#### QUESTION:

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.

1.求出行的最小值

2.求出列的最小值

3.进行乘法运算，算出总个数。

#### SOLUTION:

``````public class Solution {
public int maxCount(int m, int n, int[][] ops) {
int row = Integer.MAX_VALUE;int colum = Integer.MAX_VALUE;
for(int i= 0;i<ops.length;i++){
row = Math.min(ops[i],row);
colum = Math.min(ops[i],colum);
}
row = Math.min(m,row);
colum = Math.min(n,colum);
return row*colum;
}
}
``````
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