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

/**
 * @param {number[][]} grid
 * @return {number[][]}
 */
var onesMinusZeros = function(grid) {
    var m = grid.length;
    var n = grid[0].length;
    var colOnes = Array(n).fill(0);
    var colZeros = Array(n).fill(0);
    var rowOnes = Array(m).fill(0);
    var rowZeros = Array(m).fill(0);
    for (var i = 0; i < m; i++) {
        for (var j = 0; j < n; j++) {
            if (grid[i][j] === 1) {
                rowOnes[i]++;
                colOnes[j]++;
            }
            if (grid[i][j] === 0) {
                rowZeros[i]++;
                colZeros[j]++;
            }
        }
    }
    var diff = Array(m).fill(0).map(() => Array(n).fill(0));
    for (var i = 0; i < m; i++) {
        for (var j = 0; j < n; j++) {
            diff[i][j] = rowOnes[i] + colOnes[j] - rowZeros[i] - colZeros[j];
        }
    }
    return diff;
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(n * m).