598. Range Addition II

Problem statement:

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.

Solution:

This is the first question of leetcode weekly contest 34. Obviously, we should find the common area for all operations, which is enclosed by the min value in x/y axis. Like 221. Maximal Square.

NOTE: we should return m * n if the operation set is empty. The initialization of x_min/y_min is m/n.

Time complexity is O(n). n is the size of operation set.

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        int x_min = m;
        int y_min = n;
        for(auto op : ops){
            x_min = min(x_min, op[0]);
            y_min = min(y_min, op[1]);
        }
        return x_min * y_min;
    }
};