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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🍀Easy Java 🔥| | Python 🔥| | C++ 🔥 | find-k-pairs-with-smallest-sums | 1 | 1 | # *Code*\n```java []\nclass Pair{\n int idx1 = 0;\n int idx2 = 0;\n int sum = 0;\n Pair(int idx1,int idx2,int sum)\n {\n this.idx1 = idx1;\n this.idx2 = idx2;\n this.sum = sum;\n }\n}\nclass Solution {\n public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k)... | 6 | You are given two integer arrays `nums1` and `nums2` sorted in **ascending order** and an integer `k`.
Define a pair `(u, v)` which consists of one element from the first array and one element from the second array.
Return _the_ `k` _pairs_ `(u1, v1), (u2, v2), ..., (uk, vk)` _with the smallest sums_.
**Example 1:**... | null |
Python BFS, beating 97%. 52 ms | find-k-pairs-with-smallest-sums | 0 | 1 | Use BFS with heap. In each iteration, we get the pair with minimal sum and then push its neighboring right and bottom (if exists) into the heap. Repeat `k` or at most `m*n` times of such iterations. Time complexity `O(k lg k)`. Space complexity `O(k)` for the heap.\n\n```\n\n def kSmallestPairs(self, nums1: List[int... | 40 | You are given two integer arrays `nums1` and `nums2` sorted in **ascending order** and an integer `k`.
Define a pair `(u, v)` which consists of one element from the first array and one element from the second array.
Return _the_ `k` _pairs_ `(u1, v1), (u2, v2), ..., (uk, vk)` _with the smallest sums_.
**Example 1:**... | null |
[Python] Simple heap solution explained | find-k-pairs-with-smallest-sums | 0 | 1 | ```\nclass Solution:\n def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:\n hq = []\n heapq.heapify(hq)\n \n # add all the pairs that we can form with\n # all the (first k) items in nums1 with the first\n # item in nums2\n for i i... | 18 | You are given two integer arrays `nums1` and `nums2` sorted in **ascending order** and an integer `k`.
Define a pair `(u, v)` which consists of one element from the first array and one element from the second array.
Return _the_ `k` _pairs_ `(u1, v1), (u2, v2), ..., (uk, vk)` _with the smallest sums_.
**Example 1:**... | null |
C++ || Binary Search || Easiest Beginner Friendly Sol | guess-number-higher-or-lower | 1 | 1 | # Intuition of this Problem:\n<!-- Describe your first thoughts on how to solve this problem. -->\n**NOTE - PLEASE READ APPROACH FIRST THEN SEE THE CODE. YOU WILL DEFINITELY UNDERSTAND THE CODE LINE BY LINE AFTER SEEING THE APPROACH.**\n\n# Approach for this Problem:\n1. Initialize first to 1 and last to n.\n2. While f... | 26 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
Python Easy Solution || 100% || Binary Search || | guess-number-higher-or-lower | 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: $$O(1)$$\n<!-- Add your space complexity h... | 2 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
📌 Python3 extension of Binary Search | guess-number-higher-or-lower | 0 | 1 | ```\nclass Solution:\n def guessNumber(self, n: int) -> int:\n low = 1\n high = n\n \n while low<=high:\n mid = (low+high)//2\n gussed = guess(mid)\n \n if gussed == 0:\n return mid\n if gussed<0:\n high ... | 5 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
Easy Python Solution | guess-number-higher-or-lower | 0 | 1 | # Code\n```\nclass Solution:\n def guessNumber(self, n: int) -> int:\n low = 1\n high = n\n while low <= high:\n mid = (low + high)//2\n res = guess(mid)\n if res == 0 :\n return mid\n elif res == -1:\n high = mid - 1\n ... | 4 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
Python: Beats 99% solutions O(log(n)) (w/ comments and using Walrus Operator) | guess-number-higher-or-lower | 0 | 1 | ```\ndef guessNumber(self, n: int) -> int:\n\tlowerBound, upperBound = 1, n\n\t# Binary division faster than (lowerBound + upperBound) //2\n\tmyGuess = (lowerBound+upperBound) >> 1\n\t# walrus operator \':=\' - assigns value of the function to the variable \'res\'\n\t# and then compare res with 0\n\twhile (res := guess... | 42 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
ONE-LINER || Beats 92.63% 🔥 || SHORTEST POSSIBLE PYTHON EXPLAINED | guess-number-higher-or-lower | 0 | 1 | # \uD83D\uDD25 SOLUTION \uD83D\uDD25\n```\nclass Solution:\n def guessNumber(self, n: int) -> int:\n return bisect_left(range(n), 0, key=lambda num: -guess(num))\n \n```\n# STEP-BY-STEP\n\n- The bisect_left function is called with three arguments:\n1. The first argument is a range object created using... | 1 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
Python binary search solution | guess-number-higher-or-lower | 0 | 1 | **Python :**\n\n```\ndef guessNumber(self, n: int) -> int:\n\tlow = 0\n\thigh = n\n\twhile low <= high:\n\t\tmid = (low + high ) // 2\n\t\tres = guess(mid)\n\t\tif res < 0:\n\t\t\thigh = mid - 1\n\n\t\telif res > 0:\n\t\t\tlow = mid + 1\n\n\t\telse:\n\t\t\treturn mid\n```\n\n**Like it ? please upvote !** | 27 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
374: Solution with step by step explanation | guess-number-higher-or-lower | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. We initialize two variables left and right with 1 and n respectively, which represent the start and end points of our search range.\n\n2. We use a while loop to perform the binary search until we guess the correct number.... | 10 | We are playing the Guess Game. The game is as follows:
I pick a number from `1` to `n`. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API `int guess(int num)`, which returns three possible re... | null |
Beats 84% | Easy To Understand 🚀🔥 | guess-number-higher-or-lower-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:\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 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
[Python] DP, beat 97.52% in time, 99% in memory (with explanation) | guess-number-higher-or-lower-ii | 0 | 1 | **Appreciate if you could upvote this solution**\n\nCase 1: `n is odd`\nSelected nums = [2, 4, 6, ..., n-1]\n\nCase 2: `n is even`\nSelected nums = [1, 3, 5, ..., n-1]\n\nThe basic idea is that if we want to guess the right number, the maximum number of guessing time should be k = floor(n/2) which means we can find out... | 33 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
✅ [Python] split into groups of 3 | guess-number-higher-or-lower-ii | 0 | 1 | The most economic way of guessing a number at the tail is to pick the middle one, e.g., `... 23 24 25` (choose `24`). But this means that a split should be done at `n-4` (i.e., `22` in the previous example). The same logic applies to every possible segment, thus, the split points are `n-3, n-7, n-11, ...`. \n\nThe mini... | 2 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
python dp approach | guess-number-higher-or-lower-ii | 0 | 1 | # Intuition\nLet `dp` is the minimum cost for make sure we win in range `[l, r]` (inclusive).\nIf we choose `i` and wrong, the correct answer should be in `[l, i - 1]` or `[i + 1, r]`. So if we choose `i` and wrong, we need to get maximum cost between this 2 cases + `i` for make sure we always win even the correct answ... | 4 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
Python3 || dp, binSearch w/ explanation || T/M: 96%/ 47% | guess-number-higher-or-lower-ii | 0 | 1 | For an interval `[l,r]`, we choose a num, which if incorrect still allows us to know whether the secret is in either `[l,num-1]` or `[num+1,r]`. So, the worst-case (w-c) cost is num + max(w-c cost in`[l,num-1]`, w-c cost in `[num+1,r])`. We do this by recursion and binary search, starting with`[1,n]`.\n```\nclass Solut... | 5 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
375: Solution with step by step explanation | guess-number-higher-or-lower-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis solution uses dynamic programming to solve the problem.\n\nFirst, we define a 2D array dp of size (n+2) x (n+2), where dp[i][j] represents the minimum amount of money needed to guarantee a win when guessing a number bet... | 4 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
[Python] Very detailed explanation with examples | guess-number-higher-or-lower-ii | 0 | 1 | ```\nclass Solution:\n def getMoneyAmount(self, n: int) -> int:\n ## RC ##\n ## APPROACH : DP - MERGE INTERVALS PATTERN ##\n ## LOGIC - BOTTOM UP ##\n ## consider numbers 1,2,3,4 (here i = 1, j = 4) ## \n ## 1. we start by picking 2 or 3 ==> lets say we started by 2, player B says ... | 10 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
PYTHON 3, BOTTOM UP 2D, AND TOP DOWN W MEMOIZATION | guess-number-higher-or-lower-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst I solved using a top down with memoization approach. Then I used that to sold 2d dynamic programming approach. This made it more inuitive\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time c... | 0 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
Traditional dfs + cache solution. you will understand without explanation. | guess-number-higher-or-lower-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. $$O(n)$$ --... | 0 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
Three line python For fun | guess-number-higher-or-lower-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def getMoneyAmount(self,n): \n @cache \n def h(l,r):return 0 if r-l<1 else min([i+max(h(l,i-1),h(i+1,r)) for i in range((l+r)//2,r)])\n return h(1,n)\n``` | 0 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
Very intuitive explanation - recursive solution with Python (inefficient but helpful to understand) | guess-number-higher-or-lower-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nI really struggled to get the intuition as to how this problem can be divided into sub-problems. So if you have problems with that too, try thinking this way:\n\nWe have a range from 1 to n. Imagine that we chose some k in that range (s... | 0 | We are playing the Guessing Game. The game will work as follows:
1. I pick a number between `1` and `n`.
2. You guess a number.
3. If you guess the right number, **you win the game**.
4. If you guess the wrong number, then I will tell you whether the number I picked is **higher or lower**, and you will continue gu... | The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario. Take a small example (n = 3). What do you end up paying in the worst case? Check out this article if you're still stuck. The purel... |
376: Space 97.92%, Solution with step by step explanation | wiggle-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Check if the length of the input list nums is less than 2. If yes, return the length of nums. This is because a sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.\n\n```\n... | 6 | A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
* For ... | null |
Beats 73.3% - Simple Python Solution - Greedy | wiggle-subsequence | 0 | 1 | ```\nclass Solution:\n def wiggleMaxLength(self, nums: List[int]) -> int:\n length = 0\n curr = 0\n \n for i in range(len(nums) - 1):\n if curr == 0 and nums[i + 1] - nums[i] != 0:\n length += 1\n curr = nums[i + 1] - nums[i]\n \n ... | 3 | A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
* For ... | null |
Python/JS/Java/Go/C++ O(n) DP [w/ State machine diagram] | wiggle-subsequence | 0 | 1 | **State machine diagram**\n\n\n\n\n---\n\nPython:\n\n```\nclass Solution(object):\n def wiggleMaxLength(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n \n ... | 10 | A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
* For ... | null |
Simple O(n) Solution using Dynamic Programming | wiggle-subsequence | 0 | 1 | ```\n#####################################################################################################################\n# Problem: Wiggle Subsequence\n# Solution : Dynamic Programming\n# Time Complexity : O(n) \n# Space Complexity : O(1)\n############################################################################... | 1 | A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
* For ... | null |
[ Python / Java / C++] 🔥100% | ✅ DP | combination-sum-iv | 1 | 1 | \n```Python []\nclass Solution:\n def combinationSum4(self, nums: List[int], t: int) -> int:\n self.dp=[-1]*(t+1)\n return self.solve(nums,t)\n\t\t\t\t\n def solve(self,nums,t):\n if t==0:\n return 1\n\n if self.dp[t]!=-1:\n return self.dp[t]\n\n res=0\n\n ... | 2 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python3 Solution | combination-sum-iv | 0 | 1 | ```\nclass Solution:\n def combinationSum4(self, nums: List[int], target: int) -> int:\n dp=[0]*(target+1)\n dp[0]=1\n for t in range(target+1):\n for num in nums:\n if num<=t:\n dp[t]+=dp[t-num]\n\n return dp[target] \n``` | 2 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Simple dp. Check for every sum till target | combination-sum-iv | 0 | 1 | \n# Code\n```\nclass Solution:\n def combinationSum4(self, nums: List[int], target: int) -> int:\n nums.sort()\n dp = [0] * (target+1)\n for curr in range(target+1):\n for num in nums:\n if num > curr: continue\n elif num == curr: dp[curr] += 1\n ... | 2 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
【Video】Beats 97.25% - Python, JavaScript, Java and C++ | combination-sum-iv | 1 | 1 | # Intuition\nUse Dynamic Programming. This Python solution beats 97.25% today.\n\n\n\n\n---\n\n# Solution Video\n\n### Please subscribe to my channel from here. I have 257 videos... | 29 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python 3 || 7 lines, recursion || T/S: 97% / 100% | combination-sum-iv | 0 | 1 | ```\nclass Solution:\n def combinationSum4(self, nums: list[int], target: int) -> int:\n \n @lru_cache(None)\n def dp(target: int, res = 0)->int:\n \n for n in nums:\n \n if n == target: res+= 1\n if n < target: res+= dp(target-n)\n\n ... | 3 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
✅ 98.22% Dynamic Programming & Recursion with Memoization | combination-sum-iv | 1 | 1 | # Comprehensive Guide to Solving "Combination Sum IV": Dynamic Programming and Memoization Approaches\n\n## Introduction & Problem Statement\n\nWelcome, fellow coders! Today, we\'ll dive deep into solving an intriguing combinatorial problem\u2014**Combination Sum IV**. This problem serves as a fantastic playground to e... | 119 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python | Easy solution with Explanation | combination-sum-iv | 0 | 1 | # Intuition\nYou can solve this problem using dynamic programming. You\'ll create a table to keep track of the number of combinations for each possible target sum.\n\n# Approach\nIn this code, `dp[i]` represents the number of combinations to reach the target sum `i`. We initialize `dp[0]` to 1 because there is one way ... | 1 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
[beast 99%] with BST + DP | Python | combination-sum-iv | 0 | 1 | I have used Binary search Tree to remove elements from nums greater than target which might make program faster at cost of `nums.sort()`\nwhich might slow the program lil bit.\n\n\n\nAlthough I d... | 1 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python 1-liner. Recursion. Functional programming. | combination-sum-iv | 0 | 1 | # Approach\nThe number of `ways` to form target `k` is sum of `ways` of forming `k - x` after selecting a number `x` from `nums`. i.e,\n\n$$\nways(k) =\n\\begin{cases}\n 0 & \\text{if } k < 0 \\\\\n 1 & \\text{if } k = 0 \\\\\n \\sum_{x \\in nums} ways(k - x) & \\text{if } k > 0 \\\\\n\\end{cases}\n$$\n\n# Com... | 1 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
99.10% Best Solution || DP | combination-sum-iv | 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 an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
🚀100% || C++ || Java || Python || DP Recursive & Iterative || Commented Code🚀 | combination-sum-iv | 1 | 1 | # Problem Description\n- You are given an array of **distinct integers**, nums, and an integer **target**. \n- The **objective** is to determine the number of **possible combinations** that can be formed using elements from nums such that they add up to the given target. \n- Each element in nums can be **used multiple ... | 55 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
✔90.00% Beats C++✨ || DP || Memoization || Easy to Understand📈😇 || #Beginner😉😎 | combination-sum-iv | 1 | 1 | # Problem Understanding:\n- The code aims to solve the "Combination Sum IV" problem using dynamic programming. The problem statement can be summarized as follows:\n\n**Given:**\n\n- An array `nums` containing distinct positive integers.\n- A target integer `target`.\n\n**Task:**\nFind and return the number of possible ... | 12 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Solution with Top-Down Dynamic Programming in Python3 | combination-sum-iv | 0 | 1 | # Intuition\nThe task description is the following:\n- there\'s a list of unique `nums` and an integer `target`\n- our goal is find **how many** combinations of `nums` **sum up** to `target`\n\n```\n# Example\nnums = [1, 2], target = 3\n\n# Here\'s the list of possible combinations\n# [1, 1, 1], [1, 2], [2, 1]\n# For e... | 1 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python | Dynamic Programming | 98.87% | combination-sum-iv | 0 | 1 | # Python | Dynamic Programming | 98.87%\n```\nclass Solution:\n def combinationSum4(self, nums: List[int], target: int) -> int:\n dp = [0] * (target + 1)\n dp[0] = 1\n \n for i in range(1, target + 1):\n for num in nums:\n if i - num >= 0:\n dp... | 28 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Fastest and shortest Solution Python | combination-sum-iv | 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 an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python: easy to understand, straightforward, recursive + memo | combination-sum-iv | 0 | 1 | # Code\n```\nclass Solution:\n def combinationSum4(self, nums: List[int], target: int) -> int:\n \n @cache\n def rec(A, t):\n if t == 0:\n return 1\n res = 0\n for x in A:\n if x <= t:\n res += rec(A, t-x)\n ... | 1 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
C++/Python recursive & iterative DP||Beats 100%||why signed not long long | combination-sum-iv | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse DP to solve the problem, both of recursive version & iterative version.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n"different sequences are counted as different combinations."\nIt \'s not combination problem... | 7 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Rust 0ms/Python 50ms. Dynamic programming with explanation | combination-sum-iv | 0 | 1 | # Intuition\n\nLets assume that we know the answer for every value below $t$ and it is stored in some some array `dp`. So `dp[0]` is 1 as you can reach the value 0 only with empty set and `dp[x]` is how many possible combinations needed to sum up to `x`.\n\nNow the question is how to many times to reach the sum of `t`.... | 2 | Given an array of **distinct** integers `nums` and a target integer `target`, return _the number of possible combinations that add up to_ `target`.
The test cases are generated so that the answer can fit in a **32-bit** integer.
**Example 1:**
**Input:** nums = \[1,2,3\], target = 4
**Output:** 7
**Explanation:**
Th... | null |
Python solution beating 99.7% memory. Heapify the whole matrix inplace | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can use heap to find smallest element inside single row (`matrix[i]`) with $$O(1)$$ memory. Can we use the same approach for row?\n\n# Approach\n1. heapify each row in the matrix `heapq.heapify(matrix[i])`\n2. heapify columns so `matri... | 1 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
Superb Solution in Python3 | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | \n\n# Solution in Python3\n```\nclass Solution:\n def kthSmallest(self, matrix: List[List[int]], k: int) -> int:\n list1 = []\n for i in range(len(matrix)):\n for j in range(len(matrix)):\n list1.append(matrix[i][j])\n list1.sort()\n return list1[k-1]\n``` | 1 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
✅Python 85% Easy/Beginner Friendly [No BST/No Algo]✅ | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | # Intuition\n1) Basic Idea: The goal here is to combine all the sub-lists in the 2D array, and then find the k\'th smallest number as fast as possible. So, let\'s do that!\n\n# Approach\n1) Loop through the matrix and add each value into a new list\n2) Sort the list\n3) Return the [k-1] index of that list (Because inde... | 0 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
Python ✅✅✅|| Faster than 97.16% || Memory Beats 82.82% | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | # Code\n```\nclass Solution:\n def kthSmallest(self, matrix: List[List[int]], k: int) -> int:\n flattened = []\n for row in matrix:\n for n in row:\n flattened.append(n)\n flattened.sort()\n return flattened[k-1]\n```\n\n`.... | null |
kth Smallest Element in a sorted matrix | kth-smallest-element-in-a-sorted-matrix | 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 an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
378: Solution with step by step explanation | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nTo find the kth smallest element in a sorted matrix, we can use the binary search algorithm. First, we set the range of the search to be between the smallest element in the matrix and the largest element in the matrix. Then,... | 14 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
faster than 98.81% using python3(5 lines) | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | ```\nclass Solution:\n def kthSmallest(self, matrix: List[List[int]], k: int) -> int:\n temp_arr=[]\n for i in matrix:\n temp_arr.extend(i)\n temp_arr.sort()\n return temp_arr[k-1]\n``` | 3 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
Python Easy To Understand Solution || Beats 80.82% in Memory | kth-smallest-element-in-a-sorted-matrix | 0 | 1 | ```\nfrom itertools import chain\nclass Solution:\n def kthSmallest(self, matrix: List[List[int]], k: int) -> int:\n flatten_list = list(chain.from_iterable(matrix))\n flatten_list.sort()\n return flatten_list[k-1]\n```\n**.\n.\n.\n<<< Please Up-Vote if find this useful >>>** | 2 | Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_.
Note that it is the `kth` smallest element **in the sorted order**, not the `kth` **distinct** element.
You must find a solution with a memory complexity better than `O(n2)`.... | null |
Python | Super Efficient | Detailed Explanation | insert-delete-getrandom-o1 | 0 | 1 | In python, creating a simple api for a set() would be a perfect solution if not for the third operation, getRandom(). We know that we can retrieve an item from a set, and not know what that item will be, but that would not be actually random. (This is due to the way python implements sets. In python3, when using intege... | 296 | Implement the `RandomizedSet` class:
* `RandomizedSet()` Initializes the `RandomizedSet` object.
* `bool insert(int val)` Inserts an item `val` into the set if not present. Returns `true` if the item was not present, `false` otherwise.
* `bool remove(int val)` Removes an item `val` from the set if present. Retur... | null |
Using hashmap to track index and list to save number | insert-delete-getrandom-o1 | 0 | 1 | # Code\n```\nclass RandomizedSet:\n\n def __init__(self):\n # Dict to track index of nums in nums list \n self.toIndex = { }\n self.nums = collections.deque()\n\n def insert(self, val: int) -> bool:\n # If a number already exists then don\'t add it \n if val in self.toIndex: ret... | 1 | Implement the `RandomizedSet` class:
* `RandomizedSet()` Initializes the `RandomizedSet` object.
* `bool insert(int val)` Inserts an item `val` into the set if not present. Returns `true` if the item was not present, `false` otherwise.
* `bool remove(int val)` Removes an item `val` from the set if present. Retur... | null |
For Best Space Complexity... | insert-delete-getrandom-o1 | 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. $... | 1 | Implement the `RandomizedSet` class:
* `RandomizedSet()` Initializes the `RandomizedSet` object.
* `bool insert(int val)` Inserts an item `val` into the set if not present. Returns `true` if the item was not present, `false` otherwise.
* `bool remove(int val)` Removes an item `val` from the set if present. Retur... | null |
380: Solution with step by step explanation | insert-delete-getrandom-o1 | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe class RandomizedSet is implemented using a hash table and an array. The hash table stores the values and their indices in the array, while the array stores the actual values.\n\nThe insert method takes a value as input a... | 6 | Implement the `RandomizedSet` class:
* `RandomizedSet()` Initializes the `RandomizedSet` object.
* `bool insert(int val)` Inserts an item `val` into the set if not present. Returns `true` if the item was not present, `false` otherwise.
* `bool remove(int val)` Removes an item `val` from the set if present. Retur... | null |
in log(n) time | insert-delete-getrandom-o1-duplicates-allowed | 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:log(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:n\n<!-- Add your space complexity here, e.g. $$O(... | 2 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
381: Time 92.67%, Solution with step by step explanation | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis is an implementation of the Randomized Collection data structure in Python, which supports the following operations:\n\n1. insert(val: int) -> bool: Inserts a value to the collection. Returns true if the collection did ... | 5 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Python 3 || simple solution | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | \n```\nimport random\nfrom collections import defaultdict\n\nclass RandomizedCollection:\n\n def __init__(self):\n self.vals = []\n self.locs = defaultdict(set)\n\n def insert(self, val: int) -> bool:\n self.vals.append(val)\n self.locs[val].add(len(self.vals) - 1)\n return len(... | 1 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Simple Python easy to understand | insert-delete-getrandom-o1-duplicates-allowed | 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(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
dict and list solution. beats 57% | insert-delete-getrandom-o1-duplicates-allowed | 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. -->\nFor list, you can only pop the last element, that is the key to solve the problem.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Solution | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Clear Code #Python | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Code\n```\nimport random\nclass RandomizedCollection:\n\n def __init__(self):\n self.arr = []\n self.s_arr = {}\n def insert(self, val: int) -> bool:\n self.arr.append(val)\n if val in self.s_arr and self.s_arr[val]>0:\n self.s_arr[val]+=1\n return False\n ... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Java & Python Solution | Explained | insert-delete-getrandom-o1-duplicates-allowed | 1 | 1 | # Intuition - \n\nThe given code implements a data structure called `RandomizedCollection`, which is similar to a set but allows for duplicates and supports efficient random element retrieval. It allows inserting elements, removing elements (if they exist), and getting a random element.\n\n# Approach - \n\n1. `Randomiz... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
@SunilKing55 | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Code\n```\nclass RandomizedCollection:\n\n def __init__(self):\n self.a = []\n\n def insert(self, val: int) -> bool:\n if val in self.a:\n self.a.append(val)\n return False\n else:\n self.a.append(val)\n return True\n \n def remove(self,... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Python | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n list data structure\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(1)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n O(n)\n# Code\n```\nclass RandomizedCollection:... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Simplest Python solution ever | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
commented O(1) solution with: dict(key - set()) for insert and remove, list() for random | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nsearch for all O(1) methods. look at insert, remove and random seperately.\nThen bridge them.\n\nall lines commented\n\n# Complexity\n\n- Time complexity:\nO(1) - all operations are constant in time\n\n\n- Space complexity:\... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Solution | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | \n# Code\n```\nimport random\n\nclass RandomizedCollection:\n def __init__(self):\n self.collection = []\n self.indices = {}\n\n def insert(self, val: int) -> bool:\n if val not in self.indices:\n self.indices[val] = set()\n self.indices[val].add(len(self.collection))\n ... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Brute Fcking Force | insert-delete-getrandom-o1-duplicates-allowed | 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 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
[Python] O(1) list and dictionary solution, beats 90% | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | ```\nclass RandomizedCollection:\n\n def __init__(self):\n self.list = []\n self.dict = defaultdict(set)\n\n def insert(self, val: int) -> bool:\n indices = self.dict[val]\n indices.add(len(self.list))\n self.list.append(val)\n return len(indices) == 1\n\n def remove(s... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Python Solution that actually has O(1) for all operations | Faster than 99.8% | insert-delete-getrandom-o1-duplicates-allowed | 0 | 1 | # Intuition\nMy initial thought was to represent the set with a dictionary, with the key representing each unique item and the value acting as a counter for how many times it appears. This makes implementing insertion and removal relatively simple. For insertion you increment the counter, and for removal, you decrement... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Solution | insert-delete-getrandom-o1-duplicates-allowed | 1 | 1 | ```C++ []\nclass RandomizedCollection {\nprivate:\n unordered_map<int, vector<int>> m;\n vector<int> v;\npublic:\n RandomizedCollection() { \n }\n bool insert(int val) {\n v.push_back(val);\n bool ret = m.find(val) == m.end() ? true : false;\n m[val].push_back(v.size()-1);\n ... | 0 | `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.
Implement the `RandomizedCollection` class:
* `RandomizedCollection()` Initializes the empty `Rand... | null |
Easiest.py | linked-list-random-node | 0 | 1 | # Code\n```\nclass Solution:\n def __init__(self, head: Optional[ListNode]):\n self.ll=[]\n while head:\n self.ll.append(head.val)\n head=head.next\n def getRandom(self) -> int:\n return self.ll[randint(0, len(self.ll)-1)]\n``` | 5 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Python3 / Golang, 2 Easy and intuitive solutions | linked-list-random-node | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n## Version 1\n1. In `__init__` / constructor we initialize a list and append to it all the values from the single-linked list\n2. In `getRandom` return a random integer from the list \n\n# Complexity\n- Time complexity:\n`__init__` / constructor O(n)\... | 2 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
382: Solution with step by step explanation | linked-list-random-node | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis is an implementation of an algorithm to return a random node value from a linked list with each node having an equal probability of being chosen.\n\nThe class Solution is defined with an __init__ method that takes a sin... | 2 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Easy Python3 | linked-list-random-node | 0 | 1 | # Code\n```\nclass Solution:\n\n def __init__(self, head: Optional[ListNode]):\n self.List = []\n\n while head:\n self.List.append(head.val)\n head = head.next\n \n\n def getRandom(self) -> int:\n return self.List[random.randrange(len(self.List))]\n``` | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Simple Easiest code in python3 | linked-list-random-node | 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)$$ --... | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Python simple solution You will ever find. | linked-list-random-node | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Python3 | Faster Than 98.02% | linked-list-random-node | 0 | 1 | \n\n# Code\n```\nclass Solution:\n\n def __init__(self, head: Optional[ListNode]):\n self.lst=[]\n\n while(head):\n self.lst.append(head.val)\n head=head.next\n\n def getRandom(self) -> int:\n return random.choice(self.lst)\n\n\n``` | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Simple 🐍python solution👌 | linked-list-random-node | 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\n\nclass Solution:\n def __init__(self, head: ListNode):\n self.nodes = []\n while head:\n self.nodes.append(head.val)\n... | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Python3 || 97% fast || 2 methods || detailed explanation | linked-list-random-node | 0 | 1 | # Approach 1\n<!-- Describe your approach to solving the problem. -->\n- create array of nodes.\n- use random function to get random element from array.\n- return element.\n\n\n\n# Complexity\n- Time complex... | 2 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Basic Probability to Advanced Random Sampling | linked-list-random-node | 1 | 1 | # Intuition\nA random node has to be returned everytime, with each node having an equal probability of getting selected. If we consider the length of the linked list, generate a number between it\'s length, and return the node at that position, we will have fulfilled all the constraints\n\n# Approach\nStore the list el... | 24 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Easy python solution | linked-list-random-node | 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\n\nimport random\nclass Solution:\n\n def __init__(self, head: Optional[ListNode]):\n self.head = head \n self.arr = []\n\n def createList(s... | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
python3, using random.randrange() (70 ms, faster than 83.83%) | linked-list-random-node | 0 | 1 | Runtime: **70 ms, faster than 83.83%** of Python3 online submissions for Linked List Random Node. \nMemory Usage: 17.6 MB, less than 6.13% of Python3 online submissions for Linked List Random Node. \n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# s... | 2 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
❤ [Python3] RESERVOIR SAMPLING ໒(＾ᴥ＾)७, Explained | linked-list-random-node | 0 | 1 | In the follow-up, the problem asks us to find an algorithm for an extremely large list and with no additional space. This immediately recalls us about reservoir sampling - an online algorithm for uniformly choosing `k` random elements from a stream. In our case, `k` is equal to 1 (we need to sample just one element). W... | 42 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
easy python solution.91.90% faster..and very easy to understand. | linked-list-random-node | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nfirst step->declare a list and append data through it.\nwhen you finished to do so.\nthe use this method random.choice(self.list)\nto select the random number from the list.\n\n# Approach\n<!-- Describe your approach to solving the proble... | 1 | Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
Implement the `Solution` class:
* `Solution(ListNode head)` Initializes the object with the head of the singly-linked list `head`.
* `int getRandom()` Chooses a node randoml... | null |
Ransom Note | ransom-note | 0 | 1 | # 11/22\n\n# Code\n```\nclass Solution:\n def canConstruct(self, ransomNote: str, magazine: str) -> bool:\n for i in range(len(magazine)):\n for j in range(len(ransomNote)):\n try:\n if magazine[i] == ransomNote[j]:\n ransomNote = ransomNote.... | 3 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Very Easy || 100% || Fully Explained || C++, Java, Python, Python3 | ransom-note | 1 | 1 | # **C++ Solution:**\n```\nclass Solution {\npublic:\n bool canConstruct(string ransomNote, string magazine) {\n // Initialize an array of count with the size 26...\n int counter[26] = {0};\n // Traverse a loop through the entire String of magazine where char ch stores the char at the index of ma... | 234 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Solution Outperforms 93.93% of Users ! | ransom-note | 0 | 1 | \n\n# Approach\nThe code iterates through each character in the ransomNote. For each character, it checks if it is present in the magazine. If the character is not found in the magazine, the function immediately returns False, indicating that the ransom note cannot be constructed. If the character is found, it removes ... | 3 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Python Find/Replace - Run Time: 47ms, Memory: 16.34MB | ransom-note | 0 | 1 | # Approach\nPython provides a string function to find the first index of a substring in an input string with str().find(). This returns -1 if the substring could not be found. Python also provides string function str().replace() with the 3rd argument being which match to replace if substring found. \n\nIf a letter from... | 0 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Easiest One Linear solution in Python3 | ransom-note | 0 | 1 | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n the given code to determine if you can construct a ransomNote from the characters in a magazine is as follows:\n\n1. First, a set of unique characters in the ransomNote is created using set(ransomNote). This is done to avoid unnecessary repetitio... | 11 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
🌟Hash-table 🌟O(mxn) 🌟 Easy solution 🌟Explained | ransom-note | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst count all the character int the `ransomNote` and then also check if the same numbers of characters are present in the `magazine` or not. If so, return `True` else `False`\n\n# Approach\n<!-- Describe your approach to solving the pro... | 13 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
383: Solution with step by step explanation | ransom-note | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. We define a class Solution and a method canConstruct which takes two string arguments, ransomNote and magazine.\n\n2. We create Counter objects for both the ransomNote and magazine strings using the Counter() function fro... | 52 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
4th | ransom-note | 0 | 1 | 4\n\n# Code\n```\nclass Solution:\n def canConstruct(self, ransomNote: str, magazine: str) -> bool:\n for i in ransomNote:\n if i in magazine:\n magazine = magazine.replace(i, \'\', 1)\n ransomNote = ransomNote.replace(i, \'\', 1)\n return len(ransomNote) == 0\n... | 2 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Easy Python code for Beginners | ransom-note | 0 | 1 | # Intuition\nThe intuition behind the given code is to check if we can construct the ransomNote string using the letters from the magazine string, while adhering to the constraint that each letter in magazine can only be used once.\n\nThe code accomplishes this by using two for loops to traverse the ransomNote and maga... | 10 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Solution using python collections | ransom-note | 0 | 1 | ```\nfrom collections import defaultdict, Counter\nclass Solution:\n def canConstruct(self, ransomNote: str, magazine: str) -> bool:\n d = dict(Counter(magazine))\n for x in ransomNote:\n if x not in d:\n return False\n else:\n d[x] -= 1\n ... | 3 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
One Line: understand set concept | ransom-note | 0 | 1 | \n```\nclass Solution:\n def canConstruct(self, ransomNote: str, magazine: str) -> bool:\n return Counter(ransomNote)=Counter(magazine)\n #please upvote me it would encourage me alot\n\n \n```\n```\nclass Solution:\n def canConstruct(self, ransomNote: str, magazine: str) -> bool:\n return not C... | 14 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Python3 95.20% O(N) Solution | Beginner Friendly | ransom-note | 0 | 1 | TC: O(N), N = max(len(a), len(b))\n\n```\nclass Solution:\n def canConstruct(self, a: str, b: str) -> bool:\n letter = [0 for _ in range(26)]\n \n for c in b:\n letter[ord(c) - 97] += 1\n \n for c in a:\n letter[ord(c) - 97] -= 1\n \n return ... | 3 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
Python ||| Easy and Beginner Friendly Solution ||| | ransom-note | 0 | 1 | ```\ndef canConstruct(self, ransomNote: str, magazine: str) -> bool:\n\trn = Counter(ransomNote)\n\tmg = Counter(magazine)\n\tfor key in rn:\n\t\tif key not in mg:\n\t\t\treturn False\n\t\telse:\n\t\t\tif rn[key] > mg[key]:\n\t\t\t\treturn False\n\treturn True | 3 | Given two strings `ransomNote` and `magazine`, return `true` _if_ `ransomNote` _can be constructed by using the letters from_ `magazine` _and_ `false` _otherwise_.
Each letter in `magazine` can only be used once in `ransomNote`.
**Example 1:**
**Input:** ransomNote = "a", magazine = "b"
**Output:** false
**Example ... | null |
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