LeetCode in Kotlin

2530. Maximal Score After Applying K Operations

Medium

You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.

In one operation:

1. choose an index i such that 0 <= i < nums.length,
2. increase your score by nums[i], and
3. replace nums[i] with ceil(nums[i] / 3).

Return the maximum possible score you can attain after applying exactly k operations.

The ceiling function ceil(val) is the least integer greater than or equal to val.

Example 1:

Input: nums = [10,10,10,10,10], k = 5

Output: 50

Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.

Example 2:

Input: nums = [1,10,3,3,3], k = 3

Output: 17

Explanation: You can do the following operations:

Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10.

Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4.

Operation 3: Select i = 2, so nums becomes [1,1,1,3,3]. Your score increases by 3.

The final score is 10 + 4 + 3 = 17.

Constraints:

• 1 <= nums.length, k <= 105
• 1 <= nums[i] <= 109

Solution

import java.util.Collections
import java.util.PriorityQueue

@Suppress("NAME_SHADOWING")
class Solution {
    fun maxKelements(nums: IntArray, k: Int): Long {
        var k = k
        val p = PriorityQueue(Collections.reverseOrder<Int>())
        nums.sort()
        for (i in nums.indices) {
            p.add(nums[nums.size - i - 1])
        }
        var score: Long = 0
        while (k != 0) {
            val v1 = p.poll()
            score += v1.toLong()
            val v2 = Math.ceil(v1.toDouble() / 3).toInt()
            p.add(v2)
            k--
        }
        return score
    }
}

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