It is sufficient to track just the min a and min b

```
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
if (ops.empty()) return m*n;
int mina = m+2;
int minb = n+2;
for (int i = 0; i < ops.size(); i++) {
if (ops[i][0] < mina) {
mina = ops[i][0];
}
if (ops[i][1] < minb) {
minb = ops[i][1];
}
}
return mina*minb;
}
};
```