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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 <vector>
2#include <algorithm>
3using namespace std;
4
5int maxCount(int m, int n, vector<vector<int>>& ops) {
6 for (const auto& op : ops) {
7 m = min(m, op[0]);
8 n = min(n, op[1]);
9 }
10 return m * n;
11}The solution iterates over the operations. For each operation, it updates the minimum number of rows and columns considering the current operation's limits. Finally, it returns the product of these minimum values, which gives the area of the submatrix with 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.
The JavaScript implementation creates a matrix using nested arrays, simulates the increment operations, and determines the count of the maximum value present in the matrix.