LeetCode in Kotlin

3478. Choose K Elements With Maximum Sum

Medium

You are given two integer arrays, nums1 and nums2, both of length n, along with a positive integer k.

For each index i from 0 to n - 1, perform the following:

• Find all indices j where nums1[j] is less than nums1[i].
• Choose at most k values of nums2[j] at these indices to maximize the total sum.

Return an array answer of size n, where answer[i] represents the result for the corresponding index i.

Example 1:

Input: nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2

Output: [80,30,0,80,50]

Explanation:

• For i = 0: Select the 2 largest values from nums2 at indices [1, 2, 4] where nums1[j] < nums1[0], resulting in 50 + 30 = 80.
• For i = 1: Select the 2 largest values from nums2 at index [2] where nums1[j] < nums1[1], resulting in 30.
• For i = 2: No indices satisfy nums1[j] < nums1[2], resulting in 0.
• For i = 3: Select the 2 largest values from nums2 at indices [0, 1, 2, 4] where nums1[j] < nums1[3], resulting in 50 + 30 = 80.
• For i = 4: Select the 2 largest values from nums2 at indices [1, 2] where nums1[j] < nums1[4], resulting in 30 + 20 = 50.

Example 2:

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

Output: [0,0,0,0]

Explanation:

Since all elements in nums1 are equal, no indices satisfy the condition nums1[j] < nums1[i] for any i, resulting in 0 for all positions.

Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 1 <= nums1[i], nums2[i] <= 106
• 1 <= k <= n

Solution

import java.util.PriorityQueue

class Solution {
    fun findMaxSum(nums1: IntArray, nums2: IntArray, k: Int): LongArray {
        val n = nums1.size
        val ans = LongArray(n)
        val ps = Array<Point>(n) { i -> Point(nums1[i], nums2[i], i) }
        ps.sortWith { p1: Point, p2: Point -> p1.x.compareTo(p2.x) }
        val pq = PriorityQueue<Int>()
        var s: Long = 0
        var i = 0
        while (i < n) {
            var j = i
            while (j < n && ps[j].x == ps[i].x) {
                ans[ps[j].i] = s
                j++
            }
            for (p in i..<j) {
                val cur = ps[p].y
                if (pq.size < k) {
                    pq.offer(cur)
                    s += cur.toLong()
                } else if (cur > pq.peek()!!) {
                    s -= pq.poll()!!.toLong()
                    pq.offer(cur)
                    s += cur.toLong()
                }
            }
            i = j
        }
        return ans
    }

    private class Point(var x: Int, var y: Int, var i: Int)
}

This site is open source. Improve this page.