# LeetCode 1295. Find Numbers with Even Number of Digits

## Description

https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/

Given `n` and `m` which are the dimensions of a matrix initialized by zeros and given an array `indices` where `indices[i] = [ri, ci]`. For each pair of `[ri, ci]` you have to increment all cells in row `ri` and column `ci` by 1.

Return the number of cells with odd values in the matrix after applying the increment to all `indices`.

Example 1:

```Input: n = 2, m = 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 will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
```

Example 2:

```Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
```

Constraints:

• `1 <= n <= 50`
• `1 <= m <= 50`
• `1 <= indices.length <= 100`
• `0 <= indices[i] < n`
• `0 <= indices[i] < m`

## Explanation

Simply implement as the problem description says.

## Python Solution

``````class Solution:
def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
count = 0

matrix = [[0 for j in range(m)] for i in range(n)]

for indice in indices:
r = indice
c = indice
for i in range(len(matrix)):
for j in range(len(matrix)):
if i == r:
matrix[i][j] += 1
if j == c:
matrix[i][j] += 1

for i in range(len(matrix)):
for j in range(len(matrix)):
if matrix[i][j] % 2 != 0:
count += 1

return count``````
• Time Complexity: O(N^2)
• Space Complexity: O(N^2)