question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
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transform-to-chessboard | Transform to Chessboard | JS | just check bro | beat 100% | transform-to-chessboard-js-just-check-br-fbxd | The key here is realizing that since you cannot move individual cells, but only rows and columns, what started off in a column/row will always remain there. \nE | junyanbill | NORMAL | 2021-09-26T19:19:14.288578+00:00 | 2021-09-26T19:19:14.288619+00:00 | 252 | false | The key here is realizing that since you cannot move individual cells, but only rows and columns, what started off in a column/row will always remain there. \nE.g. in the first example case, you can view it as:\n * col 0 and col 1 swap places\n * row 1 and row 2 swap places\nAnd you can see none of the zeros and ones a... | 2 | 1 | ['JavaScript'] | 1 |
transform-to-chessboard | Python Solutions | python-solutions-by-anmol957-xlaw | \nclass Solution:\n def movesToChessboard(self, board):\n n = len(board)\n patt1 = ([0, 1]*(n//2+1))[:n]\n patt2 = ([1, 0]*(n//2+1))[:n] | anmol957 | NORMAL | 2021-09-26T17:07:09.435683+00:00 | 2021-09-26T17:07:09.435715+00:00 | 93 | false | ```\nclass Solution:\n def movesToChessboard(self, board):\n n = len(board)\n patt1 = ([0, 1]*(n//2+1))[:n]\n patt2 = ([1, 0]*(n//2+1))[:n]\n \n board_t = map(list, zip(*board))\n Cnt_r = list(Counter(tuple(row) for row in board).items())\n Cnt_c = list(Counter(tuple(... | 2 | 2 | [] | 0 |
transform-to-chessboard | [C++]Every Step Explained | cevery-step-explained-by-rajneelesh473-poqn | This isn\'t easy to get for me and I hope my code and comment is clear. I think this is a better code readability practice that algo practice.\nPlease let me kn | rajneelesh473 | NORMAL | 2021-09-26T15:42:42.633997+00:00 | 2021-09-26T15:42:42.634026+00:00 | 310 | false | This isn\'t easy to get for me and I hope my code and comment is clear. I think this is a better code readability practice that algo practice.\nPlease let me know the confusing parts :P\n```\nclass Solution {\npublic:\n // Both the column and row swap satisfies the following\n // property:\n // All the numbers... | 2 | 0 | [] | 0 |
transform-to-chessboard | Java | Counting repeating parts | java-counting-repeating-parts-by-samoyle-f2dx | It is somewhat differrent from the other solutions here. We can count repeating parts in rows and columns, like this 0111001: countOnes=3, countZeros=2 (so, min | SamoylenkoDmitry | NORMAL | 2021-09-26T11:45:33.443278+00:00 | 2021-09-26T17:23:10.179932+00:00 | 469 | false | It is somewhat differrent from the other solutions here. We can count repeating parts in rows and columns, like this `0111001`: countOnes=3, countZeros=2 (so, min=2, max=3). And we expect that `min` and `max` repeating parts are constant between the rows and the columns.\n\nAfter we proved that board is correct, we sim... | 2 | 1 | ['Java'] | 3 |
transform-to-chessboard | [C++] first check if we can transform, then count the miss match row and col. | c-first-check-if-we-can-transform-then-c-yaa2 | \nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = board.size();\n \n int cnt1 = 1, cnt2 = 0;\n | lovebaonvwu | NORMAL | 2021-09-26T08:31:12.524472+00:00 | 2021-09-26T08:31:12.524511+00:00 | 414 | false | ```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = board.size();\n \n int cnt1 = 1, cnt2 = 0;\n \n for (int i = 1; i < n; ++i) {\n \n if (board[0][0] == board[i][0]) {\n ++cnt1;\n \n ... | 2 | 0 | [] | 0 |
transform-to-chessboard | Compress rows/columns | compress-rowscolumns-by-madno-i37t | Since bounds are small, the rows/columns can be compressed into vectors.\n\nEach of the two vectors should contain exactly two type of values, \none with cardin | madno | NORMAL | 2020-12-29T14:52:43.487950+00:00 | 2021-02-06T18:03:14.377756+00:00 | 509 | false | Since bounds are small, the rows/columns can be compressed into vectors.\n\nEach of the two vectors should contain exactly two type of values, \none with cardinality n/2, the other with n-n/2. And each value has\nn/2 or n-n/2 bit set.\n\nThe rows and columns can be manipulated independently: a swap in a row\ndoesn\'t a... | 2 | 0 | ['Bit Manipulation', 'Java'] | 0 |
transform-to-chessboard | Accepted C# Solution | accepted-c-solution-by-maxpushkarev-7m7v | \n public class Solution\n {\n private int Helper(int[] rows, int[] rowPatterns, int[] columnPatterns, int n, int m)\n {\n int ro | maxpushkarev | NORMAL | 2020-04-04T09:22:17.869344+00:00 | 2020-04-04T09:22:17.869379+00:00 | 205 | false | ```\n public class Solution\n {\n private int Helper(int[] rows, int[] rowPatterns, int[] columnPatterns, int n, int m)\n {\n int rowSwaps = 0;\n\n for (int i = 0; i < n; i++)\n {\n if (rows[i] != rowPatterns[i & 1])\n {\n ... | 2 | 0 | [] | 0 |
transform-to-chessboard | [JAVA] a lot of bit operations | java-a-lot-of-bit-operations-by-tyuan73-ylru | int n = 0, mask = 0, p0 = 0xAAAAAAAA, p1 = 0x55555555;\n public int movesToChessboard(int[][] board) {\n n = board.length; // the size\n mask = | tyuan73 | NORMAL | 2018-02-12T05:08:12.303000+00:00 | 2018-02-12T05:08:12.303000+00:00 | 812 | false | int n = 0, mask = 0, p0 = 0xAAAAAAAA, p1 = 0x55555555;\n public int movesToChessboard(int[][] board) {\n n = board.length; // the size\n mask = (1 << n) - 1; // the mask, for example: 11111 for n = 5\n p0 &= mask; // the chessboard line ending with 0, for example: 01010 for n = 5\n p1... | 2 | 0 | [] | 0 |
transform-to-chessboard | [C++] Swapping rows and cols while fixing the color of first element | c-swapping-rows-and-cols-while-fixing-th-r4fs | We see that we can swap column and rows independently. So we can make sure first row and first column are in the correct order and then just verify whether the | facelessvoid | NORMAL | 2018-02-11T04:10:19.108000+00:00 | 2018-02-11T04:10:19.108000+00:00 | 775 | false | We see that we can swap column and rows independently. So we can make sure first row and first column are in the correct order and then just ```verify``` whether the rest of the board is valid. While doing this we can count minimum possible swaps in rows and cols to achieve this state.\nThe only thing that can vary is ... | 2 | 0 | [] | 0 |
transform-to-chessboard | Solution | solution-by-deleted_user-21as | C++ []\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = board.size();\n int row_counter = 0, col | deleted_user | NORMAL | 2023-04-29T08:53:19.004876+00:00 | 2023-04-29T09:09:24.322764+00:00 | 988 | false | ```C++ []\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = board.size();\n int row_counter = 0, col_counter = 0;\n for(int r = 0; r < n; r++){\n row_counter += board[r][0] ? 1 : -1;\n for(int c = 0; c < n; c++){\n ... | 1 | 0 | ['C++', 'Java', 'Python3'] | 0 |
transform-to-chessboard | JAVA Easy & Concise | java-easy-concise-by-rohitkumarsingh369-h904 | \nclass Solution {\n public int movesToChessboard(int[][] board) {\n int N = board.length, colToMove = 0, rowToMove = 0, rowOneCnt = 0, colOneCnt = 0; | rohitkumarsingh369 | NORMAL | 2021-09-27T06:39:23.283954+00:00 | 2021-09-27T06:39:23.283998+00:00 | 445 | false | ```\nclass Solution {\n public int movesToChessboard(int[][] board) {\n int N = board.length, colToMove = 0, rowToMove = 0, rowOneCnt = 0, colOneCnt = 0;\n for (int i = 0; i < N; i++) {\n for (int j = 0; j < N; j++) {\n if (((board[0][0] ^ board[i][0]) ^ (board[i][j] ^ board[0... | 1 | 0 | ['Java'] | 0 |
transform-to-chessboard | Rust bit based solution (for fun) | rust-bit-based-solution-for-fun-by-michi-wslw | I had a lot of fun with this one. Part of this relies on the fact that n <= 30. This means that the rows and columns will fit into an i32.\n\nThe hardest part | michielbaird | NORMAL | 2021-09-26T19:00:33.105060+00:00 | 2021-09-26T19:00:33.105090+00:00 | 138 | false | I had a lot of fun with this one. Part of this relies on the fact that `n <= 30`. This means that the rows and columns will fit into an `i32`.\n\nThe hardest part in my opinion is figuring out whether a given board is solvable. Going to leave out how I prove this but a solvable board has the following properties:\n\n-... | 1 | 0 | ['Bit Manipulation', 'Rust'] | 0 |
transform-to-chessboard | My C++ Solution | my-c-solution-by-muskiajbhave-sk2c | ```\nint movesToChessboard(vector>& board) {\n int n = board.size();\n int maxR = 0, maxC = 0, ones = 0, zeros = 0;\n for (int i = 0; i < n | muskiajbhave | NORMAL | 2021-09-26T16:07:23.159244+00:00 | 2021-09-26T16:07:23.159296+00:00 | 124 | false | ```\nint movesToChessboard(vector<vector<int>>& board) {\n int n = board.size();\n int maxR = 0, maxC = 0, ones = 0, zeros = 0;\n for (int i = 0; i < n; i++) {\n int rzero = 0, rone = 0, czero = 0, cone = 0;\n maxR = max(maxR, helper(-1, i, board, rzero, rone, n));\n ... | 1 | 1 | [] | 0 |
transform-to-chessboard | Simple C solution | simple-c-solution-by-linhbk93-w6k2 | \nnt movesToChessboard(int** board, int boardSize, int* boardColSize){\n int rowSum = 0, colSum = 0, row = 0, col = 0;\n for (int i = 0; i < boardSize; ++ | linhbk93 | NORMAL | 2021-09-26T08:45:45.545222+00:00 | 2021-09-26T08:45:45.545257+00:00 | 233 | false | ```\nnt movesToChessboard(int** board, int boardSize, int* boardColSize){\n int rowSum = 0, colSum = 0, row = 0, col = 0;\n for (int i = 0; i < boardSize; ++i) \n for (int j = 0; j < boardSize; ++j)\n if (board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]) return -1;\n for (int i = 0; i < ... | 1 | 0 | ['C'] | 0 |
transform-to-chessboard | (C++) 782. Transform to Chessboard | c-782-transform-to-chessboard-by-qeetcod-yls8 | \n\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = size(board); \n \n auto fn = [&](vector<in | qeetcode | NORMAL | 2021-06-29T21:33:15.805163+00:00 | 2021-06-29T21:33:15.805195+00:00 | 223 | false | \n```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int n = size(board); \n \n auto fn = [&](vector<int>& vals) {\n int total = 0, odd = 0; \n for (int i = 0; i < size(vals); ++i) {\n if (vals[0] == vals[i]) {\n ... | 1 | 0 | ['C'] | 0 |
transform-to-chessboard | Python Solution | python-solution-by-0105cs161104-cpr6 | \nclass Solution:\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n=len(board)\n for i in range(0,n):\n for j in ran | 0105CS161104 | NORMAL | 2021-06-04T09:10:59.396475+00:00 | 2021-06-04T09:10:59.396519+00:00 | 233 | false | ```\nclass Solution:\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n=len(board)\n for i in range(0,n):\n for j in range(0,n):\n if(board[0][0] ^ board[0][j] ^ board[i][0] ^ board[i][j]!=0):\n return -1\n rowSum=0\n colSum= 0\n... | 1 | 0 | [] | 0 |
transform-to-chessboard | C++ Easy to Understand Solution | c-easy-to-understand-solution-by-xiaopin-wqg4 | This solution is based on an observation by mzchen: After the moving, all rows with even numbers are the same and so are the rows with odd numbers. \nFor calcul | xiaoping3418 | NORMAL | 2021-04-22T11:52:46.409567+00:00 | 2021-04-24T23:57:43.070617+00:00 | 331 | false | This solution is based on an observation by mzchen: After the moving, all rows with even numbers are the same and so are the rows with odd numbers. \nFor calculating the moves on rows, we use a mapping table to record the index of the rows:\n1) return -1 if the mapping size is not two.\n2) return -1 if the counts of th... | 1 | 0 | [] | 1 |
transform-to-chessboard | [C++] Bit operations and explanation in comments | c-bit-operations-and-explanation-in-comm-dbsl | \nbool debug = false;\n\n/*\ntest cases\n[[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]\n[[0, 1], [1, 0]]\n[[1, 0], [1, 0]]\n[[0,1,0],[1,0,1],[0,1,0]]\n[[0,1,0],[1,0 | genxium | NORMAL | 2021-02-16T06:07:43.220289+00:00 | 2021-02-16T06:08:30.192583+00:00 | 273 | false | ```\nbool debug = false;\n\n/*\ntest cases\n[[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]\n[[0, 1], [1, 0]]\n[[1, 0], [1, 0]]\n[[0,1,0],[1,0,1],[0,1,0]]\n[[0,1,0],[1,0,1],[1,0,1]]\n[[1,1,0],[0,0,1],[0,0,1]]\n*/\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n /*\n [Lemma#1... | 1 | 0 | [] | 0 |
transform-to-chessboard | C++ Straightly and Ugly code，But, AC Solution | c-straightly-and-ugly-codebut-ac-solutio-f18f | \nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int res = min(helper(board, 0), helper(board, 1));\n return | woshiyuanlei | NORMAL | 2021-01-13T03:37:25.386361+00:00 | 2021-01-13T03:37:25.386408+00:00 | 171 | false | ```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int res = min(helper(board, 0), helper(board, 1));\n return res == INT_MAX ? -1 : res;\n }\n int helper(vector<vector<int>> board, int begin) {\n int res1 = movesToChessboardForRow(board, begin);\n ... | 1 | 0 | [] | 0 |
transform-to-chessboard | Simple C++ solution | simple-c-solution-by-caspar-chen-hku-dvth | \nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& b) {\n int N = b.size(), rowSum = 0, colSum = 0, rowSwap = 0, colSwap = 0;\n | caspar-chen-hku | NORMAL | 2020-05-18T12:51:33.290131+00:00 | 2020-05-18T12:51:33.290168+00:00 | 497 | false | ```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& b) {\n int N = b.size(), rowSum = 0, colSum = 0, rowSwap = 0, colSwap = 0;\n for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j){\n if (b[0][0]^b[i][0]^b[0][j]^b[i][j]) return -1; \n }\n for (int ... | 1 | 1 | [] | 1 |
transform-to-chessboard | java brute force...... code is very tedious and long | java-brute-force-code-is-very-tedious-an-8nj6 | \nclass Solution {\n int n;\n public int movesToChessboard(int[][] board) {\n n=board.length;int ans=-1;\n if(n%2==0){\n StringBu | CUNY-66brother | NORMAL | 2020-02-12T07:02:39.114065+00:00 | 2020-02-12T07:05:20.353516+00:00 | 495 | false | ```\nclass Solution {\n int n;\n public int movesToChessboard(int[][] board) {\n n=board.length;int ans=-1;\n if(n%2==0){\n StringBuilder s2=new StringBuilder();\n String s1=coltos(board,0);\n for(int i=0;i<s1.length();i++){\n if(s1.charAt(i)==\'0\'){\... | 1 | 0 | [] | 1 |
transform-to-chessboard | Java - 1 ms, faster than 100.00% - 43.6 MB, less than 88.89% | java-1-ms-faster-than-10000-436-mb-less-5c4k8 | \nclass Solution {\n public int movesToChessboard(int[][] board) {\n if (board == null || board.length == 0 || board[0].length == 0) {\n re | woshilxd912 | NORMAL | 2019-06-16T18:03:10.107762+00:00 | 2019-06-16T18:03:10.107809+00:00 | 807 | false | ```\nclass Solution {\n public int movesToChessboard(int[][] board) {\n if (board == null || board.length == 0 || board[0].length == 0) {\n return -1;\n }\n int N = board.length;\n for (int i = 0; i < N; ++i) {\n for (int j = 0; j < N; ++j) {\n if ((bo... | 1 | 0 | [] | 1 |
transform-to-chessboard | A detailed and easy to understand C++ version. | a-detailed-and-easy-to-understand-c-vers-3mk4 | //Feb 13 2018 //A problem that costs me a long time //Property: //1. To tranform columns, you must exchange rows and vice versa //2. The beginning board contai | debiluz | NORMAL | 2018-02-14T04:35:33.962747+00:00 | 2018-02-14T04:35:33.962747+00:00 | 599 | false | //Feb 13 2018
//A problem that costs me a long time
//Property:
//1. To tranform columns, you must exchange rows and vice versa
//2. The beginning board contains only two different kinds of rows(columns), and the two kinds are 0/1 reversed
//This is correct because after a swap of rows, columns that are the same rem... | 1 | 1 | [] | 0 |
transform-to-chessboard | 9ms short C++ solution, easy to understand | 9ms-short-c-solution-easy-to-understand-n00e5 | ``` class Solution { public: int movesToChessboard(vector>& board) { int N=board.size(),row_bal=0,col_bal=0,c=0,d=0; for(int i=0;i1||row_bal | ztc254 | NORMAL | 2018-02-14T01:30:54.169772+00:00 | 2018-02-14T01:30:54.169772+00:00 | 616 | false | ```
class Solution {
public:
int movesToChessboard(vector<vector<int>>& board) {
int N=board.size(),row_bal=0,col_bal=0,c=0,d=0;
for(int i=0;i<N;++i){
row_bal+=board[0][i]?1:-1;
col_bal+=board[i][0]?1:-1;
if(i&1){
c+=board[0][i];
d+... | 1 | 0 | [] | 0 |
transform-to-chessboard | ✅✅ Dimension Independence || C++ || Beats 100% 🔥🔥 | dimension-independence-c-beats-100-by-an-xp26 | Code | AntonioMalloc | NORMAL | 2025-01-24T01:30:24.195063+00:00 | 2025-01-24T01:30:24.195063+00:00 | 15 | false | # Code
```cpp []
class Solution {
public:
int movesToChessboard(vector<vector<int>>& board) {
int N = board.size(); // The size of the board (N x N)
// Count the frequency of each unique row in the board
unordered_map<int, int> count;
for (const auto& row : board) {
int ... | 0 | 0 | ['Array', 'Math', 'Bit Manipulation', 'Matrix', 'C++'] | 0 |
transform-to-chessboard | 782. Transform to Chessboard | 782-transform-to-chessboard-by-g8xd0qpqt-ouqu | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-01T16:26:36.857660+00:00 | 2025-01-01T16:26:36.857660+00:00 | 16 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
transform-to-chessboard | generate zebra pattern | generate-zebra-pattern-by-user5285zn-ed4l | There is no interference between row an column operation. Hence, we need to count how many operations are needed to make the first row and the first column to z | user5285Zn | NORMAL | 2024-09-05T16:08:59.248794+00:00 | 2024-09-05T16:08:59.248817+00:00 | 0 | false | There is no interference between row an column operation. Hence, we need to count how many operations are needed to make the first row and the first column to zebra pattern.\n\nIn addition, we need to check whether the whole matrix follows the same pattern as prescribed by leading row and column.\n\n\n```rust []\nfn ze... | 0 | 0 | ['Rust'] | 0 |
transform-to-chessboard | BFS and Mismatch Counting | bfs-and-mismatch-counting-by-orel12-hnnf | Intuition\nTransforming a board into a chessboard pattern involves strict conditions: alternating 0s and 1s across rows and columns. The challenge is to determi | orel12 | NORMAL | 2024-09-03T21:24:37.464827+00:00 | 2024-09-03T21:24:37.464850+00:00 | 11 | false | # Intuition\nTransforming a board into a chessboard pattern involves strict conditions: alternating `0`s and `1`s across rows and columns. The challenge is to determine if a given board can be rearranged to meet these conditions by swapping elements. Key considerations include ensuring equal or near-equal distribution ... | 0 | 0 | ['Matrix', 'Python3'] | 0 |
transform-to-chessboard | [Python] Detailed Intuition + Easy To Understand Code - Beats 98% | python-detailed-intuition-easy-to-unders-ywhg | Breakdown\n Describe your first thoughts on how to solve this problem. \n1. Check whether a solution is possible\n2. Calculate number of moves\n\n## Is a soluti | raayc | NORMAL | 2024-07-15T02:16:30.640042+00:00 | 2024-07-15T02:36:17.070448+00:00 | 36 | false | # Breakdown\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Check whether a solution is possible\n2. Calculate number of moves\n\n## Is a solution possible?\n\nFirst, we check if it is possible to turn the board into a chessboard. To do this, let\'s look at a chessboard:\n$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | cpp | cpp-by-pankajkumar101-u6qh | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | PankajKumar101 | NORMAL | 2024-05-17T00:18:35.987980+00:00 | 2024-05-17T00:18:35.988010+00:00 | 28 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
transform-to-chessboard | Kotlin with explanation | kotlin-with-explanation-by-arista_agarwa-5m2d | Approach\nValidity Check: The code starts by checking the validity of the chessboard by comparing each cell in the first row and first column with the top-left | Arista_agarwal | NORMAL | 2024-03-31T13:55:49.043171+00:00 | 2024-03-31T13:55:49.043203+00:00 | 5 | false | # Approach\nValidity Check: The code starts by checking the validity of the chessboard by comparing each cell in the first row and first column with the top-left cell (b[0][0]). It checks if the XOR of these cells equals 1. If it does, it means the chessboard cannot be formed, and -1 is returned.\nCounting Row and Colu... | 0 | 0 | ['Kotlin'] | 0 |
transform-to-chessboard | Clean & Clear & Easy to understand solution | clean-clear-easy-to-understand-solution-dl0bz | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | joelwangusa | NORMAL | 2024-02-26T02:55:51.893508+00:00 | 2024-02-26T02:55:51.893543+00:00 | 46 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | consider only the first row & col | consider-only-the-first-row-col-by-maxor-9ypt | Code\n\nclass Solution:\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n for i in range(n):\n for j | MaxOrgus | NORMAL | 2024-02-18T02:54:14.683133+00:00 | 2024-02-18T02:54:14.683155+00:00 | 14 | false | # Code\n```\nclass Solution:\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n for i in range(n):\n for j in range(n):\n for k in range(i+1,n):\n for l in range(j+1,n):\n if board[i][j] ^ board[i][l] != b... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | JS || Solution by Bharadwaj | js-solution-by-bharadwaj-by-manu-bharadw-3kst | Code\n\nvar movesToChessboard = function (board) {\n let boardSize = board.length;\n let boardSizeIsEven = boardSize % 2 === 0;\n if (!canBeTransformed | Manu-Bharadwaj-BN | NORMAL | 2024-02-08T07:24:11.406842+00:00 | 2024-02-08T07:24:11.406867+00:00 | 64 | false | # Code\n```\nvar movesToChessboard = function (board) {\n let boardSize = board.length;\n let boardSizeIsEven = boardSize % 2 === 0;\n if (!canBeTransformed(board)) return -1;\n let rowSwap = 0;\n let colSwap = 0;\n let rowSwap2 = 0;\n let colSwap2 = 0;\n\n for (let i = 0; i < boardSize; i++) {\... | 0 | 0 | ['JavaScript'] | 1 |
transform-to-chessboard | solution full comented PY3 | solution-full-comented-py3-by-borkiss-b6gk | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | borkiss | NORMAL | 2024-01-28T15:30:54.728551+00:00 | 2024-01-28T15:30:54.728582+00:00 | 16 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | Proper Explanation and easy code. Time - O(N^2) | proper-explanation-and-easy-code-time-on-f04p | Intuition\n Describe your first thoughts on how to solve this problem. \nanswer = Maximum(operations required for each row) + Maximum(operations required for ea | Jatin1358 | NORMAL | 2024-01-20T16:26:16.164187+00:00 | 2024-01-20T16:26:16.164209+00:00 | 40 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nanswer = Maximum(operations required for each row) + Maximum(operations required for each col)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Check the matrix is proper or not -\n a. In each row, count of on... | 0 | 0 | ['Java'] | 0 |
transform-to-chessboard | Python numpy solution | python-numpy-solution-by-dkoshman-6wv8 | Code\n\nimport numpy as np\n\ndef min_even_number(nums):\n return min(i for i in nums if not i & 1)\n\nclass Solution:\n def movesToChessboard(self, board | dkoshman | NORMAL | 2024-01-19T19:05:28.006867+00:00 | 2024-01-19T19:05:28.006898+00:00 | 9 | false | # Code\n```\nimport numpy as np\n\ndef min_even_number(nums):\n return min(i for i in nums if not i & 1)\n\nclass Solution:\n def movesToChessboard(self, board: List[List[int]]) -> int:\n board = np.array(board, dtype=bool)\n n = len(board)\n can_first_row_be_chessified = n // 2 <= board[0].s... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | Scala | scala-by-heavenwatcher-ee86 | scala\nobject Solution {\n def movesToChessboard(board: Array[Array[Int]]): Int = {\n def half(x: Int) = x == board.length / 2 || x == (board.length + 1) / | heavenwatcher | NORMAL | 2023-12-01T20:14:37.560997+00:00 | 2023-12-01T20:14:37.561031+00:00 | 2 | false | ```scala\nobject Solution {\n def movesToChessboard(board: Array[Array[Int]]): Int = {\n def half(x: Int) = x == board.length / 2 || x == (board.length + 1) / 2\n val vec_board = board.map(_.toVector)\n val patterns = vec_board.toSet\n Option\n .when(patterns.size == 2)(patterns.head, patterns.last)\n... | 0 | 0 | ['Scala'] | 0 |
transform-to-chessboard | Scala solution | scala-solution-by-malovig-ltcp | 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 | malovig | NORMAL | 2023-06-25T19:15:10.598551+00:00 | 2023-06-25T19:15:10.598574+00:00 | 13 | 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(n^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n^1)\n<!-- Add your space complexity here, e.... | 0 | 0 | ['Scala'] | 0 |
transform-to-chessboard | Swift Solution with Explanations and Dry Run. | swift-solution-with-explanations-and-dry-j5dr | Approach\nThe movesToChessboard function takes a 2D array (board) representing the chessboard configuration as input and returns an integer representing the min | aayushkumar20 | NORMAL | 2023-06-09T05:55:14.571259+00:00 | 2023-06-09T05:55:14.571307+00:00 | 10 | false | # Approach\nThe `movesToChessboard` function takes a 2D array (`board`) representing the chessboard configuration as input and returns an integer representing the minimum number of moves required. The code uses a dry run of the algorithm to explain the steps.\n\nHere\'s a step-by-step explanation of the code:\n\n1. Ini... | 0 | 0 | ['Swift'] | 0 |
transform-to-chessboard | Time: O(n2)O(n2) Space: O(1)O(1) | time-on2on2-space-o1o1-by-mdarafatuddin4-iepo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | mdarafatuddin481 | NORMAL | 2023-04-08T19:24:24.259260+00:00 | 2023-04-08T19:24:24.259290+00:00 | 50 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | Solid Principles | Dimension Independence | Commented and Explained | solid-principles-dimension-independence-h195o | Intuition\n Describe your first thoughts on how to solve this problem. \nOur main insight into the problem is that there must be a balance of either exactly the | laichbr | NORMAL | 2023-04-03T19:37:53.844797+00:00 | 2023-04-03T19:37:53.844838+00:00 | 49 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOur main insight into the problem is that there must be a balance of either exactly the same or alternating by 1 number of chess piece locations in two rows based on the idea that we have two rows next to each other (think of a 2 by 2 gri... | 0 | 0 | ['Python3'] | 0 |
transform-to-chessboard | [c++] Clean and Simple | O(n*n) | c-clean-and-simple-onn-by-projectcoder-csz5 | \nclass Solution {\npublic:\n int solve(int s, int h, int n) {\n int ans = 0;\n for (int i = 0; i < n; i++) ans += (((s>>i&1)==h) != (i%2==0)); | projectcoder | NORMAL | 2023-02-22T05:42:40.082688+00:00 | 2023-02-22T05:42:40.082744+00:00 | 39 | false | ```\nclass Solution {\npublic:\n int solve(int s, int h, int n) {\n int ans = 0;\n for (int i = 0; i < n; i++) ans += (((s>>i&1)==h) != (i%2==0)); \n return ans / 2;\n }\n \n \n int movesToChessboard(vector<vector<int>>& arr) { int n = arr.size();\n int row[110] = {... | 0 | 0 | [] | 0 |
transform-to-chessboard | Java working wit explanation | java-working-wit-explanation-by-obose-eudz | Intuition\n Describe your first thoughts on how to solve this problem. \nWe want to find the minimum number of swaps required to transform the given board into | Obose | NORMAL | 2023-01-26T19:06:32.603252+00:00 | 2023-01-26T19:06:32.603284+00:00 | 120 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe want to find the minimum number of swaps required to transform the given board into a "chessboard". A chessboard is defined as having alternating 1s and 0s row-wise and column-wise. For example: \n\n```\n[0, 1, 0, 1]\n[1, 0, 1, 0]\n[0,... | 0 | 0 | ['Java'] | 0 |
transform-to-chessboard | Just a runnable solution | just-a-runnable-solution-by-ssrlive-8fqr | Code\n\nimpl Solution {\n pub fn moves_to_chessboard(board: Vec<Vec<i32>>) -> i32 {\n let n = board.len();\n let mut row = 0;\n let mut | ssrlive | NORMAL | 2022-12-16T11:14:33.753148+00:00 | 2022-12-16T11:14:33.753190+00:00 | 51 | false | # Code\n```\nimpl Solution {\n pub fn moves_to_chessboard(board: Vec<Vec<i32>>) -> i32 {\n let n = board.len();\n let mut row = 0;\n let mut col = 0;\n let mut row_diff = 0;\n let mut col_diff = 0;\n for i in 0..n {\n for j in 0..n {\n if board[0][0... | 0 | 0 | ['Rust'] | 0 |
transform-to-chessboard | python3 | python3-by-swastik1712-ev4m | class Solution:\n\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n res = 0\n rowCnt = Counter(map(tuple, | swastik1712 | NORMAL | 2022-09-11T18:22:31.965020+00:00 | 2022-09-11T18:22:31.965067+00:00 | 45 | false | class Solution:\n\n def movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n res = 0\n rowCnt = Counter(map(tuple,board))\n colCnt = Counter(zip(*board))\n \n for count in (rowCnt,colCnt):\n if len(count)!=2:\n return -1\n ... | 0 | 0 | [] | 0 |
transform-to-chessboard | ✔️Python O(n) solution, O(n^2) check for possibility. With comments. No graphs. | python-on-solution-on2-check-for-possibi-gxdx | \n\tdef movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n \n # check that there are half zeroes and half ones in t | Veanules | NORMAL | 2022-04-18T15:14:53.632036+00:00 | 2022-04-18T15:18:54.023745+00:00 | 254 | false | \n\tdef movesToChessboard(self, board: List[List[int]]) -> int:\n n = len(board)\n \n # check that there are half zeroes and half ones in the first row and column\n ones = sum(board[0])\n if ones != n//2 and ones != (n+1)//2:\n return -1\n ones = sum(board[i][0] for ... | 0 | 0 | ['Python'] | 0 |
transform-to-chessboard | C++ O(n^2) | Reduce 2d problem to 2 1d problems | c-on2-reduce-2d-problem-to-2-1d-problems-zsyr | Rows and columns are orthogonal. You can make columns alternating and then make rows alternating or vice versa.\nThere can only be two unique rows and columns i | megatron10 | NORMAL | 2022-04-10T16:24:18.278269+00:00 | 2022-04-10T16:24:18.278318+00:00 | 247 | false | Rows and columns are orthogonal. You can make columns alternating and then make rows alternating or vice versa.\nThere can only be two unique rows and columns in a solvable board. Then this problem is reduced to 1d problems. Given a sequeune of the type of row/column, find steps required to make all it an alternating s... | 0 | 0 | ['C'] | 0 |
transform-to-chessboard | Java BFS + DFS but TLE | java-bfs-dfs-but-tle-by-catshaoshao-oy2b | \nHope anyone can improve my solution.\n\n...\npublic int movesToChessboard(int[][] board) {\n if (board == null || board.length == 0 || board[0] == null | catshaoshao | NORMAL | 2021-11-30T00:43:10.491833+00:00 | 2021-11-30T00:43:10.491865+00:00 | 260 | false | \nHope anyone can improve my solution.\n\n...\npublic int movesToChessboard(int[][] board) {\n if (board == null || board.length == 0 || board[0] == null || board[0].length == 0) return -1;\n \n int row = board.length;\n int col = board[0].length;\n HashSet<String> visited = new HashS... | 0 | 0 | ['Depth-First Search', 'Breadth-First Search'] | 0 |
transform-to-chessboard | [Java] exclude impossible cases, then calculate diff row/col | java-exclude-impossible-cases-then-calcu-3ook | \njava\n public int movesToChessboard(int[][] board) {\n // count how many ones in the first row and first col\n int rowOneCount = 0;\n | magiciendecode | NORMAL | 2021-09-29T18:17:41.953083+00:00 | 2021-09-29T18:17:41.953131+00:00 | 304 | false | \n```java\n public int movesToChessboard(int[][] board) {\n // count how many ones in the first row and first col\n int rowOneCount = 0;\n int colOneCount = 0;\n for (int[] ints :... | 0 | 0 | ['Java'] | 0 |
transform-to-chessboard | Java beats 98%: Long, but simple | java-beats-98-long-but-simple-by-jttux-r72k | Consider we\'re given the initial board\n\n1 1 0 0\n0 0 1 1\n1 1 0 0\n0 0 1 1\n\nThis algorithm begins by creating two lists: \n\nAdjustedRows = [1, 2, 1, 2]\nA | jttux | NORMAL | 2021-09-28T23:14:17.320555+00:00 | 2021-09-28T23:17:39.346022+00:00 | 275 | false | Consider we\'re given the initial board\n\n1 1 0 0\n0 0 1 1\n1 1 0 0\n0 0 1 1\n\nThis algorithm begins by creating two lists: \n\nAdjustedRows = [1, 2, 1, 2]\nAdjustedCols = [1, 1, 2, 2]\n\nIt then processes these lists to determine the number of swaps necessary to get into chessboard formation. Chessboard formation fo... | 0 | 0 | [] | 0 |
transform-to-chessboard | Remember it is a 1D problem (C++) | remember-it-is-a-1d-problem-c-by-tlj77-f67e | You just need to do two things:\n1. Make sure it\'s doable.\n2. find the minimum step to flip first row/column.\n\nWe can transpose the matrix, so do row equals | tlj77 | NORMAL | 2021-09-27T05:09:42.440081+00:00 | 2021-09-27T05:12:11.192193+00:00 | 103 | false | You just need to do two things:\n1. Make sure it\'s doable.\n2. find the minimum step to flip first row/column.\n\nWe can transpose the matrix, so do row equals do column.\n```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& board) {\n int row = row_min_step(board);\n if(row == ... | 0 | 0 | [] | 0 |
transform-to-chessboard | C# 100%. 100% | c-100-100-by-igor0-vhft | \npublic int MovesToChessboard(int[][] board) {\n\tif (IsBadChessboard(board)) return -1;\n\tif (IsBadChessboard2(board)) return -1;\n\tvar shiftsHor = 0;\n\tfo | igor0 | NORMAL | 2021-09-26T23:41:01.928034+00:00 | 2021-09-26T23:41:01.928066+00:00 | 65 | false | ```\npublic int MovesToChessboard(int[][] board) {\n\tif (IsBadChessboard(board)) return -1;\n\tif (IsBadChessboard2(board)) return -1;\n\tvar shiftsHor = 0;\n\tfor (int i = 0; i < board.Length; i++)\n\t{\n\t\tvar shiftsHor2 = GetShiftsHor(i, board);\n\t\tif (shiftsHor < shiftsHor2) shiftsHor = shiftsHor2;\n\t}\n\tvar ... | 0 | 0 | [] | 0 |
transform-to-chessboard | Python, short, invariants tracked | python-short-invariants-tracked-by-agaki-q19x | The invariants to check in this problem are:\n1. There should be only 2 unique rows\n2. They sum in every column must be 1\n3. Number od 1s and 0s in every row | agakishy | NORMAL | 2021-09-26T23:04:05.327987+00:00 | 2021-09-26T23:04:05.328020+00:00 | 114 | false | The invariants to check in this problem are:\n1. There should be only 2 unique rows\n2. They sum in every column must be 1\n3. Number od 1s and 0s in every row shouldn\'t be less then n//2\n4a. For odd n: if 3 above statements are true - there is only one posible solution ruled by most frequent row\n4b. For even n: the... | 0 | 0 | [] | 0 |
transform-to-chessboard | Go solution | go-solution-by-ganesh-i1p8 | https://github.com/ganeshskudva/Leetcode-Golang\n\nInspired by this solution\n\nfunc movesToChessboard(board [][]int) int {\n\tn := len(board)\n\tfor i := 0; i | ganesh | NORMAL | 2021-09-26T21:08:16.348483+00:00 | 2021-09-26T21:08:16.348529+00:00 | 31 | false | ERROR: type should be string, got "https://github.com/ganeshskudva/Leetcode-Golang\\n\\nInspired by [this](https://leetcode.com/problems/transform-to-chessboard/discuss/114847/C%2B%2BJavaPython-Solution-with-Explanation) solution\\n```\\nfunc movesToChessboard(board [][]int) int {\\n\\tn := len(board)\\n\\tfor i := 0; i < n; i++ {\\n\\t\\tfor j := 0; j < n; j++ {\\n\\t\\t\\tif (board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]) == 1 {\\n\\t\\t\\t\\treturn -1\\n\\t\\t\\t}\\n\\t\\t}\\n\\t}\\n\\n\\trowSum, colSum, rowMisplaced, colMisplaced := 0, 0, 0, 0\\n\\tfor i := 0; i < n; i++ {\\n\\t\\trowSum += board[0][i]\\n\\t\\tcolSum += board[i][0]\\n\\t\\tif board[i][0] == i%2 {\\n\\t\\t\\trowMisplaced++\\n\\t\\t}\\n\\t\\tif board[0][i] == i%2 {\\n\\t\\t\\tcolMisplaced++\\n\\t\\t}\\n\\t}\\n\\n\\tif rowSum != n/2 && rowSum != (n+1)/2 {\\n\\t\\treturn -1\\n\\t}\\n\\tif colSum != n/2 && colSum != (n+1)/2 {\\n\\t\\treturn -1\\n\\t}\\n\\n\\tif n%2 == 1 {\\n\\t\\tif colMisplaced%2 == 1 {\\n\\t\\t\\tcolMisplaced = n - colMisplaced\\n\\t\\t}\\n\\t\\tif rowMisplaced%2 == 1 {\\n\\t\\t\\trowMisplaced = n - rowMisplaced\\n\\t\\t}\\n\\t} else {\\n\\t\\tcolMisplaced = min(n-colMisplaced, colMisplaced)\\n\\t\\trowMisplaced = min(n-rowMisplaced, rowMisplaced)\\n\\t}\\n\\n\\treturn (colMisplaced + rowMisplaced) / 2\\n}\\n\\nfunc min(a, b int) int {\\n\\tif a < b {\\n\\t\\treturn a\\n\\t}\\n\\n\\treturn b\\n}\\n```" | 0 | 0 | [] | 0 |
transform-to-chessboard | {Python} Dimension independence + 2-building block for row and column | python-dimension-independence-2-building-cmxf | Approach 1[1]: Dimension independence + 2-building block for row and column\nTime/Space Complexity: O(N)\n\n\nclass Solution(object):\n def movesToChessboard | codedayday | NORMAL | 2021-09-26T20:22:23.258452+00:00 | 2021-09-26T22:41:31.031137+00:00 | 154 | false | Approach 1[1]: Dimension independence + 2-building block for row and column\nTime/Space Complexity: O(N)\n\n```\nclass Solution(object):\n def movesToChessboard(self, board):\n N = len(board)\n ans = 0\n # For each count of lines from {rows, columns}...\n for count in (collections.Counter... | 0 | 0 | ['Divide and Conquer', 'Python'] | 0 |
transform-to-chessboard | Go solution | Beats 100% | go-solution-beats-100-by-evleria-z3yf | \nfunc movesToChessboard(board [][]int) int {\n\tn, rowSum, colSum, rowSwap, colSwap := len(board), 0, 0, 0, 0\n\n\tfor i := 0; i < n; i++ {\n\t\tfor j := 0; j | evleria | NORMAL | 2021-09-26T17:57:48.895913+00:00 | 2021-09-26T17:58:38.543524+00:00 | 83 | false | ```\nfunc movesToChessboard(board [][]int) int {\n\tn, rowSum, colSum, rowSwap, colSwap := len(board), 0, 0, 0, 0\n\n\tfor i := 0; i < n; i++ {\n\t\tfor j := 0; j < n; j++ {\n\t\t\tif board[0][0]^board[i][0]^board[0][j]^board[i][j] == 1 {\n\t\t\t\treturn -1\n\t\t\t}\n\t\t}\n\t}\n\n\tfor i := 0; i < n; i++ {\n\t\trowSum... | 0 | 0 | ['Go'] | 0 |
transform-to-chessboard | Java, 95%/92%, validate then calculate, with explanation | java-9592-validate-then-calculate-with-e-226z | There are 3 necessary conditions that need to be validated. Taken together, they are sufficient to confirm that the grid is solvable.\n1. In the first row, the | mnorby26 | NORMAL | 2021-09-26T17:42:50.385745+00:00 | 2021-09-26T17:44:08.215077+00:00 | 136 | false | There are 3 necessary conditions that need to be validated. Taken together, they are sufficient to confirm that the grid is solvable.\n1. In the first row, the number of ones and zeroes must be the same, or different by one.\n2. In the first column, the number of ones and zeroes must be the same, or different by one.\... | 0 | 0 | [] | 0 |
transform-to-chessboard | [C++] Check bitmask signature of rows and columns | c-check-bitmask-signature-of-rows-and-co-p3ol | \nclass Solution {\n using sig_t = std::bitset<30>;\npublic: \n int movesToChessboard(std::vector<std::vector<int>> &board) {\n const int num_ro | phi9t | NORMAL | 2021-09-26T17:36:46.800474+00:00 | 2021-09-26T17:36:46.800526+00:00 | 77 | false | ```\nclass Solution {\n using sig_t = std::bitset<30>;\npublic: \n int movesToChessboard(std::vector<std::vector<int>> &board) {\n const int num_rows = board.size(), num_cols = board[0].size();\n if (num_rows != num_cols) {\n return -1;\n }\n\n const auto get_signature = ... | 0 | 0 | [] | 0 |
transform-to-chessboard | Rust translated | rust-translated-by-sugyan-6s65 | rust\nimpl Solution {\n pub fn moves_to_chessboard(board: Vec<Vec<i32>>) -> i32 {\n let n = board.len();\n for i in 1..n {\n if (1.. | sugyan | NORMAL | 2021-09-26T16:43:23.061215+00:00 | 2021-09-26T16:43:23.061247+00:00 | 69 | false | ```rust\nimpl Solution {\n pub fn moves_to_chessboard(board: Vec<Vec<i32>>) -> i32 {\n let n = board.len();\n for i in 1..n {\n if (1..n as i32).contains(&(0..n).map(|j| board[i][j] ^ board[0][j]).sum::<i32>()) {\n return -1;\n }\n if (1..n as i32).contai... | 0 | 0 | ['Rust'] | 0 |
transform-to-chessboard | Super Helpful Explanation | super-helpful-explanation-by-manu2710aro-0pmq | Inspired by this and this posts, but use a different approach to count swap number, IMHO, which is more intuitive.\n```\n int movesToChessboard(vector>& bo | manu2710arora | NORMAL | 2021-09-26T16:15:46.815264+00:00 | 2021-09-26T16:15:46.815333+00:00 | 86 | false | Inspired by [this](https://leetcode.com/problems/transform-to-chessboard/discuss/114847/Easy-and-Concise-Solution-with-Explanation-C%2B%2BJavaPython) and [this](https://leetcode.com/problems/transform-to-chessboard/discuss/132113/Java-Clear-Code-with-Detailed-Explanations) posts, but use a different approach to count... | 0 | 0 | [] | 0 |
transform-to-chessboard | Refer this Post | refer-this-post-by-sruti08patak-vj95 | https://leetcode.com/problems/transform-to-chessboard/discuss/114847/C%2B%2BJavaPython-Solution-with-Explanation\n\nAlso follow the comments for your learning. | sruti08patak | NORMAL | 2021-09-26T15:48:59.271965+00:00 | 2021-09-26T15:48:59.272022+00:00 | 61 | false | https://leetcode.com/problems/transform-to-chessboard/discuss/114847/C%2B%2BJavaPython-Solution-with-Explanation\n\nAlso follow the comments for your learning. | 0 | 0 | [] | 0 |
transform-to-chessboard | [Python]StraightForward Approach | pythonstraightforward-approach-by-rjvish-rzjp | ```\nfrom operator import xor\nclass Solution(object):\n def movesToChessboard(self, board):\n """\n :type board: List[List[int]]\n :rty | rjvishwa21 | NORMAL | 2021-09-26T15:40:01.470897+00:00 | 2021-09-26T15:40:13.536447+00:00 | 76 | false | ```\nfrom operator import xor\nclass Solution(object):\n def movesToChessboard(self, board):\n """\n :type board: List[List[int]]\n :rtype: int\n """\n m,n=len(board),len(board[0])\n\n # check single row\n def valid_s(r):\n n=len(r)\n rowSum=sum(... | 0 | 0 | [] | 0 |
transform-to-chessboard | Easy Explanation | easy-explanation-by-psnoy00001111-ayrg | The algorithm is based on counting. A solvable board has each row/column either the same as the first row/column or exactly cell-by-cell reversed color of the f | psnoy00001111 | NORMAL | 2021-09-26T15:36:15.347068+00:00 | 2021-09-26T15:36:15.347107+00:00 | 100 | false | The algorithm is based on counting. A solvable board has each row/column either the same as the first row/column or exactly cell-by-cell reversed color of the first row/column.\n\nIn the loop we count for rs and cs, the number of rows/columns being the same as the first row/column, and rm and cm, the number of misplace... | 0 | 0 | ['C'] | 0 |
transform-to-chessboard | Python3 | python3-by-isissifeng-wiqq | \n | isissifeng | NORMAL | 2021-09-26T15:06:25.284236+00:00 | 2021-09-26T15:06:25.284287+00:00 | 113 | false | \n | 0 | 0 | [] | 0 |
transform-to-chessboard | O(N^2) c++ code with proof of mzchen's observation in comments. | on2-c-code-with-proof-of-mzchens-observa-ypuj | This isn\'t easy to get for me and I hope my code and comment is clear. I think this is a better code readability practice that algo practice.\nPlease let me kn | consciousgaze | NORMAL | 2021-01-01T10:42:20.711946+00:00 | 2021-01-01T10:45:39.169110+00:00 | 195 | false | This isn\'t easy to get for me and I hope my code and comment is clear. I think this is a better code readability practice that algo practice.\nPlease let me know the confusing parts :P\n```\nclass Solution {\npublic:\n // Both the column and row swap satisfies the following\n // property:\n // All the numbers... | 0 | 0 | [] | 0 |
transform-to-chessboard | c++ sol | c-sol-by-66brother-xm67 | \nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& grid) {\n int N=grid.size();\n //2 cases, N==odd || N==even\n / | 66brother | NORMAL | 2020-06-06T17:04:23.192365+00:00 | 2020-06-06T17:04:42.740137+00:00 | 291 | false | ```\nclass Solution {\npublic:\n int movesToChessboard(vector<vector<int>>& grid) {\n int N=grid.size();\n //2 cases, N==odd || N==even\n //Even:0101,1010 odd: 10101 01010\n for(int i=0;i<N;i++){//basic check\n int col=0,row=0; //one\n for(int j=0;j<N;j++){\n ... | 0 | 0 | [] | 0 |
transform-to-chessboard | cpp ~0-4ms 9M bits manipulation | cpp-0-4ms-9m-bits-manipulation-by-ptache-1y7j | There are 2 properties which are invariant under transpositions of rows/columns:\n1. The set { c(i), i \u2208 {0,1,...,n-1} }, where c(i) is the count of 1s in | ptacher | NORMAL | 2019-11-27T13:25:11.729721+00:00 | 2019-11-28T11:12:56.718392+00:00 | 322 | false | There are 2 properties which are invariant under transpositions of rows/columns:\n1. The set { c(i), i \u2208 {0,1,...,n-1} }, where c(i) is the count of 1s in column *i*, as well as the counts of columns for each value *c* in the set: #{ i \u2208 {0,1,...,n-1} | c(i)=c }. Same for rows.\n2. each pair of columns are ... | 0 | 0 | [] | 0 |
transform-to-chessboard | O(N) Scala solution | on-scala-solution-by-gorokhovsky-seqo | \nimport java.lang.Integer.bitCount\n\nobject Solution {\n def movesToChessboard(board: Array[Array[Int]]): Int = {\n val N = board.size\n val mask = (1< | gorokhovsky | NORMAL | 2018-09-10T18:33:43.318794+00:00 | 2018-09-10T18:33:43.318847+00:00 | 372 | false | ```\nimport java.lang.Integer.bitCount\n\nobject Solution {\n def movesToChessboard(board: Array[Array[Int]]): Int = {\n val N = board.size\n val mask = (1<<N)-1\n val templOl = 0x55555555 & mask// 0101010101\n val templLo = ~templOl & mask;// 1010101010\n val rows = board.map(_.reduceLeft((a: Int, r: ... | 0 | 0 | [] | 0 |
transform-to-chessboard | \u301048ms Straightforward Python\u3011 | u301048ms-straightforward-pythonu3011-by-wjwr | ```\nfrom operator import xor\nclass Solution(object):\n def movesToChessboard(self, board):\n """\n :type board: List[List[int]]\n :rty | weidairpi | NORMAL | 2018-02-11T20:55:58.233000+00:00 | 2018-02-11T20:55:58.233000+00:00 | 386 | false | ```\nfrom operator import xor\nclass Solution(object):\n def movesToChessboard(self, board):\n """\n :type board: List[List[int]]\n :rtype: int\n """\n m,n=len(board),len(board[0])\n\n # check single row\n def valid_s(r):\n n=len(r)\n rowSum=sum(... | 0 | 0 | [] | 0 |
transform-to-chessboard | My Java Solution | my-java-solution-by-xuhuiwang-52kb | Only use two arrays to represent rows and columns, since they are independent, only rearrange them s.t. \n\nfor i = 1 : (N - 1)\n (states[i] & states[i + | xuhuiwang | NORMAL | 2018-02-11T19:19:22.077000+00:00 | 2018-02-11T19:19:22.077000+00:00 | 569 | false | Only use two arrays to represent rows and columns, since they are independent, only rearrange them s.t. \n\nfor i = 1 : (N - 1)\n (states[i] & states[i + 1]) == 0;\n```\npublic class Solution {\n// process independently for rows and cols, time O(N), space O(N)\n\tint N;\n\tint solve(int[] states) {\n\t\tint bm = ... | 0 | 1 | [] | 0 |
maximum-xor-after-operations | Thought process explained || Basic intuition to optimise | thought-process-explained-basic-intuitio-imjf | So during the contest, Initialy I thought about using tries(a most famous based on xor) but here doesn\'t make sense. So let\'s understand the problem first :\n | faltu_admi | NORMAL | 2022-06-25T16:24:06.612148+00:00 | 2022-06-26T02:28:09.233701+00:00 | 4,386 | false | So during the contest, Initialy I thought about using tries(a most famous based on xor) but here doesn\'t make sense. So let\'s understand the problem first :\n\nSo what we\'re going to do in each step :\n1. pick any i-th element let\'s say y\n2. pick any x\n3. make x^y\n4. and put (x&y) at i-th position \n\nSo let\'s ... | 103 | 0 | [] | 19 |
maximum-xor-after-operations | [Java/C++/Python] 1-Line Solution, OR of All Elements | javacpython-1-line-solution-or-of-all-el-ev9m | Explanation\nThe maximum possible result is res = A[0] || A[1] || A[2] ... and it\'s realisiable.\n\n# Prove\nNow we approve it\'s realisiable.\nAssume result i | lee215 | NORMAL | 2022-06-25T16:02:18.333723+00:00 | 2022-06-25T16:16:26.558999+00:00 | 8,708 | false | # **Explanation**\nThe maximum possible result is `res = A[0] || A[1] || A[2] ... ` and it\'s realisiable.\n\n# **Prove**\nNow we approve it\'s realisiable.\nAssume result is `best = XOR(A[i])` and `best < res` above.\nThere is at least one bit difference between `best` and `res`, assume it\'s `x = 1 << k`.\n\nWe can f... | 98 | 0 | ['C', 'Python', 'Java'] | 24 |
maximum-xor-after-operations | Simplest O(N) solution [WITH EXPLANATION] | simplest-on-solution-with-explanation-by-3209 | nums[i] AND (nums[i] XOR x) where x is ANY integer, basically means you can decrease that element into any number but can\'t increase it.\nReread the above stat | stardust23 | NORMAL | 2022-06-25T16:03:56.369779+00:00 | 2022-06-25T18:44:57.756539+00:00 | 2,550 | false | `nums[i] AND (nums[i] XOR x)` where *x is ANY integer*, basically means you can decrease that element into any number but can\'t increase it.\n*Reread the above statement. That\'s the key to solving the problem!*\n* AND operation can only allow a 1 to be 0 but not a 0 to become 1.\n* XOR of elements is maximised by get... | 66 | 0 | [] | 8 |
maximum-xor-after-operations | 3 lines of CODE with EXPLANATION !! | C + + | 3-lines-of-code-with-explanation-c-by-me-xly7 | nums[i] & (nums[i]^x) cannot be more than nums[i] because we are taking AND.\nSo this implies that for any number we can set the bits OFF (only the bits which a | megamind_ | NORMAL | 2022-06-25T16:01:45.512202+00:00 | 2022-06-30T11:10:15.373346+00:00 | 2,304 | false | `nums[i] & (nums[i]^x)` cannot be more than `nums[i]` because we are taking **AND.**\nSo this implies that for any number we can set the bits `OFF` ***(only the bits which are already `ON` (1))*** .\n**BUT** we cannot set any bit` ON `if it were `OFF `already .\n\nTherefore , we can say that we will set bits ON and OF... | 31 | 0 | [] | 2 |
maximum-xor-after-operations | One Liner | one-liner-by-votrubac-3kh8 | An operation can remove a bit, but cannot add one.\n \nSo, the maximum result would contain bits from all elements in the input array.\n\nPython 3\npython\nc | votrubac | NORMAL | 2022-06-25T16:00:52.601789+00:00 | 2023-09-11T18:46:08.969017+00:00 | 2,079 | false | An operation can remove a bit, but cannot add one.\n \nSo, the maximum result would contain bits from all elements in the input array.\n\n**Python 3**\n```python\nclass Solution:\n def maximumXOR(self, nums: List[int]) -> int:\n return reduce(or_, nums)\n```\n\n**C++**\n```cpp\nint maximumXOR(vector<int>& ... | 29 | 3 | ['C'] | 7 |
maximum-xor-after-operations | ✔️ EXPLAiNED || C++ || PYTHON || JAVA | explained-c-python-java-by-karan_8082-elon | UPVOTE IF HELPFuul\n\nEvery once in a while try using paper-pen. -> useful for finding pattern, cant just think about these.\n\nUpon converting to binary, there | karan_8082 | NORMAL | 2022-06-30T09:19:22.050745+00:00 | 2022-06-30T09:19:22.050771+00:00 | 1,104 | false | **UPVOTE IF HELPFuul**\n\nEvery once in a while try using paper-pen. -> useful for finding pattern, cant just think about these.\n\nUpon converting to binary, there will be a pattern.\nResult is **OR of all elements** .\n\n**OBSERvATION=>** ```nums[i] AND (nums[i] XOR x)``` \nWith this operation, we can modify each el... | 21 | 2 | ['Bit Manipulation', 'C', 'Python', 'Java'] | 3 |
maximum-xor-after-operations | Clear and Lucid Explanation along with Dry Runs | O(n) | | clear-and-lucid-explanation-along-with-d-o8k4 | Intuition\n\nWe can\'t increase nums[i] since we use AND operator with it because AND operator can\'t make \'0\' to \'1\' but it can make \'1\' to \'0\'. Did yo | KanishYathraRaj | NORMAL | 2024-01-01T09:35:00.938215+00:00 | 2024-01-01T09:35:00.938253+00:00 | 334 | false | # Intuition\n\nWe can\'t increase nums[i] since we use AND operator with it because AND operator can\'t make \'0\' to \'1\' but it can make \'1\' to \'0\'. Did you find the hint behind it? if not , don\'t worry I am here to help you!\n\nlets take first example,\n\n`nums = [ 3 , 2 , 4 , 6 ] `\n\nEvery integers has 32 bi... | 18 | 0 | ['C++'] | 3 |
maximum-xor-after-operations | Python3 || 1 line, bit operations, w/ explanation || T/S: 98% / 80% | python3-1-line-bit-operations-w-explanat-egsl | Points to consider:\n\n The problem calls for choosing an integer x, selecting an element n of the list, applying the compound operator op(n,x) = (x&n)^n, and | Spaulding_ | NORMAL | 2022-08-01T20:41:28.901062+00:00 | 2024-06-13T04:46:48.741495+00:00 | 669 | false | Points to consider:\n\n* The problem calls for choosing an integer x, selecting an element n of the list, applying the compound operator op(n,x) = (x&n)^n, and taking the bit-intersection of the modified set. Because of the associative and commutative properties of the XOR operator, it does not matter which n we choo... | 13 | 1 | ['Python', 'Python3'] | 1 |
maximum-xor-after-operations | [C++] | BIT MANIPULATION | Time-O(NlogN) | Explaination Added | c-bit-manipulation-time-onlogn-explainat-np7w | Concept- If we XOR odd number of ones then we\nget 1 and if we XOR even number of ones then\nwe will get 0.\n1^1^1^.....^1(odd number of times)=1\n1^1^1^.....^1 | niks07 | NORMAL | 2022-06-25T16:04:57.849422+00:00 | 2022-06-25T16:26:53.199151+00:00 | 937 | false | **Concept- If we XOR odd number of ones then we\nget 1 and if we XOR even number of ones then\nwe will get 0.**\n**1^1^1^.....^1(odd number of times)=1\n1^1^1^.....^1(even number of times)=0**\n**Steps-**\n* Each number has bits when we convert it into\nbinary. Eg. 4 has 3 bits which are 100.\nSo First we will find the... | 11 | 0 | ['Bit Manipulation', 'C'] | 2 |
maximum-xor-after-operations | C++ | Simple Or Operation | c-simple-or-operation-by-slow_code-l6t7 | \nlet take some example\nsuppose arr = {1,2,3,9}\nnow we will find have bit of all these\n1 -> 0001\n2 -> 0010\n3 -> 0011\n9 -> 1001\nnow to get maximum answer | Slow_code | NORMAL | 2022-06-25T16:01:24.777434+00:00 | 2022-06-25T16:25:34.142028+00:00 | 829 | false | ```\nlet take some example\nsuppose arr = {1,2,3,9}\nnow we will find have bit of all these\n1 -> 0001\n2 -> 0010\n3 -> 0011\n9 -> 1001\nnow to get maximum answer we should have as many as 1 bits possible in \nanswer so for that we need to use OR operation, \nso that we get maximum of 1 bits in every place of answer\n\... | 10 | 2 | [] | 3 |
maximum-xor-after-operations | C++ Beginner Friendly Explanation | c-beginner-friendly-explanation-by-anis2-1537 | Approach:\n\n nums[i] AND (nums[i] XOR x)\n with the above operation we can modify any nums[i] right\n as we are taking AND, we can say that none of the bits of | anis23 | NORMAL | 2022-08-03T10:35:50.844315+00:00 | 2022-08-03T10:36:59.340462+00:00 | 670 | false | **Approach:**\n\n* ```nums[i] AND (nums[i] XOR x)```\n* with the above operation we can modify any nums[i] right\n* as we are taking ```AND```, we can say that none of the bits of nums[i] can be converted from 0 to 1 \n* on the other hand, we can convert a bit from 1 to 0\n\t* as we can choose x such that nums[i]^x = 0... | 9 | 0 | ['Math', 'Bit Manipulation', 'C', 'C++'] | 2 |
maximum-xor-after-operations | [Java/Python 3] Bit manipulations w/ brief explanation and analysis. | javapython-3-bit-manipulations-w-brief-e-pk74 | Key observation: nums[i] & (nums[i] ^ x) <= nums[i]. Therefoe, we can only set off the 1 bits of each number if necessary, in order to make greatest of the xor | rock | NORMAL | 2022-06-25T16:07:38.440914+00:00 | 2022-06-25T17:52:19.226541+00:00 | 1,224 | false | **Key observation: `nums[i] & (nums[i] ^ x) <= nums[i]`**. Therefoe, we can only set off the `1` bits of each number if necessary, in order to make greatest of the xor all numbers after operations.\n\nSince 2<sup>27</sup> > 10<sup>8</sup>, we can traverse the left-most `28` bits to count how many `1`\'s on each bits.\n... | 8 | 0 | ['Bit Manipulation', 'Java', 'Python3'] | 3 |
maximum-xor-after-operations | C++ | Simple Solution | O(N) Solution | Easy to understand with Comments | Bit Manipulation | c-simple-solution-on-solution-easy-to-un-ll5h | \nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int n = nums.size();\n int ans=0;\n // for every bit check the cnt o | gargp1065 | NORMAL | 2022-06-26T05:47:41.665336+00:00 | 2022-06-26T05:47:41.665384+00:00 | 139 | false | ```\nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int n = nums.size();\n int ans=0;\n // for every bit check the cnt of the times 1 appears in the array. If the cnt is great than 1, then we can consider that bit in our ans.\n // this is becuase the given op can unset ... | 5 | 0 | ['Bit Manipulation'] | 0 |
maximum-xor-after-operations | [C++] Only Bit Count | c-only-bit-count-by-badhansen-u24m | Approach:\n1. Count number of bits in each positions\n2. Now if the position bit count is zero then we can\'t take this bit otherwise we can take this bit in ma | badhansen | NORMAL | 2022-06-25T16:08:19.436454+00:00 | 2022-06-25T16:08:28.910061+00:00 | 772 | false | **Approach:**\n1. Count number of bits in each positions\n2. Now if the position bit count is zero then we can\'t take this bit otherwise we can take this bit in mask.\n\nNow return the number \n\n**Time:** `O(N), N = Length of nums`\n**Space:** `O(32) ~ O(1)`\n```\nclass Solution {\npublic:\n int maximumXOR(vector<... | 5 | 0 | ['C', 'C++'] | 2 |
maximum-xor-after-operations | 100%FASTER || 3LINE || BITWISE-OR-ALL-ELEMENTS||C++||SHORT & SWEET CODE | 100faster-3line-bitwise-or-all-elementsc-e441 | \nint maximumXOR(vector<int>& nums) {\n int ans = 0;\n for(auto &i: nums)ans |= i;\n return ans;\n }\n | yash___sharma_ | NORMAL | 2023-03-28T12:17:09.276906+00:00 | 2023-03-28T12:17:09.276941+00:00 | 406 | false | ````\nint maximumXOR(vector<int>& nums) {\n int ans = 0;\n for(auto &i: nums)ans |= i;\n return ans;\n }\n```` | 4 | 0 | ['Bit Manipulation', 'C', 'C++'] | 0 |
maximum-xor-after-operations | Simple Solution(3 lines) ✅ || Runtime - 100% beats💥 || Memory - 100% beats💥 || With Explanation ✅ | simple-solution3-lines-runtime-100-beats-ytkq | Approach\n\nThe goal is to find the maximum possible bitwise XOR of all elements in the nums array after applying the given operation any number of times. To ac | akobirswe | NORMAL | 2023-07-30T11:54:19.774896+00:00 | 2023-07-30T11:54:19.774928+00:00 | 121 | false | # Approach\n\nThe goal is to find the maximum possible bitwise XOR of all elements in the `nums` array after applying the given operation any number of times. To achieve this, we can use the properties of the bitwise XOR operation.\n\n**Logic:**\n1. Initialize a variable `res` to 0. This variable will store the final r... | 3 | 0 | ['Array', 'Math', 'Bit Manipulation', 'Java', 'C#'] | 0 |
maximum-xor-after-operations | 100% BEATS || OR ALL ELEMENTS || C++ | 100-beats-or-all-elements-c-by-ganeshkum-5gqh | Code\n\nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int ans = 0;;\n for(auto &i: nums){\n ans |= i;\n } | ganeshkumawat8740 | NORMAL | 2023-06-19T11:25:03.136281+00:00 | 2023-06-19T11:25:03.136330+00:00 | 594 | false | # Code\n```\nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int ans = 0;;\n for(auto &i: nums){\n ans |= i;\n }\n return ans;\n }\n};\n``` | 3 | 1 | ['Math', 'Bit Manipulation', 'C++'] | 0 |
maximum-xor-after-operations | C++ || Short Simple Solution || O(N) Time Complexity | c-short-simple-solution-on-time-complexi-364l | Simple check the bit is set or not in nums[i]\nOR is the best option to maximize the ans set bits will be set for ans\n\nclass Solution {\npublic:\n int maxi | shrey0811 | NORMAL | 2023-02-13T11:44:50.665809+00:00 | 2023-02-13T11:44:50.665847+00:00 | 803 | false | **Simple check the bit is set or not in nums[i]**\n**OR is the best option to maximize the ans set bits will be set for ans**\n```\nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums)\n {\n int ans = 0;\n for(auto it : nums)\n {\n ans |=it;\n }\n return ans;... | 3 | 0 | ['Bit Manipulation', 'C', 'C++'] | 0 |
maximum-xor-after-operations | Explained with Comments | Beginner Friendly | C++ | O(n) Time O(1) Space | explained-with-comments-beginner-friendl-srzq | Explanation\nFor each bit position (0 to 31), we count the number of 1 bits for all elements in the nums array.\nWe store this in bitCounts.\n\nNow, if we take | AbhilakshSinghReen | NORMAL | 2022-08-11T14:31:52.034139+00:00 | 2022-08-11T14:31:52.034170+00:00 | 187 | false | **Explanation**\nFor each bit position (0 to 31), we count the number of 1 bits for all elements in the nums array.\nWe store this in bitCounts.\n\nNow, if we take the XOR of the entire array. In the XOR output, the bits at the positions for which the bit count is even will be 0 and for which the bit count is odd will ... | 3 | 0 | ['C'] | 1 |
maximum-xor-after-operations | 💥One Liner | one-liner-by-adityabhate-eh4t | C++\n\nint maximumXOR(vector<int>& nums) {\n return accumulate(begin(nums), end(nums), 0, bit_or<int>()); //O(n)\n}\n\n\nclass Solution {\npublic: \ | AdityaBhate | NORMAL | 2022-07-29T04:10:30.295152+00:00 | 2022-11-17T02:58:39.972166+00:00 | 678 | false | # **C++**\n```\nint maximumXOR(vector<int>& nums) {\n return accumulate(begin(nums), end(nums), 0, bit_or<int>()); //O(n)\n}\n```\n```\nclass Solution {\npublic: \n int maximumXOR(vector<int>& nums) {\n int max_XOR=nums[0];\n for(int i=1;i<nums.size();i++){ //O(n) \n... | 3 | 0 | ['Array', 'Math', 'Bit Manipulation', 'C++', 'Java'] | 2 |
maximum-xor-after-operations | ✅ O(n) | Easy Explanation. Why? | on-easy-explanation-why-by-shivamsingla-gwwt | Solution : Maximize XOR\nInitially, the question seems to be very brainstorming but when you closely look at the table below, you can actually find the solution | shivamsingla | NORMAL | 2022-06-25T16:11:41.752386+00:00 | 2022-07-12T18:53:30.544432+00:00 | 570 | false | **Solution : Maximize XOR**\nInitially, the question seems to be very brainstorming but when you closely look at the table below, you can actually find the solution like anybody:\n\n\n\n**The Real Trick:**\nSo,... | 3 | 0 | ['Bit Manipulation', 'C', 'C++', 'Java'] | 0 |
maximum-xor-after-operations | [Python] ATK++ | AGL++ | INT++ | python-atk-agl-int-by-siizoo333-kzhq | At the first glance, I really don\'t know what number should we choose at any step for nums[i] AND (nums[i] XOR x).\nHowever, recalling the theorem of XOR, if w | siizoo333 | NORMAL | 2022-06-25T16:08:33.314287+00:00 | 2022-07-23T09:57:55.242064+00:00 | 277 | false | At the first glance, I really don\'t know what number should we choose at any step for `nums[i] AND (nums[i] XOR x)`.\nHowever, recalling the theorem of XOR, if we wanna get the largest outcome by XOR-ing the array of numbers, we need odd number of 1\'s in each bit.\n\nThen, how about this `nums[i] AND (nums[i] XOR x)`... | 3 | 0 | ['Python'] | 0 |
maximum-xor-after-operations | [Python3] bit operation | python3-bit-operation-by-ye15-he8a | Please pull this commit for solutions of biweekly 81. \n\n\nclass Solution:\n def maximumXOR(self, nums: List[int]) -> int:\n return reduce(or_, nums) | ye15 | NORMAL | 2022-06-25T16:02:14.657431+00:00 | 2022-06-25T16:19:38.997391+00:00 | 105 | false | Please pull this [commit](https://github.com/gaosanyong/leetcode/commit/7a67de4e975be771355e048bf8dde4cf0906e360) for solutions of biweekly 81. \n\n```\nclass Solution:\n def maximumXOR(self, nums: List[int]) -> int:\n return reduce(or_, nums)\n``` | 3 | 0 | ['Python3'] | 0 |
maximum-xor-after-operations | c++ | OR of all elements | c-or-of-all-elements-by-aman282571-6tpv | Observation:-\n1.For any nums[i] you can\'t get more set bits then it already has because of the "and" operation in condition.\n2.So wherever you see ith bit on | aman282571 | NORMAL | 2022-06-25T16:01:16.471841+00:00 | 2022-06-25T16:01:16.471889+00:00 | 197 | false | **Observation:-**\n1.For any nums[i] you can\'t get more set bits then it already has because of the "and" operation in condition.\n2.So wherever you see ```ith``` bit on set the ith bit of answer on because you can add this bit to final answer.\n\n \n``` \nclass Solution {\npublic:\n int maximumXOR(vector<int... | 3 | 0 | [] | 0 |
maximum-xor-after-operations | C++ | O(n*31) time | O(1) space | c-on31-time-o1-space-by-vaibhavshekhawat-lwm7 | ```\nint maximumXOR(vector& nums) {\n int ans=0;\n for(int i=0;i<31;i++){\n int mask=(1<<i);\n bool f=0;\n for(au | vaibhavshekhawat | NORMAL | 2022-06-25T16:00:44.208725+00:00 | 2022-06-25T16:00:44.208764+00:00 | 340 | false | ```\nint maximumXOR(vector<int>& nums) {\n int ans=0;\n for(int i=0;i<31;i++){\n int mask=(1<<i);\n bool f=0;\n for(auto& j:nums){\n if((mask&j)!=0){f=1;break;} // break if a bit is 1 in any nums\n }\n if(f==1) ans|=mask; ... | 3 | 0 | ['C'] | 0 |
maximum-xor-after-operations | [C++] 3 lines, Bitwise OR of all elements | c-3-lines-bitwise-or-of-all-elements-by-8ih2i | \nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int ans = nums[0];\n for(int i = 1; i < nums.size(); ++i) ans |= nums[i];\n | aniket_jain_ | NORMAL | 2022-06-25T16:00:39.260747+00:00 | 2022-06-25T16:00:39.260790+00:00 | 233 | false | ```\nclass Solution {\npublic:\n int maximumXOR(vector<int>& nums) {\n int ans = nums[0];\n for(int i = 1; i < nums.size(); ++i) ans |= nums[i];\n return ans;\n }\n};\n``` | 3 | 0 | [] | 0 |
maximum-xor-after-operations | C++ || 100% Beats || Easy Solution | c-100-beats-easy-solution-by-ankit_jha_6-2kyt | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ankit_jha_6 | NORMAL | 2025-01-22T05:31:59.635496+00:00 | 2025-01-22T05:31:59.635496+00:00 | 67 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
AND operation can only clear bits (change 1 to 0) but never set bits (change 0 to 1).
The OR operation ensures that any bit that is 1 in any number in the array remains 1 in the final result.
XOR is maximized when t... | 2 | 0 | ['Bit Manipulation', 'C++'] | 0 |
maximum-xor-after-operations | Python. few lines code but a bit more explanation provided. | python-few-lines-code-but-a-bit-more-exp-71lh | IntuitionAll bit manupulations tasks with XOR requeres a bit attention.
In this particular, key in in
R = "nums[i] AND (nums[i] XOR x)"
lets' dissolve this expr | dev_lvl_80 | NORMAL | 2025-01-16T02:09:08.645469+00:00 | 2025-01-16T02:09:25.346321+00:00 | 41 | false | # Intuition
All bit manupulations tasks with XOR requeres a bit attention.
In this particular, key in in
R = "nums[i] AND (nums[i] XOR x)"
lets' dissolve this expression
1. R cannot be more than nums[i], because of AND
2. nums[i] XOR x, can result in any number
Looking at 2 we can clarify 1 as R can be any number whe... | 2 | 0 | ['Python3'] | 0 |
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