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.

Code

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

simply brute force it.

class Solution {
public:
    vector<vector<int>> onesMinusZeros(vector<vector<int>>& grid) {
       int n = grid.size();
       int m = grid[0].size();
 
       vector<int> rows(n, 0);
       vector<int> cols(m, 0);
 
       for(int i = 0; i < grid.size(); i++) {
           for(int j = 0; j < grid[i].size(); j++) {
               rows[i] += grid[i][j];
               cols[j] += grid[i][j];
           }
       }
 
       for(int i = 0; i < grid.size(); i++) {
           for(int j = 0; j < grid[i].size(); j++) {
               grid[i][j] = (rows[i] + cols[j]) - (grid.size() - rows[i]) - (grid[i].size() - cols[j]);
           }
       }
        return grid;
    }
};

Source

• Difference Between Ones and Zeros in Row and Column - LeetCode