LeetCode 598: Range Addition II Solution

Master LeetCode problem 598 (Range Addition II), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

598. Range Addition II

Easy

Array Math

Problem Explanation

Explanation

To solve this problem efficiently, we can observe that the maximum integers in the matrix will be found at the intersection of the maximum values of ai and bi from all the operations. Therefore, the number of maximum integers in the matrix will be the product of the maximum values of ai and bi.

Initialize minAi and minBi to m and n respectively.
Iterate through all the operations and update minAi and minBi to the minimum ai and bi values encountered.
The number of maximum integers in the matrix will be minAi * minBi.

Time Complexity: O(n), where n is the number of operations. Space Complexity: O(1)

Solution Code

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

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