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 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 $O(mn)$.
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 $O(m+n)$.
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 $O(m+n)$.