RangeAdditionII [source code]

public class RangeAdditionII {
    public int maxCount(int m, int n, int[][] ops) {
        if (m * n == 0) return 0;
        if (ops.length == 0) return m * n;
        int minM = 40000, minN = 40000;
        for (int[] op : ops) {
            if (op[0] < minM) minM = op[0];
            if (op[1] < minN) minN = op[1];
        }
        return minM * minN;
    }
}

算法整体比较简单, 没什么好说的. 速度是9ms, 13%.

翻了一下, submission 好像没有找到更好的答案, discussion 的最优解跟我这个基本差不多的意思; editorial 的
最优解也是类似的思路;

Problem Description

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.
Difficulty:Easy
Category:Algorithms
Acceptance:48.66%
Contributors:
fallcreek
Topics:
Math
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(M) Range Addition
Company:
IXL