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2482. Difference Between Ones and Zeros in Row and Column

Description

You are given a 0-indexed m x n binary matrix grid.

A 0-indexed m x n difference matrix diff is created with the following procedure:

  • Let the number of ones in the ith row be onesRowi.
  • Let the number of ones in the jth column be onesColj.
  • Let the number of zeros in the ith row be zerosRowi.
  • Let the number of zeros in the jth column be zerosColj.
  • diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj

Return the difference matrix diff.

 

Example 1:

Input: grid = [[0,1,1],[1,0,1],[0,0,1]]
Output: [[0,0,4],[0,0,4],[-2,-2,2]]
Explanation:
- diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 2 + 1 - 1 - 2 = 0 
- diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 2 + 1 - 1 - 2 = 0 
- diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 2 + 3 - 1 - 0 = 4 
- diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 2 + 1 - 1 - 2 = 0 
- diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 2 + 1 - 1 - 2 = 0 
- diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 2 + 3 - 1 - 0 = 4 
- diff[2][0] = onesRow2 + onesCol0 - zerosRow2 - zerosCol0 = 1 + 1 - 2 - 2 = -2
- diff[2][1] = onesRow2 + onesCol1 - zerosRow2 - zerosCol1 = 1 + 1 - 2 - 2 = -2
- diff[2][2] = onesRow2 + onesCol2 - zerosRow2 - zerosCol2 = 1 + 3 - 2 - 0 = 2

Example 2:

Input: grid = [[1,1,1],[1,1,1]]
Output: [[5,5,5],[5,5,5]]
Explanation:
- diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 3 + 2 - 0 - 0 = 5
- diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 3 + 2 - 0 - 0 = 5
- diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 3 + 2 - 0 - 0 = 5
- diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 3 + 2 - 0 - 0 = 5
- diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 3 + 2 - 0 - 0 = 5
- diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 3 + 2 - 0 - 0 = 5

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 105
  • 1 <= m * n <= 105
  • grid[i][j] is either 0 or 1.

Solutions

  • class Solution {
        public int[][] onesMinusZeros(int[][] grid) {
            int m = grid.length, n = grid[0].length;
            int[] rows = new int[m];
            int[] cols = new int[n];
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    int v = grid[i][j];
                    rows[i] += v;
                    cols[j] += v;
                }
            }
            int[][] diff = new int[m][n];
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j]);
                }
            }
            return diff;
        }
    }
    
  • class Solution {
    public:
        vector<vector<int>> onesMinusZeros(vector<vector<int>>& grid) {
            int m = grid.size(), n = grid[0].size();
            vector<int> rows(m);
            vector<int> cols(n);
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    int v = grid[i][j];
                    rows[i] += v;
                    cols[j] += v;
                }
            }
            vector<vector<int>> diff(m, vector<int>(n));
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j]);
                }
            }
            return diff;
        }
    };
    
  • class Solution:
        def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
            m, n = len(grid), len(grid[0])
            rows = [0] * m
            cols = [0] * n
            for i, row in enumerate(grid):
                for j, v in enumerate(row):
                    rows[i] += v
                    cols[j] += v
            diff = [[0] * n for _ in range(m)]
            for i, r in enumerate(rows):
                for j, c in enumerate(cols):
                    diff[i][j] = r + c - (n - r) - (m - c)
            return diff
    
    
  • func onesMinusZeros(grid [][]int) [][]int {
    	m, n := len(grid), len(grid[0])
    	rows := make([]int, m)
    	cols := make([]int, n)
    	diff := make([][]int, m)
    	for i, row := range grid {
    		diff[i] = make([]int, n)
    		for j, v := range row {
    			rows[i] += v
    			cols[j] += v
    		}
    	}
    	for i, r := range rows {
    		for j, c := range cols {
    			diff[i][j] = r + c - (n - r) - (m - c)
    		}
    	}
    	return diff
    }
    
  • function onesMinusZeros(grid: number[][]): number[][] {
        const m = grid.length;
        const n = grid[0].length;
        const rows = new Array(m).fill(0);
        const cols = new Array(n).fill(0);
        for (let i = 0; i < m; i++) {
            for (let j = 0; j < n; j++) {
                if (grid[i][j]) {
                    rows[i]++;
                    cols[j]++;
                }
            }
        }
        const ans = Array.from({ length: m }, () => new Array(n).fill(0));
        for (let i = 0; i < m; i++) {
            for (let j = 0; j < n; j++) {
                ans[i][j] = rows[i] + cols[j] - (m - rows[i]) - (n - cols[j]);
            }
        }
        return ans;
    }
    
    
  • impl Solution {
        pub fn ones_minus_zeros(grid: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
            let m = grid.len();
            let n = grid[0].len();
            let mut rows = vec![0; m];
            let mut cols = vec![0; n];
            for i in 0..m {
                for j in 0..n {
                    if grid[i][j] == 1 {
                        rows[i] += 1;
                        cols[j] += 1;
                    }
                }
            }
            let mut ans = vec![vec![0; n]; m];
            for i in 0..m {
                for j in 0..n {
                    ans[i][j] = (rows[i] + cols[j] - (m - rows[i]) - (n - cols[j])) as i32;
                }
            }
            ans
        }
    }
    
    

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