LeetCode-in-Java

1252. Cells with Odd Values in a Matrix

Easy

There is an m x n matrix that is initialized to all 0’s. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix.

For each location indices[i], do both of the following:

1. Increment all the cells on row ri.
2. Increment all the cells on column ci.

Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.

Example 1:

Input: m = 2, n = 3, indices = [[0,1],[1,1]]

Output: 6

Explanation: Initial matrix = [[0,0,0],[0,0,0]].

After applying first increment it becomes [[1,2,1],[0,1,0]].

The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.

Example 2:

Input: m = 2, n = 2, indices = [[1,1],[0,0]]

Output: 0

Explanation: Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.

Constraints:

• 1 <= m, n <= 50
• 1 <= indices.length <= 100
• 0 <= ri < m
• 0 <= ci < n

Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

Solution

public class Solution {
    public int oddCells(int n, int m, int[][] indices) {
        int[][] matrix = new int[n][m];
        for (int[] index : indices) {
            addOneToRow(matrix, index[0]);
            addOneToColumn(matrix, index[1]);
        }
        int oddNumberCount = 0;
        for (int[] ints : matrix) {
            for (int j = 0; j < matrix[0].length; j++) {
                if (ints[j] % 2 != 0) {
                    oddNumberCount++;
                }
            }
        }
        return oddNumberCount;
    }

    private void addOneToColumn(int[][] matrix, int columnIndex) {
        for (int i = 0; i < matrix.length; i++) {
            matrix[i][columnIndex] += 1;
        }
    }

    private void addOneToRow(int[][] matrix, int rowIndex) {
        for (int j = 0; j < matrix[0].length; j++) {
            matrix[rowIndex][j] += 1;
        }
    }
}

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