# 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?

## Solutions

Solution 1: Simulation

We create a matrix $g$ to store the results of the operations. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to all elements in the $r_i$th row and the $c_i$th column of the matrix.

After the simulation, we traverse the matrix and count the number of odd numbers.

The time complexity is $O(\text{indices.length} \times (m+n) + mn)$, and the space complexity is $O(mn)$.

Solution 2: Space Optimization

We use row array $row$ and column array $col$ to record the number of times each row and column are increased. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to $row[r_i]$ and $col[c_i]$ respectively.

After the operation, we can calculate that the count at position $(i, j)$ is $row[i] + col[j]$. We traverse the matrix and count the number of odd numbers.

The time complexity is $O(\text{indices.length} + mn)$, and the space complexity is $O(m+n)$.

Solution 3: Mathematical Optimization

We notice that only when exactly one of $row[i]$ and $col[j]$ is odd, the number at position $(i, j)$ in the matrix will be odd.

We count the number of odd numbers in $row$, denoted as $cnt1$; the number of odd numbers in $col$, denoted as $cnt2$. Then the final number of odd numbers is $cnt1 \times (n-cnt2) + cnt2 \times (m-cnt1)$.

The time complexity is $O(\text{indices.length} + m + n)$, and the space complexity is $O(m+n)$.

• class Solution {
public int oddCells(int m, int n, int[][] indices) {
int[][] g = new int[m][n];
for (int[] e : indices) {
int r = e[0], c = e[1];
for (int i = 0; i < m; ++i) {
g[i][c]++;
}
for (int j = 0; j < n; ++j) {
g[r][j]++;
}
}
int ans = 0;
for (int[] row : g) {
for (int v : row) {
ans += v % 2;
}
}
return ans;
}
}

• class Solution {
public:
int oddCells(int m, int n, vector<vector<int>>& indices) {
vector<vector<int>> g(m, vector<int>(n));
for (auto& e : indices) {
int r = e[0], c = e[1];
for (int i = 0; i < m; ++i) ++g[i][c];
for (int j = 0; j < n; ++j) ++g[r][j];
}
int ans = 0;
for (auto& row : g)
for (int v : row) ans += v % 2;
return ans;
}
};

• class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
g = [[0] * n for _ in range(m)]
for r, c in indices:
for i in range(m):
g[i][c] += 1
for j in range(n):
g[r][j] += 1
return sum(v % 2 for row in g for v in row)


• func oddCells(m int, n int, indices [][]int) int {
g := make([][]int, m)
for i := range g {
g[i] = make([]int, n)
}
for _, e := range indices {
r, c := e[0], e[1]
for i := 0; i < m; i++ {
g[i][c]++
}
for j := 0; j < n; j++ {
g[r][j]++
}
}
ans := 0
for _, row := range g {
for _, v := range row {
ans += v % 2
}
}
return ans
}