# 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:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won’t exceed 10,000.
这道题看上去很唬人，不过纸上比划一下发现其实非常简单 —— 就是求最小的范围（注意是范围，我一开始简单的想成求最小的区块）

```public class Solution {
public int maxCount(int m, int n, int[][] ops) {
if(ops.length == 0) return m*n;
int r = ops.length-1;
int minrow = Integer.MAX_VALUE;
int mincol = Integer.MAX_VALUE;
while(r>=0){
if(ops[r] < minrow){
minrow = ops[r];
}
if(ops[r] < mincol){
mincol = ops[r];
}
r--;
}
return mincol * minrow;
}
}```

[0人评了分，平均: 0/5]