# [LeetCode]#1252. Cells with Odd Values in a Matrix

**Environment: Python 3.7**

**Key technique:** for

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][0] < n`

`0 <= indices[i][1] < m`

**Analysis:**

- Greate m x n matrix.
- add 1 based on in indices.
- Sum all old number.

**Solution:**

`class Solution:`

def oddCells(self, n, m, indices):

count=0

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

for c in indices:

x,y=c[0],c[1]

for i in range(m):

matrix[x][i]+=1

for j in range(n):

matrix[j][y]+=1

for i in range(n):

for j in range(m):

if matrix[i][j] % 2 !=0:

count+=1

return count

**Submitted result:**

**Lesson learn:**

“matrix=[[0]*m for i in range(n)]” is a short method to create m x n matrix.

**Reference:**