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number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Using Hash Map || Easy || C++ | using-hash-map-easy-c-by-amit5446tiwari-w7c3 | 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 | AMIT5446TIWARI | NORMAL | 2023-12-22T12:35:59.452218+00:00 | 2023-12-22T12:35:59.452241+00:00 | 30 | false | # 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 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++11 STL std::unordered_map -- Map with Square as Key & Repeat Squares as Value | c11-stl-stdunordered_map-map-with-square-qn9d | Description\n\nEssentially, this solution breaks the problem into:\n1. Creating an unordered_map with squares as key and the value storing the number of times t | magnetrwn | NORMAL | 2023-12-07T21:58:33.422821+00:00 | 2023-12-07T21:58:33.422850+00:00 | 10 | false | # Description\n\nEssentially, this solution breaks the problem into:\n1. Creating an `unordered_map` with squares as key and the value storing the number of times the same square has been found (this is why we couldn\'t use an `unordered_set` here).\n2. Searching across the keys of the squares maps for matches with the... | 0 | 0 | ['Math', 'Hash Function', 'C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | hash c solution | hash-c-solution-by-nameldi-ecq1 | \nint cal(int* nums1, int nums1Size, int* nums2, int nums2Size){\n int* hash2 = (int*)calloc(1e5+1, sizeof(int));\n int ans = 0;\n long long tmp;\n | nameldi | NORMAL | 2023-11-19T12:40:31.183051+00:00 | 2023-11-19T12:40:31.183067+00:00 | 0 | false | ```\nint cal(int* nums1, int nums1Size, int* nums2, int nums2Size){\n int* hash2 = (int*)calloc(1e5+1, sizeof(int));\n int ans = 0;\n long long tmp;\n for(int i = 0; i < nums2Size; i++){\n for(int j = 0; j < nums1Size; j++){\n tmp = nums1[j];\n tmp *= nums1[j];\n if(t... | 0 | 0 | ['C'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Map/Hash Table Solution in TS/JS | maphash-table-solution-in-tsjs-by-soccer-jch2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nPrecalculated all the s | soccermaganti | NORMAL | 2023-11-05T19:43:56.311786+00:00 | 2023-11-05T19:43:56.311803+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nPrecalculated all the squared values and getting their frequencies. Then using two pointers iterated through both arrays checking for type 1 and type 2 and added the f... | 0 | 0 | ['Array', 'Hash Table', 'Ordered Map', 'TypeScript'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | count the frequency of every product formed | easy to understand | count-the-frequency-of-every-product-for-9acr | Code\n\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) \n {\n int type1 = 0;\n unordered_map<long long in | akshat0610 | NORMAL | 2023-10-30T17:56:21.134271+00:00 | 2023-10-30T17:56:21.134291+00:00 | 25 | false | # Code\n```\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) \n {\n int type1 = 0;\n unordered_map<long long int,int>mp;\n for(int i = 0 ; i < nums1.size() ; i++) mp[1LL * nums1[i]*nums1[i]]++;\n for(int j = 0 ; j < nums2.size() ; j++)\n {\n ... | 0 | 0 | ['Array', 'Hash Table', 'Math', 'Two Pointers', 'C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | EASY CPP MAPPING | easy-cpp-mapping-by-divyanshojha99-53pa | Intuition\nThe problem requires finding the number of triplets that satisfy either Type 1 or Type 2 conditions. Type 1 triplets have the property that nums1[i]^ | divyanshojha99 | NORMAL | 2023-10-11T19:03:15.790517+00:00 | 2023-10-11T19:03:15.790534+00:00 | 11 | false | # Intuition\nThe problem requires finding the number of triplets that satisfy either Type 1 or Type 2 conditions. Type 1 triplets have the property that `nums1[i]^2` is equal to `nums2[j] * nums2[k]`, and Type 2 triplets have the property that `nums2[i]^2` is equal to `nums1[j] * nums1[k]`.\n\n# Approach\nTo solve this... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Easy Solution | java-easy-solution-by-madhurigali-wn6v | 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 | MadhuriGali | NORMAL | 2023-10-11T06:32:32.482829+00:00 | 2023-10-11T06:32:32.482849+00:00 | 25 | false | # 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:\nO(n2)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n ... | 0 | 0 | ['Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Reducing arrays into a sum of the helper function results | reducing-arrays-into-a-sum-of-the-helper-17xz | Approach\nCreate a helper function counting possible products of a number in an array. Return the sum of both arrays being mapped to the power of 2 and reduced | melihov | NORMAL | 2023-09-15T09:00:03.808161+00:00 | 2023-09-15T09:13:25.226730+00:00 | 3 | false | # Approach\nCreate a helper function counting possible products of a number in an array. Return the sum of both arrays being mapped to the power of 2 and reduced with the helper function.\n\n# Code\n```\nclass Solution {\n\tfunc countProducts(of: Int, inArray: [Int]) -> Int {\n\t\tguard inArray.count > 1 else { return ... | 0 | 0 | ['Swift'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Two Pointer Approch Using HashMap | 100%Beats | two-pointer-approch-using-hashmap-100bea-2j1z | 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 | Shree_Govind_Jee | NORMAL | 2023-09-13T17:46:07.256208+00:00 | 2023-09-13T17:46:07.256237+00:00 | 40 | false | # 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(n2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$... | 0 | 0 | ['Array', 'Hash Table', 'Math', 'Two Pointers', 'Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy to understand JavaScript solution | easy-to-understand-javascript-solution-b-cbmn | Complexity\n- Time complexity:\nO(n^2)\n\n- Space complexity:\nO(n)\n\n# Code\n\nvar numTriplets = function(nums1, nums2) {\n const triplets = (numsA, numsB) | tzuyi0817 | NORMAL | 2023-08-27T06:22:19.836434+00:00 | 2023-08-27T06:23:23.795834+00:00 | 4 | false | # Complexity\n- Time complexity:\n$$O(n^2)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nvar numTriplets = function(nums1, nums2) {\n const triplets = (numsA, numsB) => {\n const squareMap = numsA.reduce((map, num) => {\n const square = num ** 2;\n const count = map.get(square) ?? 0... | 0 | 0 | ['JavaScript'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Only Using Map Simple Solution ✅✅ | only-using-map-simple-solution-by-nandun-hd8d | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | nandunk | NORMAL | 2023-08-25T14:40:06.652043+00:00 | 2023-08-25T14:40:06.652066+00:00 | 34 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n$$O(n2)$$ \n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python3 Two Pointer Solution | O(1) Space, O(n*m) Time with Indepth Explanation | python3-two-pointer-solution-o1-space-on-aqan | Intuition\n Describe your first thoughts on how to solve this problem. \nThere are two types of triples we\'re looking for: \n\n Type 1: Triplet (i, j, k) if | colewinfield | NORMAL | 2023-07-12T23:37:04.976835+00:00 | 2023-07-12T23:39:03.820890+00:00 | 55 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThere are two types of triples we\'re looking for: \n\n Type 1: Triplet (i, j, k) if nums1[i]^2 == nums2[j] * nums2[k] \n Type 2: Triplet (i, j, k) if nums2[i]^2 == nums1[j] * nums1[k]\n\nNotice the symmetry. It\'s saying we must lo... | 0 | 0 | ['Two Pointers', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python3 solution || easy to understand || beginner's friendly | python3-solution-easy-to-understand-begi-tiv1 | Code\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n Map1 = Counter([n * n for n in nums1])\n Map2 = C | khushie45 | NORMAL | 2023-07-02T06:21:39.031739+00:00 | 2023-07-02T06:21:39.031765+00:00 | 27 | false | # Code\n```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n Map1 = Counter([n * n for n in nums1])\n Map2 = Counter([n * n for n in nums2])\n\n res = 0\n for i in range(len(nums1) - 1):\n for j in range(i + 1, len(nums1)):\n ... | 0 | 0 | ['Hash Table', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy understand python solution | easy-understand-python-solution-by-peter-8o3c | Code\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n Count1 = Counter(nums1)\n Count2 = Counter(nums2) | petersing001 | NORMAL | 2023-06-21T09:52:40.480390+00:00 | 2023-06-21T09:52:40.480408+00:00 | 36 | false | # Code\n```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n Count1 = Counter(nums1)\n Count2 = Counter(nums2)\n res = 0\n for i in Count1:\n db = Count2.copy()\n while db:\n item = db.popitem()\n Q,... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | unique and easy solution | unique-and-easy-solution-by-adityaclearz-vm2r | 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 | adityaclearzultra | NORMAL | 2023-06-21T07:06:10.386036+00:00 | 2023-06-21T07:06:10.386064+00:00 | 41 | false | # 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)$$ --... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python| pass with accumulate frequencys of nums^2 | python-pass-with-accumulate-frequencys-o-0z4n | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | MikeShieh | NORMAL | 2023-06-07T06:09:15.821123+00:00 | 2023-06-07T06:09:15.821167+00:00 | 15 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\... | 0 | 0 | ['Python'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ Solution || Easy Approach | c-solution-easy-approach-by-vedantk1003-fmoe | \nclass Solution {\npublic:\n #define ll long long int\n \n int count(vector<int> & A,vector<int>& B)\n {\n map<ll,int> m;\n int c=0;\ | vedantk1003 | NORMAL | 2023-05-25T17:09:43.842099+00:00 | 2023-05-25T17:09:43.842154+00:00 | 9 | false | ```\nclass Solution {\npublic:\n #define ll long long int\n \n int count(vector<int> & A,vector<int>& B)\n {\n map<ll,int> m;\n int c=0;\n for(auto a :A) m[(ll)a*a]++;\n\n for(int i=0;i<B.size()-1;i++)\n {\n for(int j{i+1};j<B.size();j++)\n {\n ... | 0 | 0 | ['C'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Brute Force Slowish | brute-force-slowish-by-stolenfuture-zpe1 | Code\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def count_triplets(nums1: List[int], nums2: List[int]) - | StolenFuture | NORMAL | 2023-05-19T00:42:22.922665+00:00 | 2023-05-19T00:42:22.922690+00:00 | 20 | false | # Code\n```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def count_triplets(nums1: List[int], nums2: List[int]) -> int:\n count = 0\n product_map = defaultdict(int)\n\n for i in range(len(nums1)):\n for j in range(i + 1,... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | PYTHON3 SIMPLEST SOLUTION. BEATS 96% USERS. SELF EXPLANATORY. | python3-simplest-solution-beats-96-users-y9fn | \nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n nums1SquareFrequency=defaultdict(int)\n nums2SquareFreq | pchakhilkumar | NORMAL | 2023-03-05T15:42:18.251667+00:00 | 2023-03-05T15:42:18.251704+00:00 | 8 | false | ```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n nums1SquareFrequency=defaultdict(int)\n nums2SquareFrequency=defaultdict(int)\n for num in nums1:\n nums1SquareFrequency[num**2]+=1\n for num in nums2:\n nums2SquareFrequency[n... | 0 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [LC-1577-M | Python3] A Plain Solution | lc-1577-m-python3-a-plain-solution-by-dr-equi | Use collections.Counter and iterations to count.\n\npython3 []\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n | drsv | NORMAL | 2023-02-18T02:01:14.716064+00:00 | 2023-02-18T02:11:23.839462+00:00 | 26 | false | Use `collections.Counter` and iterations to count.\n\n```python3 []\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n sq_cnter1 = Counter([num**2 for num in nums1])\n sq_cnter2 = Counter([num**2 for num in nums2])\n\n def count(nums, sq_cnter):\n n... | 0 | 0 | ['Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | [Python3, Java, C++] Easy Sliding Window O(n) | python3-java-c-easy-sliding-window-on-by-eyf1 | Intuition: \nWhenever you are faced with a circular array question, you can just append the array to itself to get rid of the circular array part of the proble | tojuna | NORMAL | 2022-01-09T04:01:51.703446+00:00 | 2022-01-09T09:12:54.378181+00:00 | 17,196 | false | *Intuition*: \nWhenever you are faced with a circular array question, you can just append the array to itself to get rid of the circular array part of the problem\n\n***Explanation***:\n* Count number of ones in nums, let the number of ones be stored in the variable `ones`\n* Append nums to nums because we have to loo... | 262 | 6 | ['C', 'Java', 'Python3'] | 25 |
minimum-swaps-to-group-all-1s-together-ii | Same Concept of Sliding Window || O(N) in 3 Langs. || Margin(94%) | same-concept-of-sliding-window-on-in-3-l-ms67 | \n# Intuition Behind the Code\n### Fools yell from bot accounts, we just need to move on.. This is my soln with submission history attached below..After Several | Aim_High_212 | NORMAL | 2024-08-02T00:32:52.087485+00:00 | 2024-08-02T03:55:56.604735+00:00 | 35,104 | false | \n# Intuition Behind the Code\n### Fools yell from bot accounts, we just need to move on.. This is my soln with submission history attached below..After Several attempts was able to provide soln with acceptable margin\n\n1. Counting 1s (`k`):\n - First, the code calculates the total number of 1s in the array, denoted... | 123 | 3 | ['Array', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3'] | 24 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window | sliding-window-by-votrubac-b335 | We move the fixed-size sliding window, wrapping it around the array. The lenght of the sliding window is the number of ones.\n\nWe track and return the minimum | votrubac | NORMAL | 2022-01-09T04:02:09.487035+00:00 | 2022-01-09T06:51:08.297220+00:00 | 11,269 | false | We move the fixed-size sliding window, wrapping it around the array. The lenght of the sliding window is the number of `ones`.\n\nWe track and return the minimum number of "holes" in our sliding window. Each of those "holes" needs to be swapped.\n\n**Java**\n```java\npublic int minSwaps(int[] nums) {\n int ones = Ar... | 98 | 6 | ['C', 'Java'] | 16 |
minimum-swaps-to-group-all-1s-together-ii | O(n) | Sliding Window | 2-ms Beats 100% Java | C++ | Python | Rust | Go | JavaScript | on-sliding-window-2-ms-beats-100-java-c-r0xy5 | Appraoch 1\n# Intuition\n\nThis problem of finding the minimum number of swaps to group all 1\'s in a circular binary array presents a unique challenge due to i | kartikdevsharma_ | NORMAL | 2024-08-02T04:22:06.594261+00:00 | 2024-08-20T06:50:09.856272+00:00 | 14,093 | false | # Appraoch 1\n# Intuition\n\nThis problem of finding the minimum number of swaps to group all 1\'s in a circular binary array presents a unique challenge due to its circular nature. The key insight is to recognize that we\'re essentially looking for a contiguous subarray of a specific length (equal to the total number ... | 72 | 0 | ['Array', 'Sliding Window', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript'] | 11 |
minimum-swaps-to-group-all-1s-together-ii | Easy understanding C++ code with approach | easy-understanding-c-code-with-approach-4ie5x | We are given a binary circular array (0s and 1s) and we are required to group all the 1s together with minimum no. of swaps.\n\nFirst we count the total no. of | venkatasaitanish | NORMAL | 2022-01-10T08:20:37.346382+00:00 | 2022-01-10T09:58:32.716288+00:00 | 4,606 | false | We are given a binary circular array (0s and 1s) and we are required to group all the 1s together with minimum no. of swaps.\n\nFirst we count the total no. of 1s in the given array. If the no. of 1s in the array is less than or equal to 1, this means all the 1s are grouped together, so no swaps are required and we ret... | 57 | 1 | ['C', 'Sliding Window', 'C++'] | 7 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Sliding window || O(n) || Full exaplanation | c-sliding-window-on-full-exaplanation-by-mste | 1)Count the total number of 1s. Let m be that number\n2)Find contiguous region of length m that has the most 1s in it\n3)The number of 0s in this region is the | abhi_ak1012 | NORMAL | 2022-01-09T05:28:05.004852+00:00 | 2022-01-09T05:51:51.708530+00:00 | 2,973 | false | 1)Count the total number of 1s. Let **m** be that number\n2)Find **contiguous region** of length **m** that has the most 1s in it\n3)The number of **0s** in this region is the minimum required swaps.*Each swap will move one 1 into the region and **one 0** to the remainder.*\n4)Finally we will use modulo operation for h... | 26 | 1 | ['Array', 'C', 'Sliding Window'] | 5 |
minimum-swaps-to-group-all-1s-together-ii | Sliding window with comments, Python | sliding-window-with-comments-python-by-k-xrd8 | First, by appending nums to nums, you can ignore the effect of split case.\nThen, you look at the window whose width is width. Here, width = the number of 1\'s | kryuki | NORMAL | 2022-01-09T06:05:02.445830+00:00 | 2022-01-17T07:38:56.491118+00:00 | 2,032 | false | First, by appending nums to nums, you can ignore the effect of split case.\nThen, you look at the window whose width is `width`. Here, `width` = the number of 1\'s in the original nums. This is because you have to gather all 1\'s in this window at the end of some swap operations. Therefore, the number of swap operation... | 23 | 1 | ['Sliding Window', 'Python', 'Python3'] | 4 |
minimum-swaps-to-group-all-1s-together-ii | Sliding window||45ms Beats 99.61% | sliding-window45ms-beats-9961-by-anwende-ird1 | Intuition\n Describe your first thoughts on how to solve this problem. \nUse sliding window. Since the array is circular, the right pointer r should go to til n | anwendeng | NORMAL | 2024-08-02T01:57:01.885902+00:00 | 2024-08-02T12:13:17.327589+00:00 | 5,936 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse sliding window. Since the array is circular, the right pointer r should go to til `n+n0` or `n+n1` to find the circular subinterval of length `n0` (`n1`) all 1s (0s) with minimal swaps.\n\nLater try optimization. 2nd C++ tries to redu... | 18 | 1 | ['Array', 'Sliding Window', 'C++', 'Python3'] | 10 |
minimum-swaps-to-group-all-1s-together-ii | Java - Sliding window - O(n) - Easy understanding | java-sliding-window-on-easy-understandin-6w4z | To solve this problem lets understand the basic sliding window technique :\n\t\n\t1. We should go for sliding window when we want to find the minimum or maximum | Surendaar | NORMAL | 2022-01-09T05:30:44.900299+00:00 | 2022-01-09T05:32:16.709408+00:00 | 2,900 | false | To solve this problem lets understand the basic sliding window technique :\n\t\n\t1. We should go for sliding window when we want to find the minimum or maximum range in a given array\n\t2. Sliding window has 2 methods as fixed and dynamic range\n\nLets understand the problem more clearly:\n\t1. first we need to group ... | 18 | 1 | ['Sliding Window', 'Java'] | 8 |
minimum-swaps-to-group-all-1s-together-ii | Easy Solution Beats 100% | easy-solution-beats-100-by-sachinonly-rcnf | Sliding Window\nPython []\nclass Solution(object):\n def minSwaps(self, nums):\n total = sum(nums) # Total number of 1\'s in the array\n n = l | Sachinonly__ | NORMAL | 2024-08-02T02:00:59.880435+00:00 | 2024-08-02T09:58:14.926300+00:00 | 4,140 | false | # Sliding Window\n```Python []\nclass Solution(object):\n def minSwaps(self, nums):\n total = sum(nums) # Total number of 1\'s in the array\n n = len(nums)\n \n # If there are no 1\'s, no swaps are needed\n if total == 0:\n return 0\n \n # Extend the array... | 16 | 1 | ['Array', 'Sliding Window', 'Prefix Sum', 'Python', 'C++', 'Java'] | 6 |
minimum-swaps-to-group-all-1s-together-ii | ✅💯🔥Simple Code🚀📌|| 🔥✔️Easy to understand🎯 || 🎓🧠Beats 100%🔥|| Beginner friendly💀💯 | simple-code-easy-to-understand-beats-100-7ugx | Beats 100%\n\n\n# Code for Java\n\nclass Solution {\n public int minSwaps(int[] nums) {\n int s = nums.length;\n int ones = 0;\n for (in | atishayj4in | NORMAL | 2024-08-02T07:35:41.772274+00:00 | 2024-08-02T07:45:08.882925+00:00 | 1,555 | false | # Beats 100%\n\n\n# Code for Java\n```\nclass Solution {\n public int minSwaps(int[] nums) {\n int s = nums.length;\n int ones = 0;\n for (int n: nums) {\n ... | 15 | 0 | ['Array', 'C', 'Sliding Window', 'Python', 'C++', 'Java', 'JavaScript'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | C++ || PYTHON || Sliding Window (approach) || Easy to understand || O(1) space | c-python-sliding-window-approach-easy-to-k5tx | Intuition - It\'s a circular array problem, but to handle circular array contraint just append the array to itself or use modulo for this purpose and it will be | typical_soul | NORMAL | 2022-01-09T05:30:54.413025+00:00 | 2022-01-25T05:29:27.719028+00:00 | 1,217 | false | **Intuition - It\'s a circular array problem, but to handle circular array contraint just append the array to itself or use modulo for this purpose and it will become a linear array type problem.**\n\n**-** find the total number of 1\'s in whole array. like **totalOnes** in the below code.\n**-** count of number of 1\'... | 14 | 0 | ['Sliding Window'] | 6 |
minimum-swaps-to-group-all-1s-together-ii | [Java/Python 3] Sliding Window O(n) code w/ brief explanation and analysis. | javapython-3-sliding-window-on-code-w-br-wqoh | Denote as winWidth the total number of 1\'s in the input array, then the goal of swaps is to get a window of size winWidth full of 1\'s. Therefore, we can maint | rock | NORMAL | 2022-01-09T04:32:18.550171+00:00 | 2022-01-12T16:00:38.518308+00:00 | 1,219 | false | Denote as `winWidth` the total number of `1`\'s in the input array, then the goal of swaps is to get a window of size `winWidth` full of `1`\'s. Therefore, we can maintain a sliding window of size `winWidth` to find the maximum `1`\'s inside, and accordingly the minimum number of `0`\'s inside the sliding window is the... | 14 | 6 | ['Java', 'Python3'] | 3 |
minimum-swaps-to-group-all-1s-together-ii | O(N) || EASY- EXPLAINED ||JAVA C++ PYTHON | on-easy-explained-java-c-python-by-abhis-91dk | HEY , IF THE HELPS YOU THEN PLS UPVOTE IT . THANKS @aryan43 for C++ code. Hope to see your knight badge soon.\n\n### Approach\n\n1. Count the total number of 1s | Abhishekkant135 | NORMAL | 2024-08-02T00:28:50.444047+00:00 | 2024-08-02T00:28:50.444075+00:00 | 3,871 | false | # HEY , IF THE HELPS YOU THEN PLS UPVOTE IT . THANKS @aryan43 for C++ code. Hope to see your knight badge soon.\n\n### Approach\n\n1. **Count the total number of 1s** in the array. This count will be the size of the window we need to consider.\n2. **Form an extended array**: To handle the circular nature, concatenate t... | 12 | 1 | ['Array', 'Sliding Window', 'Python', 'C++', 'Java'] | 7 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Sliding Window | c-sliding-window-by-abhay5349singh-tizg | Connect with me on LinkedIn: https://www.linkedin.com/in/abhay5349singh/\n\nApproach: let count of 1s be \'k\', so after rearranging, we will be having a window | abhay5349singh | NORMAL | 2022-01-09T04:07:23.980364+00:00 | 2023-07-14T04:07:11.097329+00:00 | 1,628 | false | **Connect with me on LinkedIn**: https://www.linkedin.com/in/abhay5349singh/\n\n**Approach**: let count of 1s be \'k\', so after rearranging, we will be having a window of size \'k\'\n \n```\nclass Solution {\npublic:\n\n int minSwaps(vector<int> &a) {\n int n=a.size();\n \n int k=0;\n fo... | 11 | 1 | ['C++'] | 5 |
minimum-swaps-to-group-all-1s-together-ii | 🔥 Java O(N) Explanation with Diagrams ✍🏻 | java-on-explanation-with-diagrams-by-yas-kj36 | \uD83D\uDCCC Sliding Window \n\n\n\n\n\n\n\n\nTime Complexity : O(N) || Space Complexity : O(N)\n\n\n public int minSwaps(int[] nums) {\n \n in | Yash_kr | NORMAL | 2022-03-09T10:33:57.301327+00:00 | 2022-03-12T16:21:39.199004+00:00 | 635 | false | # \uD83D\uDCCC Sliding Window \n<img width="50%" src="https://assets.leetcode.com/users/images/74c6717f-82e1-49ef-ac59-0def9f630ae6_1646821665.903593.png">\n<img width="50%" src="https://assets.leetcode.com/users/images/55496339-0d86-4821-b376-f7c1eb100d99_1646821665.8353672.png">\n<img width="50%" src="https://assets.... | 10 | 1 | ['Sliding Window'] | 5 |
minimum-swaps-to-group-all-1s-together-ii | [C++] Sliding window | c-sliding-window-by-sanjeev1709912-aexu | This problem is extension of previous problem Minimum Swaps to Group All 1\'s Together\nSo, I treat it in same way i treat previous problem using sliding windo | sanjeev1709912 | NORMAL | 2022-01-09T04:03:21.426102+00:00 | 2022-01-09T04:38:53.347707+00:00 | 379 | false | This problem is extension of previous problem **[Minimum Swaps to Group All 1\'s Together](https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together/)**\nSo, I treat it in same way i treat previous problem using sliding window.\n\nQ1. What must we find?\nAns. we must find portion in which the number of one ... | 9 | 6 | [] | 2 |
minimum-swaps-to-group-all-1s-together-ii | EASY AND SIMPLE APPROACH | easy-and-simple-approach-by-abb333-dqr9 | Complexity\n- Time complexity: O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | abb333 | NORMAL | 2024-08-02T05:26:53.565047+00:00 | 2024-08-02T05:26:53.565079+00:00 | 1,299 | false | # Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& v) {\n int ones = 0;\n for (auto& x : v) {\n ... | 8 | 0 | ['Array', 'Sliding Window', 'C++'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Python 3 || 7 lines, extend , zip || T/S: 99% / 27% | python-3-7-lines-extend-zip-ts-99-27-by-i4bkj | \nclass Solution:\n def minSwaps(self, nums: list[int]) -> int:\n\n n, k = len(nums), sum(nums)\n mx = tally = sum(nums[:k])\n nums.exte | Spaulding_ | NORMAL | 2024-08-02T01:40:39.672844+00:00 | 2024-08-02T01:41:20.713420+00:00 | 56 | false | ```\nclass Solution:\n def minSwaps(self, nums: list[int]) -> int:\n\n n, k = len(nums), sum(nums)\n mx = tally = sum(nums[:k])\n nums.extend(nums[:k])\n \n for num1, num2 in zip(nums[k:], nums):\n tally+= num1 - num2\n if tally > mx: mx = tally\n \n ... | 8 | 0 | ['Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Classic Sliding Window Question | classic-sliding-window-question-by-heman-pgh4 | Basic Idea: Find the total no. of ones in the given array. Now make a sliding window of that size. Window with maximum no. of ones will require the minimum numb | hemantsingh_here | NORMAL | 2024-08-02T15:52:27.592382+00:00 | 2024-08-02T15:52:27.592412+00:00 | 135 | false | **Basic Idea:** Find the total no. of ones in the given array. Now make a sliding window of that size. Window with maximum no. of ones will require the minimum number of swaps.\n\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int n = nums.size();\n int count_1 = count(nums.begin(), nums.... | 7 | 0 | ['C'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window | O(n) | C++ | Easy approach | Simple | sliding-window-on-c-easy-approach-simple-4u5t | Intuition\n Describe your first thoughts on how to solve this problem. \nI started thinking that, eventually, all the ones in the array will be grouped together | gavnish_kumar | NORMAL | 2024-08-02T05:32:01.975843+00:00 | 2024-08-02T05:32:01.975904+00:00 | 947 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI started thinking that, eventually, all the ones in the array will be grouped together to form a subarray whose size is equal to the total number of ```1\'s``` in the array. From this, I considered how to handle a subarray of that specif... | 7 | 0 | ['C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Easy Sliding window | with vid explanation | easy-sliding-window-with-vid-explanation-3t70 | Vid Explanation\nhttps://youtu.be/xsH9DvzkRPY\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires finding the minim | Atharav_s | NORMAL | 2024-08-02T05:40:18.416124+00:00 | 2024-08-02T05:40:18.416158+00:00 | 565 | false | # Vid Explanation\nhttps://youtu.be/xsH9DvzkRPY\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding the minimum number of swaps to group all 1s together in a circular array. Intuitively, we can focus on a window of size equal to the total number of 1s. We need... | 6 | 1 | ['C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Sliding Window || easy to understand | c-sliding-window-easy-to-understand-by-v-iiic | Just check the number of zeroes in each window of size equal to the number of 1\'s present in the array.\n\n\nclass Solution {\npublic:\n int minSwaps(vector | VineetKumar2023 | NORMAL | 2022-01-09T04:02:08.686352+00:00 | 2022-01-09T04:02:08.686398+00:00 | 521 | false | Just check the number of zeroes in each window of size equal to the number of 1\'s present in the array.\n\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int count=0;\n for(int i=0;i<nums.size();i++)\n if(nums[i]==1)\n count++;\n int result=0;\n ... | 6 | 0 | ['C', 'Sliding Window'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Easy C++ Solution || Sliding Window | easy-c-solution-sliding-window-by-jitena-jisx | 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 | jitenagarwal20 | NORMAL | 2024-08-02T05:28:14.793776+00:00 | 2024-08-02T05:28:14.793807+00:00 | 75 | false | # 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 | 0 | ['C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Best Approach in Easy words (Javascript, C++, Java and Python) | best-approach-in-easy-words-javascript-c-hg0l | Approach\nSince we have to group all 1s in the nums, we will first calculate the total number of 1s. Which will be the windowSize.\n\nAfter this we will count t | Nabeel7801 | NORMAL | 2024-08-02T00:39:41.475974+00:00 | 2024-08-02T13:21:23.171211+00:00 | 1,516 | false | # Approach\nSince we have to group all 1s in the nums, we will first calculate the total number of 1s. Which will be the `windowSize`.\n\nAfter this we will count the 1s in that sliding window. And then iterate through the entire length of the nums array. By doing this we can get the maximum number of 1s in the window ... | 5 | 0 | ['Array', 'Sliding Window', 'Python', 'C++', 'Java', 'JavaScript'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window + Comments | sliding-window-comments-by-xxvvpp-fksb | We just want to get all the 1\'s together irrespective of the places where we put them, we just want them together.\n\n Maintain a window of size of number of o | xxvvpp | NORMAL | 2022-01-09T04:42:23.664787+00:00 | 2022-01-09T07:04:31.007469+00:00 | 574 | false | We just want to get all the 1\'s together irrespective of the places where we put them, we just want them together.\n\n* Maintain a window of size of number of ones in whole array.\n* Now check in every consecutive window of same size,keep on counting number of zeroes present.\n* Number of Zeroes present in the current... | 5 | 0 | ['C'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | ✅ One Line Solution | one-line-solution-by-mikposp-731w | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\nTime complexity: O(n). Space complexity: O(1).\n | MikPosp | NORMAL | 2024-08-02T09:38:27.100381+00:00 | 2024-08-02T09:38:27.100416+00:00 | 497 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\nTime complexity: $$O(n)$$. Space complexity: $$O(1)$$.\n```\nclass Solution:\n def minSwaps(self, a: List[int]) -> int:\n return (n:=len(a),o:=sum(a),w:=sum(a[:o])) and o-max(w:=w-a[i]+a[(i+o)%n] fo... | 4 | 0 | ['Array', 'Sliding Window', 'Python', 'Python3'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | C++ | Efficient Minimum Swaps to Group All 1's Together II | 49ms Beats [99.02%] | c-efficient-minimum-swaps-to-group-all-1-q2i7 | Intuition\nTo minimize the number of swaps needed to group all 1\'s in a circular binary array, we need to find the optimal position where all 1\'s can be group | user4612MW | NORMAL | 2024-08-02T03:05:23.362198+00:00 | 2024-08-02T03:05:23.362255+00:00 | 18 | false | # Intuition\nTo minimize the number of swaps needed to group all $$1$$\'s in a circular binary array, we need to find the optimal position where all $$1$$\'s can be grouped together with the fewest swaps. The challenge is to consider the circular nature of the array, which means that the end of the array is adjacent to... | 4 | 0 | ['C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Best for beginners✅ | best-for-beginners-by-ratneshh-rm8t | Intuition\n\n Describe your first thoughts on how to solve this problem. \nUnderstanding the Problem: Recognize that we need to rearrange the array such that al | ratneshh | NORMAL | 2024-08-02T02:09:04.093550+00:00 | 2024-08-02T02:09:04.093575+00:00 | 1,262 | false | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\nUnderstanding the Problem: Recognize that we need to rearrange the array such that all 1s are group... | 4 | 0 | ['Array', 'Sliding Window', 'Java'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Fixed sliding Window✔️✔️✅ | fixed-sliding-window-by-dixon_n-tw3j | Code\n\nclass Solution {\n public int minSwaps(int[] nums) {\n int ones = 0;\n for (int num : nums) {\n ones += num;\n }\n\n | Dixon_N | NORMAL | 2024-06-06T13:28:35.450205+00:00 | 2024-06-06T13:28:35.450236+00:00 | 211 | false | # Code\n```\nclass Solution {\n public int minSwaps(int[] nums) {\n int ones = 0;\n for (int num : nums) {\n ones += num;\n }\n\n int n = nums.length;\n int res = nums.length;\n int left = 0, right = 0, cnt = 0;\n\n // Create a sliding window of size \'ones... | 4 | 0 | ['Java'] | 4 |
minimum-swaps-to-group-all-1s-together-ii | [JAVA] O(N) Sliding Window Solution || Read the commented code if you have time. | java-on-sliding-window-solution-read-the-4236 | \n public int minSwaps(int[] arr) {\n int count = 0;\n int n=arr.length;\n for (int i = 0; i < n; ++i)\n count+=arr[i];\n\n | bharghavsaip | NORMAL | 2022-05-04T17:54:34.973398+00:00 | 2022-05-04T18:15:01.517326+00:00 | 597 | false | \n public int minSwaps(int[] arr) {\n int count = 0;\n int n=arr.length;\n for (int i = 0; i < n; ++i)\n count+=arr[i];\n\n // Find unwanted elements in current\n // window of size \'count\'\n int bad = 0;\n for (int i = 0; i < count; ++i)\n if (arr[i] =... | 4 | 0 | ['Array', 'Sliding Window', 'Java'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | C++ solution(sliding window) | c-solutionsliding-window-by-sethiyashris-lzjt | Here we use the concept of sliding window..we count the number of 1\'s in vector and make a window of that size...and we once traverse in the window of that siz | sethiyashristi20 | NORMAL | 2022-03-27T14:41:25.782675+00:00 | 2022-03-27T14:42:29.752148+00:00 | 176 | false | **Here we use the concept of sliding window..we count the number of 1\'s in vector and make a window of that size...and we once traverse in the window of that size and count no of 1\'s in that window(count)...and assign maxcount as count and noww we simply traverse the complete vector in that window size(k) and decrea... | 4 | 0 | ['Sliding Window'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | C++ Simple and Clean Sliding-Window Solution, Detailed Explanation | c-simple-and-clean-sliding-window-soluti-11u5 | Idea:\nWe want to find the "window" of size ones with the most ones.\nFirst, we count the ones in the original array.\nNow, because the array is circular, the e | yehudisk | NORMAL | 2022-01-13T14:18:05.036676+00:00 | 2022-01-13T14:18:05.036728+00:00 | 344 | false | **Idea:**\nWe want to find the "window" of size `ones` with the most ones.\nFirst, we count the ones in the original array.\nNow, because the array is circular, the easiest way is to concatenate two arrays and then we can search the window regularly.\nWe count the ones in the initial window in the left - from index 0 t... | 4 | 0 | ['C'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Intuition and Thought Process with code explained | intuition-and-thought-process-with-code-dq17t | The questions asks for the minimum number of swaps to group all the 1\'s together at any location.\n\n\nThought process:\n\n I initially started thinking in ter | khsquare | NORMAL | 2022-01-09T09:09:58.133268+00:00 | 2022-01-09T09:09:58.133312+00:00 | 426 | false | The questions asks for the minimum number of swaps to group all the 1\'s together at **any location**.\n\n\nThought process:\n\n* I initially started thinking in terms of sorting the array in non-increasing order (**this meant brining all the 1\'s to the front of the array**). I then thought of using xor (**with bitset... | 4 | 0 | ['C', 'Sliding Window', 'C++'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Prefix Sum | c-prefix-sum-by-kapooryash713-c4ff | Approach:\n1- We will calculate the number of 1\'s in the array.\n2- We will create Prefix sum array.\n3- Since, we know we can swap any distinct elements of an | kapooryash713 | NORMAL | 2022-01-09T07:55:49.492852+00:00 | 2022-01-09T07:57:23.107986+00:00 | 219 | false | Approach:\n1- We will calculate the number of 1\'s in the array.\n2- We will create Prefix sum array.\n3- Since, we know we can swap any distinct elements of any indices. Example : [1,0,1,0,0,1] . Here if we swap the 1 index with the last index. array becomes [1,1,1,0,0,0]. So we know that that we need to check the req... | 4 | 1 | [] | 1 |
minimum-swaps-to-group-all-1s-together-ii | ✅ [C++] Easy Code with inline Explaination. O(n) | c-easy-code-with-inline-explaination-on-9hgbq | \nint minSwaps(vector<int>& nums) {\n int one=0,n=nums.size(),ans=INT_MAX;\n for(int i=0;i<n;i++) if(nums[i]) one++;\n int start=0,end=0,cn | Yogaboi | NORMAL | 2022-01-09T04:02:36.851319+00:00 | 2022-01-09T04:33:10.759012+00:00 | 380 | false | ```\nint minSwaps(vector<int>& nums) {\n int one=0,n=nums.size(),ans=INT_MAX;\n for(int i=0;i<n;i++) if(nums[i]) one++;\n int start=0,end=0,cnt=0; //cnt -> keep count of one\'s inside window //Window look like this [start ,,,....,,, end] \n while(end<n+one){ // check... | 4 | 1 | ['C', 'Sliding Window', 'C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Minimum Swaps to Group All 1's together - sliding window technique | minimum-swaps-to-group-all-1s-together-s-96c4 | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve the problem of finding the minimum number of swaps required to group all 1s in | sudharshm2005 | NORMAL | 2024-08-02T16:23:11.157912+00:00 | 2024-08-02T16:23:11.157943+00:00 | 22 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve the problem of finding the minimum number of swaps required to group all 1s in a binary array, we need to find the smallest window size containing the maximum number of 1s and then calculate how many 0s are in that window. This n... | 3 | 0 | ['Array', 'Sliding Window', 'Python3'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Full Explanation(image) easy 8 line with Bonus code.....👻 | full-explanationimage-easy-8-line-with-b-47oc | Bonus Code at the end of explanation\n\n\nProblem Understanding\nThe goal is to find the minimum number of swaps required to group all 1s together in a circular | Kunj_2803 | NORMAL | 2024-08-02T13:06:38.242444+00:00 | 2024-08-02T13:50:39.211762+00:00 | 307 | false | **Bonus Code at the end of explanation**\n\n\n**Problem Understanding**\nThe goal is to find the minimum number of swaps required to group all 1s together in a circular array.\n\n**Solution :**\n**1.Get targeted ones window size**\n by `window=nums.count(1)` or `window=sum(nums)`\n\n**2.ans and count_1 to store answ... | 3 | 0 | ['Sliding Window', 'C++', 'Java', 'Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Basic yet highly efficient solution. | basic-yet-highly-efficient-solution-by-n-750x | Complexity\n- Time complexity: (O(n)+O(windowSize)+O(n))~O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity | NehaSinghal032415 | NORMAL | 2024-08-02T11:36:49.882124+00:00 | 2024-08-02T11:36:49.882146+00:00 | 68 | false | # Complexity\n- Time complexity: (O(n)+O(windowSize)+O(n))~O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int minSwaps(int[] nums) {\n int windowSize=0;\n int minimu... | 3 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-afjw | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook | shishirRsiam | NORMAL | 2024-08-02T10:00:22.016691+00:00 | 2024-08-02T10:00:22.016720+00:00 | 524 | false | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add ... | 3 | 0 | ['Array', 'Sliding Window', 'C++'] | 3 |
minimum-swaps-to-group-all-1s-together-ii | Python | Sliding Window | python-sliding-window-by-khosiyat-ycqe | see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n def minSwaps(self, nums: List[int]) -> int:\n n = len(nums)\n k = sum(n | Khosiyat | NORMAL | 2024-08-02T08:21:59.305345+00:00 | 2024-08-02T08:21:59.305372+00:00 | 543 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together-ii/submissions/1341619703/?envType=daily-question&envId=2024-08-02)\n\n# Code\n```\nclass Solution:\n def minSwaps(self, nums: List[int]) -> int:\n n = len(nums)\n k = sum(nums) \n if ... | 3 | 0 | ['Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Use Silding Window And Find the Window Needing Min. Swaps | Java | C++ | [Vidoeo Solution] | use-silding-window-and-find-the-window-n-xube | Intuition, approach, and complexity discussed in video solution in detail\nhttps://youtu.be/LRjOvAslzCg\n# Code\nJava\n\nclass Solution {\n public int minSwa | Lazy_Potato_ | NORMAL | 2024-08-02T05:49:44.219180+00:00 | 2024-08-02T05:49:44.219207+00:00 | 393 | false | # Intuition, approach, and complexity discussed in video solution in detail\nhttps://youtu.be/LRjOvAslzCg\n# Code\nJava\n```\nclass Solution {\n public int minSwaps(int[] nums) {\n return Math.min(findMinSwaps(nums, 0), findMinSwaps(nums, 1));\n }\n private int findMinSwaps(int nums[], int val){\n ... | 3 | 0 | ['Array', 'Sliding Window', 'C++', 'Java'] | 15 |
minimum-swaps-to-group-all-1s-together-ii | Swift💯fast🛝🪟 | swiftfast-by-upvotethispls-5bpf | Sliding Window (accepted answer)\n\nclass Solution {\n func minSwaps(_ nums: [Int]) -> Int {\n let windowSize = nums.reduce(0,+)\n var onesInWi | UpvoteThisPls | NORMAL | 2024-08-02T00:46:20.793526+00:00 | 2024-08-02T01:41:19.936519+00:00 | 104 | false | **Sliding Window (accepted answer)**\n```\nclass Solution {\n func minSwaps(_ nums: [Int]) -> Int {\n let windowSize = nums.reduce(0,+)\n var onesInWindow = nums.prefix(windowSize).reduce(0,+)\n let maxOnesInWindow = nums.indices.reduce(0) { maxOnes, i in\n onesInWindow -= nums[i] // ... | 3 | 0 | ['Swift', 'Sliding Window'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Python sliding window solution | python-sliding-window-solution-by-vincen-dr4u | \ndef minSwaps(self, nums: List[int]) -> int:\n\tk, n = sum(nums), len(nums)\n\ts, mx = sum(nums[:k]), 0\n\tfor i in range(k, 2*n):\n\t\ts += (nums[i%n]-nums[i% | vincent_great | NORMAL | 2022-08-16T20:35:47.161154+00:00 | 2022-08-16T20:35:47.161179+00:00 | 295 | false | ```\ndef minSwaps(self, nums: List[int]) -> int:\n\tk, n = sum(nums), len(nums)\n\ts, mx = sum(nums[:k]), 0\n\tfor i in range(k, 2*n):\n\t\ts += (nums[i%n]-nums[i%n-k])\n\t\tmx = max(s, mx)\n\treturn k-mx\n``` | 3 | 0 | [] | 2 |
minimum-swaps-to-group-all-1s-together-ii | 2 Solutions | O(1) Space | O(N) Time | Similar to Minimum Swaps to Group All 1's Together | 2-solutions-o1-space-on-time-similar-to-mdn67 | Count the minimum swaps required to group all ones in non circular Array. Question: https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together/\n2. Si | rahul_107 | NORMAL | 2022-01-31T18:41:33.120289+00:00 | 2022-01-31T18:50:48.160511+00:00 | 341 | false | 1. Count the minimum swaps required to group all ones in non circular Array. Question: https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together/\n2. Similarly, Count the minimum swaps required to group all zeros in non circular Array.\n3. Return minimum of both. (By seeing the example attached with code we ... | 3 | 0 | ['Sliding Window', 'Prefix Sum'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | a few solutions | a-few-solutions-by-claytonjwong-7685 | Let K be the quantity of 1s in the input array A. Then we can append A onto itself (to consider the wrap-around case) and use a sliding window i..j of size K t | claytonjwong | NORMAL | 2022-01-15T21:31:33.786840+00:00 | 2022-01-15T21:36:20.997289+00:00 | 113 | false | Let `K` be the quantity of 1s in the input array `A`. Then we can append `A` onto itself (to consider the wrap-around case) and use a sliding window `i..j` of size `K` to count the 1s within the window. The "cost" of swapping 1s into the 0s within the window is then the size of the window minus the count of 1s within... | 3 | 0 | [] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Counting Ones || Sliding Window Approach || C++ Clean Code | counting-ones-sliding-window-approach-c-3o71c | Code: \n\n\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int cntOnes = 0;\n int n = nums.size();\n \n for(auto | i_quasar | NORMAL | 2022-01-09T06:14:12.151873+00:00 | 2022-01-09T06:14:12.151915+00:00 | 145 | false | # Code: \n\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int cntOnes = 0;\n int n = nums.size();\n \n for(auto& num : nums) if(num) cntOnes++;\n \n if(cntOnes == 0 || cntOnes == n) return 0;\n \n int maxOnes = 0, ones = 0;\n int w... | 3 | 0 | ['Sliding Window', 'Counting'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ | SLIDING WINDOW | 0-1 COUNT | c-sliding-window-0-1-count-by-chikzz-5mct | PLEASE UPVOTE IF U LIKE MY SOLUTION\n\n\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int n=nums.size();\n int co=0,minn=n+1 | chikzz | NORMAL | 2022-01-09T05:26:55.412895+00:00 | 2022-01-09T05:26:55.412940+00:00 | 175 | false | **PLEASE UPVOTE IF U LIKE MY SOLUTION**\n\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int n=nums.size();\n int co=0,minn=n+1;\n for(auto &x:nums)\n {if(x==1)co++;}\n int cz=n-co;\n int coo=0,czz=0;\n for(int i=0;i<co;i++)\n {\n ... | 3 | 1 | [] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Python Solution Sliding Window with Explaination | python-solution-sliding-window-with-expl-y9jf | After swaps , the length of the sequence of 1 is equal to the total number of 1. Use sliding window with the size of the total number of 1 to find the interval | yuyingji | NORMAL | 2022-01-09T04:16:13.875605+00:00 | 2022-01-09T05:34:39.456483+00:00 | 181 | false | After swaps , the length of the sequence of 1 is equal to the total number of 1. Use sliding window with the size of the total number of 1 to find the interval which has the most number of 1. The number of 0 in this interval will be the minimum swap needed.\n\nk is the number of 1.\ns represent the number of 1 in [i-k ... | 3 | 1 | ['Sliding Window'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | BEATS 98% || 4STEPS XPLANATION || JAVA|| EASY BEGINNER | beats-98-4steps-xplanation-java-easy-beg-odt5 | BOISS N PRETTIES UPVOTE!! KRDE NA BRO ;() TUJHE MOMOS KI KASAM.\n\n1. Prefix Sum Calculation:\n - Well No explanation for this.\n\n2. Initial Conditions:\n | Abhishekkant135 | NORMAL | 2024-08-26T19:21:33.106345+00:00 | 2024-08-26T19:21:33.106374+00:00 | 12 | false | # BOISS N PRETTIES UPVOTE!! KRDE NA BRO ;() TUJHE MOMOS KI KASAM.\n\n1. **Prefix Sum Calculation:**\n - Well No explanation for this.\n\n2. **Initial Conditions:**\n - `i` and `j` are initialized to 0 and `totalOnes`.\n - `totalOnes` represents the total number of 1s in the array.\n\n3. **Sliding Window and Swaps... | 2 | 0 | ['Sliding Window', 'Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Easiest Explanation You Will Find | easiest-explanation-you-will-find-by-hri-ywzy | Approach\nIf you have premium account, solve this problem first: 1151. Minimum Swaps to Group All 1\'s Together. This is a similar problem where the array is no | hridoy100 | NORMAL | 2024-08-03T19:53:49.858060+00:00 | 2024-08-03T19:53:49.858090+00:00 | 14 | false | # Approach\nIf you have premium account, solve this problem first: [1151. Minimum Swaps to Group All 1\'s Together](https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together/). This is a similar problem where the array is not circular.\n\nTo solve that problem you will use a basic sliding window technique:\n... | 2 | 0 | ['Array', 'Sliding Window', 'Java'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Easy Solution || Beats 96.66% || Sliding Window || O(n) Time complexity || O(1) Space Complexity | easy-solution-beats-9666-sliding-window-0uo0q | Intuition\n\n####### SLIDING WINDOW ###########\nPLEASE upVOTE \uD83D\uDE4F\uD83D\uDE4F\n\n\n# Approach\n\nApproach of this question is so simple, just w | vector_08 | NORMAL | 2024-08-02T17:53:36.695358+00:00 | 2024-08-02T17:53:36.695388+00:00 | 48 | false | # Intuition\n```\n####### SLIDING WINDOW ###########\nPLEASE upVOTE \uD83D\uDE4F\uD83D\uDE4F\n```\n\n# Approach\n```\nApproach of this question is so simple, just we have to think more about circular array and subarrays.\nStep 1 : Declare a variable totalOnes and store total number of ones in it.\nStep 2 : Decla... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Answer lelo!! Sliding window approach for the code | answer-lelo-sliding-window-approach-for-he5nc | \n# Approach\n Describe your approach to solving the problem. \nInitialization:\n\n- k is the total number of 1s in the array. This is because we want to group | AmanShukla07 | NORMAL | 2024-08-02T14:18:05.364488+00:00 | 2024-08-02T16:29:46.267628+00:00 | 61 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Initialization:**\n\n- k is the total number of 1s in the array. This is because we want to group all 1s together.\n- mx (maximum number of 1s in any window of length k) and cnt (current count of 1s in the first window) are initialized to the numb... | 2 | 1 | ['Array', 'Sliding Window', 'Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | easy C++ O(n) approach | easy-c-on-approach-by-aayush8910sh-avur | \n\n# Approach\n1. Count the number of 1s in the array:\n\n2. Traverse the array to count the total number of 1s, denoted as k.\nIf k is 0 or equal to the lengt | aayush8910sh | NORMAL | 2024-08-02T12:25:25.588810+00:00 | 2024-08-02T12:25:25.588844+00:00 | 125 | false | \n\n# Approach\n1. Count the number of 1s in the array:\n\n2. Traverse the array to count the total number of 1s, denoted as k.\nIf k is 0 or equal to the length of the array (n), return 0 because no swaps are needed.\n3. Initialize the first window:\n\n4. Use a sliding window of size k to count the number of 0s in the... | 2 | 0 | ['C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 💯 | easiestfaster-lesser-cpython3javacpython-tqc8 | Intuition\n\n\n\n\n\n\n\njavascript []\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar minSwaps = function (nums) {\n let n = nums.length, one | Edwards310 | NORMAL | 2024-08-02T11:33:31.336516+00:00 | 2024-08-02T11:33:31.336539+00:00 | 232 | false | # Intuition\n\n\n\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | Shree_Govind_Jee | NORMAL | 2024-08-02T07:44:32.012874+00:00 | 2024-08-02T07:44:32.012900+00:00 | 204 | false | # Complexity\n- Time complexity:$$O(n*X)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n// class Solution {\n// public int minSwaps(int[] nums) {\n// int one = 0;\n// for (int num : nums)... | 2 | 0 | ['Array', 'Math', 'Greedy', 'Sliding Window', 'Java'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Kotlin. Beats 100% (355 ms). Doubled array (System.arraycopy) + sliding window | kotlin-beats-100-355-ms-doubled-array-sy-4mxq | \n\n# Code\n\nclass Solution {\n fun minSwaps(nums: IntArray): Int {\n var ones = 0\n for (n in nums) {\n ones += n\n }\n\n | mobdev778 | NORMAL | 2024-08-02T07:36:44.898248+00:00 | 2024-08-02T15:24:49.020291+00:00 | 37 | false | \n\n# Code\n```\nclass Solution {\n fun minSwaps(nums: IntArray): Int {\n var ones = 0\n for (n in nums) {\n ones += n\n }\n\n var windowOnes = 0\n for (i in... | 2 | 0 | ['Sliding Window', 'Kotlin'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Simple C++ solution | Sliding Window | simple-c-solution-sliding-window-by-saur-dvzf | Intuition\nSince we need to find the minimum swaps, we need to count the zeros in a fixed window of size equal to the total number of ones.\n\n# Approach\n1. Co | saurabhdamle11 | NORMAL | 2024-08-02T05:34:57.401928+00:00 | 2024-08-02T05:34:57.401961+00:00 | 379 | false | # Intuition\nSince we need to find the minimum swaps, we need to count the zeros in a fixed window of size equal to the total number of ones.\n\n# Approach\n1. Count the number of ones which will be the size of the sliding window.\n2. Append the array to itself at the end as it is a circular array.\n3. Count the number... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | 93% Beats in C++ || Easy understandable Sliding window code 💯✅ | 93-beats-in-c-easy-understandable-slidin-ok63 | \n\n\n# Approach\n Describe your approach to solving the problem. \nWe can observe that after grouping the number of ones in the group is number of ones in the | srimukheshsuru | NORMAL | 2024-08-02T05:17:18.851453+00:00 | 2024-08-02T05:17:18.851481+00:00 | 63 | false | \n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can observe that after grouping the number of ones in the group is number of ones in the complete vector. now apply a sliding wi... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Easy to understand C++ solution || Sliding window || 🔥Beats 91%, O(n) | easy-to-understand-c-solution-sliding-wi-eaqa | Intuition\nProblem can be approached using a sliding window technique to find the segment with the maximum number of 1\'s, as this will result in the minimum sw | omkarsalunkhe3597 | NORMAL | 2024-08-02T05:14:25.887441+00:00 | 2024-08-02T05:14:25.887466+00:00 | 158 | false | # Intuition\nProblem can be approached using a sliding window technique to find the segment with the maximum number of 1\'s, as this will result in the minimum swaps required to group all the 1\'s together.\n\n\n# Approach\n1. **Calculate Total Ones**:\n - Count the total number of 1\'s in the array, which determines... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | ✅✅👌👌GREAT SOLUTION ,UNDERSTAND BETTER, EASY APPROACH💯💯 | great-solution-understand-better-easy-ap-3b0a | PRETTY EASY APPROACH FOR BETTER UNDERSTANDING THE SLIDING WINDOW CONCEPT (BEGINNER FRIENDLY)\n\n\npublic int minSwaps(int[] a) {\n\tint windowSize = 0;\n\tfor(i | _dc | NORMAL | 2024-08-02T04:49:52.898535+00:00 | 2024-08-02T12:55:42.740359+00:00 | 86 | false | ***_PRETTY EASY APPROACH FOR BETTER UNDERSTANDING THE SLIDING WINDOW CONCEPT (BEGINNER FRIENDLY)_***\n\n```\npublic int minSwaps(int[] a) {\n\tint windowSize = 0;\n\tfor(int x : a){\n\t\twindowSize = x == 1 ? windowSize+1 : windowSize; \n\t} \n\tif (windowSize ==a.length)return 0;\n\n\tint back= 0 ;int front=windowSize... | 2 | 1 | ['Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window || Fixed Size || C++ | sliding-window-fixed-size-c-by-oreocraz_-wpra | 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 | oreocraz_273 | NORMAL | 2024-08-02T04:43:45.602849+00:00 | 2024-08-02T04:43:45.602883+00:00 | 183 | false | # 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 | 0 | ['Sliding Window', 'C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | ✅💪🔥 Simplest Python code || Sliding Window method || Beats 93% 🔥 || Easy understanding | simplest-python-code-sliding-window-meth-iu8m | Intuition\nAdd the same array and use sliding window approach.\n\n# Approach\n- Count the total number of 1\'s: We count how many 1\'s are in the original array | 1MB | NORMAL | 2024-08-02T04:17:28.822723+00:00 | 2024-08-02T04:17:28.822757+00:00 | 10 | false | # Intuition\nAdd the same array and use sliding window approach.\n\n# Approach\n- **Count the total number of 1\'s**: We count how many 1\'s are in the original array. This count determines the size of the window we need to consider for grouping all 1\'s together. **Edge case :** If the count is equal to the length of ... | 2 | 0 | ['Array', 'Sliding Window', 'Python'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | 2 min read Python solution || Beginner friendly | 2-min-read-python-solution-beginner-frie-00t7 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe will solve this problem using a sliding window approach.\n\n# Approach\n Describe yo | satwika-55 | NORMAL | 2024-08-02T04:03:59.105178+00:00 | 2024-08-02T04:04:42.207907+00:00 | 592 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will solve this problem using a sliding window approach.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSince the question describes a circular array, we create a new array doubling the original array.\n```\n a... | 2 | 0 | ['Python3'] | 3 |
minimum-swaps-to-group-all-1s-together-ii | "Optimal Code | TC O(n) | SC O(1) | Beats 93% | Easy to understand" | optimal-code-tc-on-sc-o1-beats-93-easy-t-fdau | Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to find the minimum number of swaps required to group all the 1s together i | coder_for_sure | NORMAL | 2024-08-02T03:09:48.253492+00:00 | 2024-08-02T03:19:02.863527+00:00 | 32 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to find the minimum number of swaps required to group all the 1s together in a given binary array. The approach revolves around the idea of using a sliding window to find the position where the group of 1s can be moved to mini... | 2 | 0 | ['Sliding Window', 'Python3'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Easy Sliding Window Solution | Golang | easy-sliding-window-solution-golang-by-j-xzse | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can use a sliding window to check the number of ones missing from each interval. The | jonalfarlinga | NORMAL | 2024-08-02T02:32:27.399160+00:00 | 2024-08-02T02:32:27.399199+00:00 | 91 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can use a sliding window to check the number of ones missing from each interval. The fact that the list is circular, means that we need to ensure we check the window starting before index `0`.\n\nThe size of the window, $$k$$, is the n... | 2 | 0 | ['Array', 'Sliding Window', 'Go'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Easy Video Solution || Faster than 97% | c-easy-video-solution-faster-than-97-by-i4l8i | The video soltuion for the below code is\nhttps://youtu.be/ow8SNSXYepc\n\n# Code\n\nclass Solution {\nprivate:\n int minCount(vector<int>& nums, int check){\ | __SAI__NIVAS__ | NORMAL | 2024-08-02T02:26:17.708464+00:00 | 2024-08-02T02:26:17.708494+00:00 | 584 | false | The video soltuion for the below code is\nhttps://youtu.be/ow8SNSXYepc\n\n# Code\n```\nclass Solution {\nprivate:\n int minCount(vector<int>& nums, int check){\n int sz = nums.size();\n int windowSize = 0;\n for(auto &num : nums) windowSize += (num == check);\n int maxWindowCheckerCnt = I... | 2 | 0 | ['C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | C# Solution for Minimum Swaps to Group All 1's Together II Problem | c-solution-for-minimum-swaps-to-group-al-otzq | Intuition\n Describe your first thoughts on how to solve this problem. \n1.\tCalculate Total Ones: Count the total number of 1\u2019s (totalOnes) in the array.\ | Aman_Raj_Sinha | NORMAL | 2024-08-02T00:57:04.956955+00:00 | 2024-08-02T00:57:04.956982+00:00 | 220 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.\tCalculate Total Ones: Count the total number of 1\u2019s (totalOnes) in the array.\n2.\tPrefix Sums: Use a prefix sum array to quickly calculate the number of 1\u2019s in any subarray. This helps in efficiently computing the number of... | 2 | 0 | ['C#'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Group all 1s and 0s together, Sliding Window | group-all-1s-and-0s-together-sliding-win-gbbn | Channel Link\n\n# Approach\nGroup all 1s together\nGroup all 0s together ( it will result in 1s placed at corners )\n\n# Complexity\n- Time complexity: O(n)\n A | cs_iitian | NORMAL | 2024-08-02T00:42:33.839589+00:00 | 2024-08-02T01:00:01.755349+00:00 | 914 | false | ## [Channel Link](https://www.youtube.com/channel/UCuxmikkhqbmBOUVxf-61hxw)\n\n# Approach\nGroup all 1s together\nGroup all 0s together ( it will result in 1s placed at corners )\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 sp... | 2 | 0 | ['Java'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | JS | js-by-manu-bharadwaj-bn-2tbt | Code\n\nvar minSwaps = function (nums) {\n let n = nums.length;\n let x = nums.reduce((x, e) => x + e, 0);\n\n let max = -Infinity, y = 0;\n for (le | Manu-Bharadwaj-BN | NORMAL | 2024-03-28T06:01:26.873321+00:00 | 2024-03-28T06:01:26.873353+00:00 | 90 | false | # Code\n```\nvar minSwaps = function (nums) {\n let n = nums.length;\n let x = nums.reduce((x, e) => x + e, 0);\n\n let max = -Infinity, y = 0;\n for (let z = 0, a = -x; z < n + x; z++, a++) {\n y += nums[z % n];\n if (a >= 0) {\n y -= nums[a % n];\n }\n max = Math.max... | 2 | 0 | ['JavaScript'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Easy implementation || beats 97.42% | easy-implementation-beats-9742-by-jeet_s-p7w3 | Intuition\n\n Describe your first thoughts on how to solve this problem. \nUse fixed size sliding window technique in which size of the window is total number o | jeet_sankala | NORMAL | 2024-02-03T11:58:06.680872+00:00 | 2024-02-03T12:02:42.382042+00:00 | 424 | false | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse fixed size sliding window technique in which size of the window is total number of 1\'s(k) in t... | 2 | 0 | ['Sliding Window', 'C++'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | Java/Python/Typescript Sliding window technique Time complexity: O(n)/Space complexity: O(1) | javapythontypescript-sliding-window-tech-gyk8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nThe approach to solving this problem involves using a sliding window tech | akashgond3112 | NORMAL | 2023-10-02T20:59:18.642390+00:00 | 2023-10-02T20:59:18.642412+00:00 | 54 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe approach to solving this problem involves using a sliding window technique to find the minimum number of swaps required to group all the ones together in a circular binary array.\n\n1. First, count the total number of on... | 2 | 0 | ['Python', 'Java', 'TypeScript'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ Solution | Sliding Window | Easy to Understand | c-solution-sliding-window-easy-to-unders-pw6z | \nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n\n int n = nums.size();\n int target = 0;\n\n for(int i=0;i<n;i++) {\n | Adi_217 | NORMAL | 2023-08-03T14:06:58.315265+00:00 | 2023-08-03T14:06:58.315285+00:00 | 164 | false | ```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n\n int n = nums.size();\n int target = 0;\n\n for(int i=0;i<n;i++) {\n if(nums[i]==1) {\n target++;\n }\n }\n\n if(target == 0) {\n return 0;\n }\n\n fo... | 2 | 0 | ['Two Pointers', 'Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window Easy , C++ Beats 90% ✅✅ | sliding-window-easy-c-beats-90-by-deepak-98lf | Approach\n Describe your approach to solving the problem. \nwhat do you think about Window Size.......???\n\nWindow Size = No. of 1 in the given array\nNow try | Deepak_5910 | NORMAL | 2023-06-26T07:53:10.387234+00:00 | 2023-06-26T07:53:47.367901+00:00 | 421 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nwhat do you think about Window Size.......???\n\n**Window Size = No. of 1 in the given array**\nNow try to find out the window size subarray that contains maximum 1.\n\n**Note:-to Handle the circular array problem append the array itself.**\n\n# Compl... | 2 | 0 | ['Sliding Window', 'C++'] | 2 |
minimum-swaps-to-group-all-1s-together-ii | ❇ Crisp n Clear👌 🏆O(N)❤️ Javascript❤️ Runtime👀90%🕕 Meaningful Vars✍️ 🔴❇ ✅ 👉 💪🙏 | crisp-n-clear-on-javascript-runtime90-me-1dyk | Intuition\n Describe your first thoughts on how to solve this problem. \nCount how many total ones are there(totalOnes)\n\n# Approach\n Describe your approach t | anurag-sindhu | NORMAL | 2023-01-08T15:22:15.319988+00:00 | 2023-01-08T15:22:15.320029+00:00 | 622 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCount how many total ones are there(totalOnes)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nselect first initial block and how many ones are there\nfind a block of totalOnes length which has maxOnesInBlock\nretu... | 2 | 0 | ['JavaScript'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Short & Concise Sliding window | C++ | short-concise-sliding-window-c-by-tushar-s94k | \nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int l = 0, s = 0, i = 0, ans = INT_MAX;\n for(int i : nums) l += i;\n | TusharBhart | NORMAL | 2022-10-14T16:25:20.587945+00:00 | 2022-10-14T16:25:20.587991+00:00 | 527 | false | ```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int l = 0, s = 0, i = 0, ans = INT_MAX;\n for(int i : nums) l += i;\n nums.insert(nums.end(), nums.begin(), nums.end());\n \n for(int j=0; j<nums.size(); j++) {\n s += nums[j];\n if(j - i + 1... | 2 | 0 | ['Sliding Window', 'C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | 📌📌Sliding Window || Easy To Understand || C++ Code | sliding-window-easy-to-understand-c-code-e9s8 | Using Sliding Window\n\n Time Complexity :- O(N)\n\n Space Complexity :- O(N)\n\n\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n\n i | __KR_SHANU_IITG | NORMAL | 2022-10-10T10:30:03.521165+00:00 | 2022-10-10T10:30:03.521202+00:00 | 754 | false | * ***Using Sliding Window***\n\n* ***Time Complexity :- O(N)***\n\n* ***Space Complexity :- O(N)***\n\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n\n int n = nums.size();\n\n // count the no. of ones in nums\n\n int count_1 = 0;\n\n for(int i = 0; i < n; i++)\n ... | 2 | 0 | ['C', 'Sliding Window', 'C++'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | Interview Ready Solution | Easy To Understand | Sliding Window | interview-ready-solution-easy-to-underst-yecg | I understand this might not be the best leetcode runtime or memory usuage solution but it is the most understandble for interview purpose.\n\n```\nclass Solutio | Sim1337 | NORMAL | 2022-07-23T15:38:27.333007+00:00 | 2022-07-23T15:38:27.333032+00:00 | 256 | false | I understand this might not be the best leetcode runtime or memory usuage solution but it is the most understandble for interview purpose.\n\n```\nclass Solution {\n //Time: O(n), where n represent the length of nums\n //Space: O(1), no need to store any data in any DS\n public int minSwaps(int[] nums) {\n ... | 2 | 0 | ['Sliding Window', 'Java'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | JavaScript Solution - Sliding Window | javascript-solution-sliding-window-by-de-1ce1 | \nvar minSwaps = function(nums) {\n const MAX = Number.MAX_SAFE_INTEGER;\n \n let ones = nums.reduce((acc, bit) => acc + bit, 0);\n \n const doub | Deadication | NORMAL | 2022-05-30T02:49:22.212997+00:00 | 2022-05-30T02:51:12.827251+00:00 | 258 | false | ```\nvar minSwaps = function(nums) {\n const MAX = Number.MAX_SAFE_INTEGER;\n \n let ones = nums.reduce((acc, bit) => acc + bit, 0);\n \n const doubledNums = nums.concat(nums.slice(0, nums.length - 1));\n \n let minSwap = MAX;\n let left = 0;\n let onesWithinWindow = 0;\n \n for (let i ... | 2 | 0 | ['Sliding Window', 'JavaScript'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ || Sliding Window approach || Removing circularity | c-sliding-window-approach-removing-circu-ti5d | ,,,\n\n int minSwaps(vector& nums) {\n \n int window=count(nums.begin(), nums.end(), 1); //determining window size by counting number of occu | mrigank_2003 | NORMAL | 2022-02-27T05:21:07.063692+00:00 | 2022-02-27T05:34:27.691457+00:00 | 107 | false | ,,,\n\n int minSwaps(vector<int>& nums) {\n \n int window=count(nums.begin(), nums.end(), 1); //determining window size by counting number of occurance of 1.\n nums.insert(nums.end(), nums.begin(), nums.end()); //in order to convert circular array to linear one appending the same array at the ... | 2 | 0 | ['C', 'Sliding Window'] | 0 |
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