2482. Difference Between Ones and Zeros in Row and Column

Problem

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.

Solution (Java)

class Solution {
    public int[][] onesMinusZeros(int[][] arr) {
        int m = arr.length;
        int n = arr[0].length;
        int ans [][] = new int[m][n];
        int row = 0;
        for(int i = 0 ; i < m ; i++) {
            int one = 0;
            for(int j = 0 ; j < n ; j++) {
                if(arr[i][j] == 1)one++;
            }
            ans[i][n-1] = one;
            if(i == m-1) row = one;
        }

        for(int i = 0 ; i < n ; i++) {
            int one = 0;
            for(int j = 0 ; j < m ; j++) {
                if(arr[j][i] == 1)one++;
            }
            ans[m-1][i] = one;
        }

        for(int i = 0 ; i < m-1 ; i++) {
            for(int j = 0 ; j < n ; j++) {
                int put = ans[m-1][j] + ans[i][n-1] - (m-ans[m-1][j]+n-ans[i][n-1]);
                ans[i][j] = put;
            }
        }

        for(int i = m-1 ; i < m ; i++) {
            for(int j = 0 ; j < n ; j++) {
                int put = ans[m-1][j] + row - (m-ans[m-1][j]+n-row);
                ans[i][j] = put;
            }
        }

        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).