java.lang.Object
g1201_1300.s1252_cells_with_odd_values_in_a_matrix.Solution

public class Solution extends Object
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:** ![](https://assets.leetcode.com/uploads/2019/10/30/e1.png) **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:** ![](https://assets.leetcode.com/uploads/2019/10/30/e2.png) **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?
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • oddCells

      public int oddCells(int n, int m, int[][] indices)