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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).

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