LeetCode 598. Range Addition II

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.

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        if(ops.size()==0) return m*n;
        int mincol=99999, minrow=99999;
        for(int i=0; i<ops.size(); i++){
            minrow=min(ops[i][0],minrow);
            mincol=min(ops[i][1],mincol);
        }
        return mincol*minrow;
    }
};