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.
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 result of operations. For each pair $(r_i, c_i)$ in $\textit{indices}$, we add $1$ to all numbers in the $r_i$-th row of the matrix and add $1$ to all elements in the $c_i$-th column.
After the simulation ends, we traverse the matrix and count the number of odd numbers.
The time complexity is $O(k \times (m + n) + m \times n)$, and the space complexity is $O(m \times n)$. Here, $k$ is the length of $\textit{indices}$.
We can use a row array $\textit{row}$ and a column array $\textit{col}$ to record the number of times each row and column is incremented. For each pair $(r_i, c_i)$ in $\textit{indices}$, we add $1$ to $\textit{row}[r_i]$ and $\textit{col}[c_i]$ respectively.
After the operations are completed, the count at position $(i, j)$ can be calculated as $\textit{row}[i] + \textit{col}[j]$. We traverse the matrix and count the number of odd numbers.
The time complexity is $O(k + m \times n)$, and the space complexity is $O(m + n)$. Here, $k$ is the length of $\textit{indices}$.
We notice that a number at position $(i, j)$ in the matrix will be odd only when exactly one of $\textit{row}[i]$ and $\textit{col}[j]$ is odd and the other is even.
We count the number of odd numbers in $\textit{row}$, denoted as $\textit{cnt1}$, and the number of odd numbers in $\textit{col}$, denoted as $\textit{cnt2}$. Therefore, the final count of odd numbers is $\textit{cnt1} \times (n - \textit{cnt2}) + \textit{cnt2} \times (m - \textit{cnt1})$.
The time complexity is $O(k + m + n)$, and the space complexity is $O(m + n)$. Here, $k$ is the length of $\textit{indices}$.