LeetCode in Kotlin

598. Range Addition II

Easy

You are given an m x n matrix M initialized with all 0’s and an array of operations ops, where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi.

Count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3, ops = [[2,2],[3,3]]

Output: 4

Explanation: The maximum integer in M is 2, and there are four of it in M. So return 4.

Example 2:

Input: m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]

Output: 4

Example 3:

Input: m = 3, n = 3, ops = []

Output: 9

Constraints:

• 1 <= m, n <= 4 * 104
• 0 <= ops.length <= 104
• ops[i].length == 2
• 1 <= ai <= m
• 1 <= bi <= n

Solution

class Solution {
    /*
     * Since the incrementing starts from zero to op[0] and op[1], we only need to find the range that
     * has the most overlaps. Thus we keep finding the minimum of both x and y.
     */
    fun maxCount(m: Int, n: Int, ops: Array<IntArray>): Int {
        var x = m
        var y = n
        for (op in ops) {
            x = x.coerceAtMost(op[0])
            y = y.coerceAtMost(op[1])
        }
        return x * y
    }
}

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