LeetCode 2530 - Maximal Score After Applying K Operations

The problem asks us to maximize a score after performing exactly k operations on an array of integers. Each operation allows us to choose any element nums[i], add its value to the score, and then replace it with its ceiling division by 3.

leetcodemediumarraygreedyheap-(priority-queue)

LeetCode Problem 2530

Difficulty: 🟡 Medium
Topics: Array, Greedy, Heap (Priority Queue)

Solution

Problem Understanding

The problem asks us to maximize a score after performing exactly k operations on an array of integers. Each operation allows us to choose any element nums[i], add its value to the score, and then replace it with its ceiling division by 3. Essentially, every time we pick a number, it decreases for future operations, so the greedy choice is to always pick the largest available number at that moment to maximize immediate gain.

The input consists of an integer array nums of length up to 100,000 and a number k up to 100,000. Each array element can be as large as 1 billion. This indicates that a naive approach iterating over the array to find the maximum in each operation would be too slow, as it could lead to O(n * k) time complexity, which may reach 10^10 operations.

Important edge cases include arrays with all identical numbers, arrays with a single very large number, and small arrays where k exceeds the number of elements. The problem guarantees at least one element and at least one operation, so we do not have to handle empty inputs.

The output is a single integer representing the maximum possible score achievable after exactly k operations.

Approaches

The brute-force approach would involve iterating through the array to find the maximum element at each operation, updating it, and repeating k times. This approach works correctly because picking the largest element maximizes immediate gain. However, for large arrays and large k, this approach is too slow, with worst-case time complexity O(n * k).

The optimal approach uses a max-heap (priority queue). By pushing all elements into a max-heap, we can extract the current largest element in O(log n) time, add it to the score, apply the ceiling division, and push the updated value back into the heap. Repeating this k times ensures that we always pick the largest available element efficiently.

The key insight is that greedily selecting the largest element at each operation is always optimal because the value of elements only decreases after selection.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * k) O(1) Scan array each time to find max element
Optimal O(k * log n) O(n) Use max-heap to efficiently track largest element

Algorithm Walkthrough

  1. Initialize a max-heap with all elements of nums. Since Python only has a min-heap, store negative values to simulate a max-heap.
  2. Initialize score to 0. This will accumulate the total score across operations.
  3. Repeat k times:

a. Extract the maximum element from the heap (the most negative in Python's heap).

b. Add this value to the score.

c. Compute the updated value as ceil(value / 3) and push it back into the heap. 4. After completing k operations, return the accumulated score.

Why it works: At each step, we always pick the largest available element, ensuring the maximum possible addition to the score. The heap guarantees we can efficiently find and update the maximum element each operation. This greedy choice is optimal because any other element selection yields a smaller immediate score and cannot be better for subsequent operations, as all elements only decrease over time.

Python Solution

from typing import List
import heapq
import math

class Solution:
    def maxKelements(self, nums: List[int], k: int) -> int:
        # Convert nums into a max-heap using negative values
        max_heap = [-num for num in nums]
        heapq.heapify(max_heap)
        
        score = 0
        
        for _ in range(k):
            # Pop the largest element
            current = -heapq.heappop(max_heap)
            score += current
            
            # Apply ceil division and push back
            updated = math.ceil(current / 3)
            heapq.heappush(max_heap, -updated)
        
        return score

Implementation walkthrough: We first convert nums into a max-heap by negating values. For k operations, we pop the largest element, add it to score, compute the ceiling division by 3, and push it back. Using a heap ensures that extracting and inserting elements takes O(log n), making the algorithm efficient.

Go Solution

package main

import (
    "container/heap"
    "math"
)

type IntHeap []int

func (h IntHeap) Len() int            { return len(h) }
func (h IntHeap) Less(i, j int) bool  { return h[i] > h[j] } // Max-heap
func (h IntHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *IntHeap) Push(x any)         { *h = append(*h, x.(int)) }
func (h *IntHeap) Pop() any {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[0 : n-1]
    return x
}

func maxKelements(nums []int, k int) int64 {
    maxHeap := &IntHeap{}
    heap.Init(maxHeap)
    for _, num := range nums {
        heap.Push(maxHeap, num)
    }

    var score int64 = 0
    for i := 0; i < k; i++ {
        current := heap.Pop(maxHeap).(int)
        score += int64(current)
        updated := int(math.Ceil(float64(current) / 3.0))
        heap.Push(maxHeap, updated)
    }

    return score
}

Go-specific notes: Go requires defining a heap interface. The Less function is reversed to implement a max-heap. Integer overflow is handled by using int64 for the score, since individual elements can be up to 10^9 and k up to 10^5.

Worked Examples

Example 1: nums = [10,10,10,10,10], k = 5

Step Heap state (max to min) Popped Updated Score
1 [10,10,10,10,10] 10 4 10
2 [10,10,10,10,4] 10 4 20
3 [10,10,4,4,10] 10 4 30
4 [10,10,4,4,4] 10 4 40
5 [10,4,4,4,4] 10 4 50

Example 2: nums = [1,10,3,3,3], k = 3

Step Heap state Popped Updated Score
1 [10,3,3,3,1] 10 4 10
2 [4,3,3,1,3] 4 2 14
3 [3,3,2,1,2] 3 1 17

Complexity Analysis

Measure Complexity Explanation
Time O(k * log n) Each of the k operations requires extracting and inserting into the heap, each O(log n)
Space O(n) Max-heap stores all n elements

The algorithm is efficient because it avoids scanning the array repeatedly. Even for the largest constraints, 10^5 operations on a heap of 10^5 elements is feasible.

Test Cases

# Provided examples
assert Solution().maxKelements([10,10,10,10,10], 5) == 50  # all elements picked once
assert Solution().maxKelements([1,10,3,3,3], 3) == 17      # greedy picks largest elements

# Edge cases
assert Solution().maxKelements([1], 1) == 1                 # single element
assert Solution().maxKelements([9,27,81], 5) == 183        # multiple updates
assert Solution().maxKelements([1000000000]*5, 5) == 5000000000  # large numbers

# Stress case
nums = [i for i in range(1, 100001)]
assert Solution().maxKelements(nums, 100000) > 0           # large input
Test Why
[10,10,10,10,10], 5 verifies simple repeated picks
[1,10,3,3,3], 3 ensures greedy choice is applied
[1], 1 single element edge case
[9,27,81], 5 multiple reductions per element
[1000000000]*5, 5