[LeetCode] 2482. Difference Between Ones and Zeros in Row and Column

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.

行和列中一和零的差值。

给你一个下标从 0 开始的 m x n 二进制矩阵 grid 。
我们按照如下过程，定义一个下标从 0 开始的 m x n 差值矩阵 diff ：
令第 i 行一的数目为 onesRowi 。
令第 j 列一的数目为 onesColj 。
令第 i 行零的数目为 zerosRowi 。
令第 j 列零的数目为 zerosColj 。
diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj
请你返回差值矩阵 diff 。

思路

创建两个hashmap分别记录每一行和每一列上 1 的个数，然后按照题目给的公式返回二维数组。这里我用数组代替 hashmap 加快速度。

复杂度

时间O(mn)
空间O(mn) - output matrix

代码

Java实现

class Solution {
    public int[][] onesMinusZeros(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[] rowMap = new int[m];
        int[] colMap = new int[n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    rowMap[i]++;
                    colMap[j]++;
                }
            }
        }

        int[][] diff = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                diff[i][j] = rowMap[i] + colMap[j] - (m - rowMap[i]) - (n - colMap[j]);
            }
        }
        return diff;
    }
}