leetcode-598-easy

Range Addition II

You are given an m x n matrix M initialized with all 0's and an array of operations ops,
 where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi.

Count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3, ops = [[2,2],[3,3]]
Output: 4
Explanation: The maximum integer in M is 2, and there are four of it in M. So return 4.
Example 2:

Input: m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]
Output: 4
Example 3:

Input: m = 3, n = 3, ops = []
Output: 9
Constraints:

1 <= m, n <= 4 * 104
0 <= ops.length <= 104
ops[i].length == 2
1 <= ai <= m
1 <= bi <= n

思路一：对数组的操作是累加操作，所以只要找到数组哪块区域进行的累加操作最多，就可以算出结果。

    public int maxCount(int m, int n, int[][] ops) {
        if (ops.length == 0) {
            return m * n;
        }

        int minX = Integer.MAX_VALUE;
        int minY = Integer.MAX_VALUE;

        for (int[] op : ops) {
            minX = Math.min(minX, op[0]);
            minY = Math.min(minY, op[1]);
        }

        return minX * minY;
    }