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Formatted question description: https://leetcode.ca/all/2482.html

# 2482. Difference Between Ones and Zeros in Row and Column

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: 01 Matrix, Special Positions in a Binary Matrix, Remove All Ones With Row and Column Flips.

## 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, C++, Python)

• 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
}
}



Explain:

nope.

Complexity:

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