LeetCode in Kotlin

646. Maximum Length of Pair Chain

Medium

You are given an array of n pairs pairs where pairs[i] = [lefti, righti] and lefti < righti.

A pair p2 = [c, d] follows a pair p1 = [a, b] if b < c. A chain of pairs can be formed in this fashion.

Return the length longest chain which can be formed.

You do not need to use up all the given intervals. You can select pairs in any order.

Example 1:

Input: pairs = [[1,2],[2,3],[3,4]]

Output: 2

Explanation: The longest chain is [1,2] -> [3,4].

Example 2:

Input: pairs = [[1,2],[7,8],[4,5]]

Output: 3

Explanation: The longest chain is [1,2] -> [4,5] -> [7,8].

Constraints:

• n == pairs.length
• 1 <= n <= 1000
• -1000 <= lefti < righti <= 1000

Solution

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        if (pairs.size == 1) {
            return 1
        }
        pairs.sortWith { a: IntArray, b: IntArray ->
            a[1] - b[1]
        }
        var min = pairs[0][1]
        var max = 1
        for (i in 1 until pairs.size) {
            if (pairs[i][0] > min) {
                max++
                min = pairs[i][1]
            }
        }
        return max
    }
}

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