LeetCode in Kotlin

3033. Modify the Matrix

Easy

Given a 0-indexed m x n integer matrix matrix, create a new 0-indexed matrix called answer. Make answer equal to matrix, then replace each element with the value -1 with the maximum element in its respective column.

Return the matrix answer.

Example 1:

Input: matrix = [[1,2,-1],[4,-1,6],[7,8,9]]

Output: [[1,2,9],[4,8,6],[7,8,9]]

Explanation: The diagram above shows the elements that are changed (in blue).

• We replace the value in the cell [1][1] with the maximum value in the column 1, that is 8.
• We replace the value in the cell [0][2] with the maximum value in the column 2, that is 9.

Example 2:

Input: matrix = [[3,-1],[5,2]]

Output: [[3,2],[5,2]]

Explanation: The diagram above shows the elements that are changed (in blue).

Constraints:

• m == matrix.length
• n == matrix[i].length
• 2 <= m, n <= 50
• -1 <= matrix[i][j] <= 100
• The input is generated such that each column contains at least one non-negative integer.

Solution

class Solution {
    fun modifiedMatrix(matrix: Array<IntArray>): Array<IntArray> {
        for (i in matrix.indices) {
            for (j in matrix[0].indices) {
                if (matrix[i][j] == -1) {
                    var y = 0
                    for (ints in matrix) {
                        if (ints[j] > y) {
                            y = ints[j]
                        }
                    }
                    matrix[i][j] = y
                }
            }
        }
        return matrix
    }
}

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