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:
Increment all the cells on row ri. 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.
Solution
class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
rows = [0] * m
cols = [0] * n
for row, col in indices:
rows[row] ^= 1
cols[col] ^= 1
cnt = 0
for i in range(m):
for j in range(n):
if rows[i] ^ cols[j] == 1:
cnt += 1
return cnt