# leetcode\_598

Given an m \* n matrix M initialized with all 0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M\[i]\[j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1: Input: m = 3, n = 3 operations = \[\[2,2],\[3,3]] Output: 4 Explanation: Initially, M = \[\[0, 0, 0], \[0, 0, 0], \[0, 0, 0]]

After performing \[2,2], M = \[\[1, 1, 0], \[1, 1, 0], \[0, 0, 0]]

After performing \[3,3], M = \[\[2, 2, 1], \[2, 2, 1], \[1, 1, 1]]

So the maximum integer in M is 2, and there are four of it in M. So return 4. Note: The range of m and n is \[1,40000]. The range of a is \[1,m], and the range of b is \[1,n]. The range of operations size won't exceed 10,000.

## Solutions

1. **straight forward**
2. All ops will cause regions within the former minimum op increse by 1.

```cpp
class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        int minx = INT_MAX, miny = INT_MAX;
        for (auto & v : ops) {
            minx = min(minx, v[0]);
            miny = min(miny, v[1]);
        }

        return minx != INT_MAX  ? minx * miny : m * n;
    }
};
```