LeetCode: Range Addition II

LeetCode: 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:

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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.
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public class Solution {
    public int maxCount(int m, int n, int[][] ops) {

         if (ops.length == 0) {
            return m * n;
        }
        int x = Integer.MAX_VALUE;
        int y = Integer.MAX_VALUE;

        for (int[] op : ops) {
            x = Math.min(x, op[0]);
            y = Math.min(y, op[1]);
        }
        return x * y;
    }
}