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To solve this problem efficiently, observe that each operation affects a submatrix starting from the top-left corner. Thus, the final result of the operations is determined by the smallest intersected submatrix affected by all operations. Find the minimum value of "a" and "b" from all operations in ops. This will give you the dimensions of the area where the maximum integers will exist after applying all operations.
The key insight is that the size of the submatrix affected by all operations determines the count of maximum numbers in the final matrix.
Time Complexity: O(k), where k is the number of operations.
Space Complexity: O(1), as we are using only a few extra variables.
1#include <stdio.h>
2
3int maxCount(int m, int n, int ops[][2], int opsSize) {
4 int minRow = m;
5 int minCol = n;
6 for (int i = 0; i < opsSize; i++) {
7 if (ops[i][0] < minRow) {
8 minRow = ops[i][0];
9 }
10 if (ops[i][1] < minCol) {
11 minCol = ops[i][1];
12 }
13 }
14 return minRow * minCol;
15}
16
17int main() {
18 int ops[][2] = {{2, 2}, {3, 3}};
19 int m = 3;
20 int n = 3;
21 printf("%d\n", maxCount(m, n, ops, 2));
22 return 0;
23}This solution iterates over all operations provided and finds the minimum values of rows and columns from the operations. These minimums define the dimensions of the submatrix that will have the maximum value after all operations. We simply calculate the area of this submatrix to get the count of maximum integers.
This approach simulates the operations directly on the matrix. For each operation, we iterate through the specified submatrix and increment the values. Finally, we determine the number of times the maximum value occurs in the matrix. Although this method is less efficient due to its higher computational cost, it reinforces understanding of the problem.
Time Complexity: O(m * n * k), where m and n are the dimensions of the matrix, and k is the number of operations.
Space Complexity: O(m * n), for the matrix.
This Python solution utilizes a two-dimensional list to maintain the values affected by each operation. The matrix is updated according to each operation, and the maximum value and its count are tracked and returned.