ... | 4 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Easy solution using recursion, backtracking and math | sudoku-solver | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n bool isValid(vector<vector<char>> &board, int row, int col, char c) {\n for (int i = 0; i < 9; i++) {\n if (board[i][col] == c)\n return false;\n if (board[row][i] == c)\n return false;\n if (board[3 * (row /... | 6 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Python Iterative backtracking solution.. Easy to understand | sudoku-solver | 0 | 1 | \n# Code\n```\nclass Solution:\n \n def solveSudoku(self, board: List[List[str]]) -> None:\n row = { i:[0]*10 for i in range(0,9) }\n col = { i:[0]*10 for i in range(0,9) }\n box = { i:[0]*10 for i in range(0,9) }\n n,m = len(board),len(board[0])\n for i in range(0,n):\n ... | 3 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
[Python3] - Straightforward and Readable Solution | sudoku-solver | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe easiest Method is just trial and error. We do this by finding and empty cell and checking which numbers are available for this cell. Then we input these numbers one by one and traverse deeper continuing this approach recursively until... | 6 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
🐍 97% faster || Clean & Concise || Well-Explained 📌📌 | sudoku-solver | 0 | 1 | ## IDEA:\n*The simple idea here is to fill a position and take the updated board as the new board and check if this new board has a possible solution.\nIf there is no valid solution for our new board we revert back to our previous state of the board and try by filling the next possible solution.*\n\nChecking if a value... | 33 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
[Fastest Solution Explained] O(n)time complexity O(n)space complexity | sudoku-solver | 1 | 1 | (Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, ***please upvote*** this post.)\n***Take care brother, peace, love!***\n\n```\n```\n\nThe best result for the code below is ***0ms / 38.2MB*** (beats 92.04% / 24.00%).\n* *** Java ***\n\n```\n\nclass Solution... | 8 | Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy **all of the following rules**:
1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly onc... | null |
Easy python solution: recursive approach | count-and-say | 0 | 1 | ```\nclass Solution:\n def countAndSay(self, n: int) -> str:\n if n==1:\n return "1"\n x=self.countAndSay(n-1)\n s=""\n y=x[0]\n ct=1\n for i in range(1,len(x)):\n if x[i]==y:\n ct+=1\n else:\n s+=str(ct)\n ... | 24 | The **count-and-say** sequence is a sequence of digit strings defined by the recursive formula:
* `countAndSay(1) = "1 "`
* `countAndSay(n)` is the way you would "say " the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.
To determine how you "say " a digit string, spli... | The following are the terms from n=1 to n=10 of the count-and-say sequence:
1. 1
2. 11
3. 21
4. 1211
5. 111221
6. 312211
7. 13112221
8. 1113213211
9. 31131211131221
10. 13211311123113112211 To generate the nth term, just count and say the n-1th term. |
Easy python solution: recursive approach | count-and-say | 0 | 1 | ```\nclass Solution:\n def countAndSay(self, n: int) -> str:\n if n==1:\n return "1"\n x=self.countAndSay(n-1)\n s=""\n y=x[0]\n ct=1\n for i in range(1,len(x)):\n if x[i]==y:\n ct+=1\n else:\n s+=str(ct)\n ... | 24 | The **count-and-say** sequence is a sequence of digit strings defined by the recursive formula:
* `countAndSay(1) = "1 "`
* `countAndSay(n)` is the way you would "say " the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.
To determine how you "say " a digit string, spli... | The following are the terms from n=1 to n=10 of the count-and-say sequence:
1. 1
2. 11
3. 21
4. 1211
5. 111221
6. 312211
7. 13112221
8. 1113213211
9. 31131211131221
10. 13211311123113112211 To generate the nth term, just count and say the n-1th term. |
My Python Solution | count-and-say | 0 | 1 | # Code\n```\nclass Solution:\n def countAndSay(self, n: int) -> str:\n if n==1:return "1"\n s="1"\n while n>1:\n tmp=[]\n st=[]\n cnt=0\n for i in s:\n if st and st[-1]==i:\n cnt+=1\n else:\n ... | 1 | The **count-and-say** sequence is a sequence of digit strings defined by the recursive formula:
* `countAndSay(1) = "1 "`
* `countAndSay(n)` is the way you would "say " the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.
To determine how you "say " a digit string, spli... | The following are the terms from n=1 to n=10 of the count-and-say sequence:
1. 1
2. 11
3. 21
4. 1211
5. 111221
6. 312211
7. 13112221
8. 1113213211
9. 31131211131221
10. 13211311123113112211 To generate the nth term, just count and say the n-1th term. |
My Python Solution | count-and-say | 0 | 1 | # Code\n```\nclass Solution:\n def countAndSay(self, n: int) -> str:\n if n==1:return "1"\n s="1"\n while n>1:\n tmp=[]\n st=[]\n cnt=0\n for i in s:\n if st and st[-1]==i:\n cnt+=1\n else:\n ... | 1 | The **count-and-say** sequence is a sequence of digit strings defined by the recursive formula:
* `countAndSay(1) = "1 "`
* `countAndSay(n)` is the way you would "say " the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.
To determine how you "say " a digit string, spli... | The following are the terms from n=1 to n=10 of the count-and-say sequence:
1. 1
2. 11
3. 21
4. 1211
5. 111221
6. 312211
7. 13112221
8. 1113213211
9. 31131211131221
10. 13211311123113112211 To generate the nth term, just count and say the n-1th term. |
Finding Combination Sum - Backtracking Approach | (Runtime - beats 80.99%) | 2 Way Decision Tree | combination-sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code aims to find all unique combinations of numbers from the candidates list that sum up to the given target. It uses depth-first search (DFS) to explore different combinations, and it backtracks whenever a combination doesn\'t lead ... | 1 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Beat 95% 48 ms with this solution||Backtracking and Recursive. Don't like and upvote ! | combination-sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo solve this problem, we need to list all possible combination\n==> Use backtracking and recursive\nWe can decrease runtime further by remove some unused branch e... | 2 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
General Backtracking questions solutions in Python for reference : | combination-sum | 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... | 265 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Python 3 | DFS/Backtracking & Two DP methods | Explanations | combination-sum | 0 | 1 | ### Approach \\#1. DFS (Backtracking)\n- Straight forward backtracking\n- `cur`: current combination, `cur_sum` current combination sum, `idx` index current at (to avoid repeats)\n- Time complexity: `O(N^(M/min_cand + 1))`, `N = len(candidates)`, `M = target`, `min_cand = min(candidates)`\n- Space complexity: `O(M/min_... | 145 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Backtraking 4 connected problems | combination-sum | 0 | 1 | # 1.combination sum\n```\nclass Solution:\n def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:\n res=[]\n def dfs(candidates,target,path,res):\n if target==0:\n res.append(path)\n return\n for i in range(len(candidates)):... | 12 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Most optimal solution using recursion and backtracking with complete explanation | combination-sum | 1 | 1 | \n\n# Approach\n- The findAns function is a recursive helper function that takes an index, the remaining target value, the candidates array, the answer vector (ans), and the current helper vector (helper).\n\n- The base case of the recursion is when the current index is equal to the size of the candidates array. If the... | 4 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Backtracking with Python3 well Explantation (Easy & Clear) | combination-sum | 0 | 1 | # Intuition:\nThe problem requires finding all possible combinations of elements from a given list of integers such that their sum is equal to a target value. The approach is to use backtracking to generate all possible combinations and keep track of the valid ones.\n\n# Approach:\nThe given code defines a class Soluti... | 11 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Python3 Recursive Solution | combination-sum | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:\n ans=[]\n n=len(candidates)\n def solve(idx,sum,lst):\n if idx>=n:\n if sum==target:\n ans.append(lst)\n return \n ... | 1 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Python || BackTracking || Easy TO Understand 🔥 | combination-sum | 0 | 1 | # Code\n```\nclass Solution:\n def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:\n def comb(i, target, ds, ans):\n if i == len(candidates):\n if target == 0 and ds not in ans:\n ans.append(ds.copy())\n return\n ... | 2 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
✅ Python short and clean solution explained | combination-sum | 0 | 1 | ```\nclass Solution:\n def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:\n self.ans = [] # for adding all the answers\n def traverse(candid, arr,sm): # arr : an array that contains the accused combination; sm : is the su... | 40 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Easy to understand python recursion!!! | combination-sum | 0 | 1 | \n# Code\n```\nclass Solution:\n def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:\n res = []\n\n def solve(cand, t, pre, arr):\n if pre == t:\n res.append(sorted(arr.copy()))\n \n if pre > t:\n return\n ... | 2 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
🚀 [VIDEO] Mastering A Journey into Recursion & DFS | combination-sum | 0 | 1 | # Intuition\nFor this problem, the initial thoughts revolve around the concept of recursive computation. Since we can choose a number multiple times, and we need to find all possible combinations, we can start by choosing a number, subtract it from the target, and then recursively repeat this process. If we hit the tar... | 3 | Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
The **same** number may be chosen from `candidates` an **unlimited number of... | null |
Bad Code! - Recursion in Py | combination-sum-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSort the array and try out all possible subsequence, thinking in terms of whatif my subsequence starts with this particular element. How it will progress? till where it go?\n# Approach\n<!-- Describe your approach to solving the problem. ... | 0 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Easy Explanation using Images and Dry run - Using Set & Without Set🐍👍 | combination-sum-ii | 1 | 1 | # Understanding the Question\nHELLO! Lets analyze how to solve this problem. well if you are solving this problem then i am assuming that you already solved **Combination Sum 1 problem** , Where we could pick any value multiple times. \nThis problem is a little different. Here we cant choose a single value multiple tim... | 17 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
[Python3] DFS solutions/templates to 6 different classic backtracking problems & more | combination-sum-ii | 0 | 1 | I have compiled solutions for all the 6 classic backtracking problems, you can practise them together for better understanding. Good luck with your preparation/interviews! \n\n[39. Combination Sum](https://leetcode.com/problems/combination-sum/)\n```\nclass Solution:\n def combinationSum(self, candidates: List[int],... | 149 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Backtracking concept | combination-sum-ii | 0 | 1 | # Backtracking Logic\n```\nclass Solution:\n def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:\n list1=[]\n candidates.sort()\n def dfs(candidates,target,path,list1):\n if target==0:\n list1.append(path)\n return\n ... | 7 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
✅ Clean and Easy solution - backtracking based Python | combination-sum-ii | 0 | 1 | # Intuition\nThe problem asks for all unique combinations of numbers in the candidates list that sum up to the target. Since each number can only be used once and we want unique combinations, **sorting** the candidates array can be advantageous. \n\nIt allows us to easily `skip over duplicate elements and make the summ... | 2 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Python3, Backtracking, mega easy to understand | combination-sum-ii | 0 | 1 | # Code\n```\nclass Solution:\n def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:\n ans = []\n candidates.sort()\n\n def backtracking(start, total, path):\n if total == target:\n ans.append(path)\n return\n \n ... | 3 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Most optimal solution using advanced backtracking and recursion | Beats 100% | Solution Explained | combination-sum-ii | 1 | 1 | \n# Approach\n- The `findAns` function is a recursive helper function that takes an index, the remaining target value, the candidates array, the answer vector (`ans`), and the current helper vector (`helper`).\n\n- The base case of the recursion is when the target value becomes 0. This means a valid combination has bee... | 7 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Runtime (Beats 97.33%) Combination Sum II with step by step explanation | combination-sum-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nSort the candidates list in ascending order.\nDefine a backtracking function that takes in 3 parameters: start index, a list path to store each combination, and the target value.\nIf the target value is 0, append the path li... | 2 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Python || BackTracking || Easy To Understand 🔥 | combination-sum-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def combinationSum2(self, candidates, target):\n candidates.sort() \n result = []\n def combine_sum_2(nums, start, path, result, target):\n if not target:\n result.append(path)\n return\n fo... | 2 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Python || 99.53% Faster || Backtracking || Easy | combination-sum-ii | 0 | 1 | ```\nclass Solution:\n def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:\n def solve(ind,target):\n if target==0:\n ans.append(op[:])\n return \n for i in range(ind,len(candidates)):\n if i>ind and candidates[i]... | 3 | Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.
Each number in `candidates` may only be used **once** in the combination.
**Note:** The solution set must not contain duplicate combinations.... | null |
Easy Solution | first-missing-positive | 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 unsorted integer array `nums`, return the smallest missing positive integer.
You must implement an algorithm that runs in `O(n)` time and uses constant extra space.
**Example 1:**
**Input:** nums = \[1,2,0\]
**Output:** 3
**Explanation:** The numbers in the range \[1,2\] are all in the array.
**Example 2:*... | Think about how you would solve the problem in non-constant space. Can you apply that logic to the existing space? We don't care about duplicates or non-positive integers Remember that O(2n) = O(n) |
BEATS 90% of the user ! Simplest answer . | first-missing-positive | 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 | Given an unsorted integer array `nums`, return the smallest missing positive integer.
You must implement an algorithm that runs in `O(n)` time and uses constant extra space.
**Example 1:**
**Input:** nums = \[1,2,0\]
**Output:** 3
**Explanation:** The numbers in the range \[1,2\] are all in the array.
**Example 2:*... | Think about how you would solve the problem in non-constant space. Can you apply that logic to the existing space? We don't care about duplicates or non-positive integers Remember that O(2n) = O(n) |
✅ 100% Solved | Both - Very Simple way and Optimized way to solve Trapping rain water problem. | trapping-rain-water | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code calculates trapped rainwater by iterating through each element in the input array. For each element, it determines the minimum height of barriers on both sides and calculates the trapped water if the minimum height exceeds the cu... | 30 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Easy Solution | trapping-rain-water | 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 `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Easy Python Solution) | trapping-rain-water | 0 | 1 | \n# Code\n```\nclass Solution:\n def trap(self, height: List[int]) -> int:\n left = 0\n right = len(height) - 1\n leftMax = 0\n rightMax = 0\n bucket = 0\n\n while left < right:\n if height[left] < height[right]:\n if leftMax < height[left]:\n ... | 0 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Algorithm in Python (Beats 80%) | trapping-rain-water | 0 | 1 | ## Trapping Rain Water\n\nGiven an array of non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.\n\n### Approach:\n\nWe can solve this problem by maintaining two arrays, `l` and `r`, which represent the maximum height of bars to the left... | 7 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Easy to understand 99% fast method (Python) | trapping-rain-water | 0 | 1 | # Intuition\n\nTo solve this problem, imagine that the map is a pyramid.\nWe have steps up to the column with the highest height.\nAfter the peak with the maximum height, we have steps going down.\nRight.\nNow we need to calculating total sum of bars with water.\n\n# Approach\n\nFirst step - for i in range(1,listLen):\... | 5 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Easy solution using 2 lists in python | trapping-rain-water | 0 | 1 | ```\nclass Solution:\n def trap(self, height: List[int]) -> int:\n data=[0 for i in range(len(height))]\n c=0\n front=[height[0]]\n back=[height[-1]]\n r=len(height)\n for i in range(1,len(height)):\n front.append(max(front[i-1],height[i]))\n back.appen... | 2 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Python3- Easy to understand | trapping-rain-water | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def trap(self, arr: List[int]) -> int:\n left,right=[],[]\n mx,mx1,n=0,0,len(arr)\n for i in arr:\n mx=max(mx,i)\n left.append(mx)\n for i in range(n-1,-1,-1):\n mx1=max(mx1,arr[i])\n right.append(mx1)\n ... | 1 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
Short solution that beats 80% (python) | trapping-rain-water | 0 | 1 | # Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def trap(self, height: List[int]) -> int:\n curr_max = 0\n max_left = [curr_max:= max(curr_max, h) for h in height]\n curr_max = 0\n max_right = [curr_max:= max(curr_max, h) for h... | 3 | Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
**Example 1:**
**Input:** height = \[0,1,0,2,1,0,1,3,2,1,2,1\]
**Output:** 6
**Explanation:** The above elevation map (black section) is represented by array \[0,1,0,2,1,... | null |
this problem solve | multiply-strings | 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 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Readable Python solution | multiply-strings | 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 two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Python3 || Beats 10% :( | multiply-strings | 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 two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Python one-liner top 5% | multiply-strings | 0 | 1 | # Approach\nSince we\'re using dynamically-typed language - we can convert a string into an integer, perform our calculations and convert the solution back to string in just one line, using built-in library :D\n\n\n# Code\n```\nclass Solution:\n def multiply(self, num1: str, num2: str) -> str:\n return str(in... | 0 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Easy Python Solution without Using Dictionary | multiply-strings | 0 | 1 | ```\nclass Solution:\n def multiply(self, num1: str, num2: str) -> str:\n if num1 == \'0\' or num2 == \'0\':\n return \'0\'\n \n def decode(num):\n ans = 0\n for i in num:\n ans = ans*10 +(ord(i) - ord(\'0\'))\n return ans\n\n def... | 47 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Python3 || Beats 96.57% || One liner beginner solution. | multiply-strings | 0 | 1 | # Code\n```\nclass Solution:\n def multiply(self, num1: str, num2: str) -> str:\n return str(int(num1)*int(num2))\n```\n# Please do upvote if you like the solution. | 4 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Python3 | multiply-strings | 0 | 1 | Ru\n## \u0418\u0434\u0435\u044F\n\u041D\u0443\u0436\u043D\u043E \u0441\u043E\u0437\u0434\u0430\u0442\u044C \u043C\u0430\u0441\u0441\u0438\u0432 \u0434\u043B\u0438\u043D\u043E\u0439 n+m, \u0433\u0434\u0435 n \u0438 m - \u0434\u043B\u0438\u043D\u044B \u0441\u0442\u0440\u043E\u043A num1 \u0438 num2 \u0441\u043E\u043E\u044... | 1 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
Simple Python Solution beats 90% | multiply-strings | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def multiply(self, num1: str, num2: str) -> str:\n dic={\n \'1\':1,\n \'2\':2,\n \'3\':3,\n \'4\':4,\n \'5\':5,\n \'6\':6,\n \'7\':7,\n \'8\':8,\n \'9\':9,\n \'0\':0... | 2 | Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
**Example 1:**
**Input:** num1 = "2", num2 = "3"
**Output:** "6"
**Ex... | null |
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