forked from sureshmangs/Code
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path1252. Cells with Odd Values in a Matrix.cpp
More file actions
63 lines (37 loc) · 1.34 KB
/
1252. Cells with Odd Values in a Matrix.cpp
File metadata and controls
63 lines (37 loc) · 1.34 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
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
class Solution {
public:
int oddCells(int n, int m, vector<vector<int>>& indices) {
vector<vector<int> > mat(n, vector<int> (m, 0));
for(auto &it: indices){
int x=it[0];
int y=it[1];
for(int i=0;i<n;i++) mat[i][y]++;
for(int j=0;j<m;j++) mat[x][j]++;
}
int res=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++)
if(mat[i][j]%2!=0)
res++;
}
return res;
}
};