LeetCode in Kotlin

73. Set Matrix Zeroes

Medium

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0’s.

You must do it in place.

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]

Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]

Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

• m == matrix.length
• n == matrix[0].length
• 1 <= m, n <= 200
• -231 <= matrix[i][j] <= 231 - 1

Follow up:

• A straightforward solution using O(mn) space is probably a bad idea.
• A simple improvement uses O(m + n) space, but still not the best solution.
• Could you devise a constant space solution?

Solution

class Solution {
    // Approach: Use first row and first column for storing whether in future
    //           the entire row or column needs to be marked 0
    fun setZeroes(matrix: Array<IntArray>) {
        val m = matrix.size
        val n: Int = matrix[0].size
        var row0 = false
        var col0 = false
        // Check if 0th col needs to be market all 0s in future
        for (ints in matrix) {
            if (ints[0] == 0) {
                col0 = true
                break
            }
        }
        // Check if 0th row needs to be market all 0s in future
        for (i in 0 until n) {
            if (matrix[0][i] == 0) {
                row0 = true
                break
            }
        }
        // Store the signals in 0th row and column
        for (i in 1 until m) {
            for (j in 1 until n) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = 0
                    matrix[0][j] = 0
                }
            }
        }
        // Mark 0 for all cells based on signal from 0th row and 0th column
        for (i in 1 until m) {
            for (j in 1 until n) {
                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
                    matrix[i][j] = 0
                }
            }
        }
        // Set 0th column
        for (i in 0 until m) {
            if (col0) {
                matrix[i][0] = 0
            }
        }
        // Set 0th row
        for (i in 0 until n) {
            if (row0) {
                matrix[0][i] = 0
            }
        }
    }
}

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