1252. Cells with Odd Values in a Matrix

Description

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

cells-with-odd-values-in-a-matrix.py

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        grid = [[0]*m for _ in range(n)]

        for r,c in indices:
            for i in range(n):
                for j in range(m):
                    if j == c:
                        grid[i][j] += 1
                    if i == r: 
                        grid[i][j] += 1

        return sum(1 for num in sum(grid,[]) if num & 1)