LeetCode in Kotlin

2482. Difference Between Ones and Zeros in Row and Column

Medium

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

class Solution {
    fun onesMinusZeros(grid: Array<IntArray>): Array<IntArray> {
        val rowOne = IntArray(grid.size)
        val colOne = IntArray(grid[0].size)
        val m = grid.size
        val n = grid[0].size
        for (i in 0 until m) {
            var c = 0
            for (j in 0 until n) {
                if (grid[i][j] == 1) {
                    c++
                }
            }
            rowOne[i] = c
        }
        for (i in 0 until n) {
            var c = 0
            for (ints in grid) {
                if (ints[i] == 1) {
                    c++
                }
            }
            colOne[i] = c
        }
        for (i in 0 until m) {
            for (j in 0 until n) {
                grid[i][j] = rowOne[i] + colOne[j] - (m - rowOne[i]) - (n - colOne[j])
            }
        }
        return grid
    }
}

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