LeetCode 598 - Range Addition II

https://leetcode.com/problems/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.

https://discuss.leetcode.com/topic/90547/java-solution-find-min
思路1: brute force - O(m * n * k), k = number of ops
思路2: 从 ops 中找到被操作次数最多的 row 和 column，这可以通过遍历 ops 数组，分别找到 row 和 column 的最小值来完成

You don't need the first if statement if you init row = m and col = n :)

    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length == 0) {
            return m * n;
        }
        
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int[] op : ops) {
            row = Math.min(row, op[0]);
            col = Math.min(col, op[1]);
        }
        
        return row * col;
    }

https://discuss.leetcode.com/topic/90540/c-java-clean-code
https://www.nowtoshare.com/zh/Article/Index/70563

    public int maxCount(int m, int n, int[][] ops) {
        for (int[] op : ops) {
            m = Math.min(op[0], m);
            n = Math.min(op[1], n);
        }
        return m * n;        
    }