LeetCode in Kotlin

2542. Maximum Subsequence Score

Medium

You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k. You must choose a subsequence of indices from nums1 of length k.

For chosen indices i0, i1, …, ik - 1, your score is defined as:

• The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
• It can defined simply as: (nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 1]).

Return the maximum possible score.

A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.

Example 1:

Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3

Output: 12

Explanation:

The four possible subsequence scores are:

• We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.

• We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.

• We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.

• We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.

Therefore, we return the max score, which is 12.

Example 2:

Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1

Output: 30

Explanation: Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.

Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 0 <= nums1[i], nums2[j] <= 105
• 1 <= k <= n

Solution

import java.util.PriorityQueue

class Solution {
    private class PairInfo(var val1: Int, var val2: Int)

    fun maxScore(nums1: IntArray, nums2: IntArray, k: Int): Long {
        val n = nums1.size
        val nums = arrayOfNulls<PairInfo>(n)
        for (i in 0 until n) {
            nums[i] = PairInfo(nums1[i], nums2[i])
        }

        nums.sortWith sort@{ a: PairInfo?, b: PairInfo? ->
            if (a!!.val2 == b!!.val2) {
                return@sort a.val1 - b.val1
            }
            a.val2 - b.val2
        }
        var sum: Long = 0
        var ans: Long = 0
        val pq = PriorityQueue<Int>()
        for (i in n - 1 downTo 0) {
            val minVal = nums[i]!!.val2
            while (pq.size > k - 1) {
                sum -= pq.poll().toLong()
            }
            sum += nums[i]!!.val1.toLong()
            pq.add(nums[i]!!.val1)
            if (pq.size == k) {
                ans = Math.max(ans, sum * minVal)
            }
        }
        return ans
    }
}

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