LeetCode 2558 - Take Gifts From the Richest Pile

The problem presents a scenario where we have multiple piles of gifts, represented as an integer array gifts, with each element indicating the number of gifts in that pile.

leetcodeeasyarrayheap-(priority-queue)simulation

LeetCode Problem 2558

Difficulty: 🟢 Easy
Topics: Array, Heap (Priority Queue), Simulation

Solution

Problem Understanding

The problem presents a scenario where we have multiple piles of gifts, represented as an integer array gifts, with each element indicating the number of gifts in that pile. Every second, you must choose the pile with the most gifts and reduce its count to the floor of its square root. The operation is repeated k times, and the goal is to calculate the total number of gifts remaining after all operations.

In other words, the input array gifts tells us the initial number of gifts in each pile, and k tells us how many seconds (iterations) we perform the "take from the richest pile" operation. The expected output is a single integer representing the sum of all gifts left after performing k reductions. The constraints 1 <= gifts.length <= 10^3 and 1 <= k <= 10^3 indicate that the input is small enough to allow iterative operations, but gifts[i] can be as large as 10^9, meaning we cannot afford to iterate linearly over piles to find the maximum every time without optimization.

Important edge cases include situations where all piles have the same number of gifts, when all piles contain only 1 gift (square root does not reduce it), and when k is larger than the number of piles.

Approaches

The simplest approach is brute force. In each of the k seconds, we scan the entire gifts array to find the pile with the maximum gifts, then replace it with its floor square root. While this works correctly, finding the maximum each time takes O(n) time, and with k iterations, the total time complexity becomes O(n * k). For the given constraints (up to 1000 piles and 1000 seconds), this could be up to 10^6 operations, which is acceptable but not optimal.

A better approach uses a max heap (priority queue). By storing all piles in a max heap, we can efficiently extract the pile with the maximum gifts in O(log n) time. After taking the gifts, we compute the floor of the square root and push it back into the heap, maintaining the heap property. Repeating this k times gives an overall time complexity of O(k * log n), which is faster for larger inputs. This approach leverages the heap data structure to optimize repeated maximum selection.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * k) O(1) Scan array each iteration to find max
Optimal (Heap) O(k * log n) O(n) Use max heap to efficiently get max pile

Algorithm Walkthrough

  1. Initialize a max heap and insert all gift counts as negative values (to simulate a max heap with Python's min heap).

  2. Repeat the following k times:

  3. Pop the largest element from the heap (this represents the pile with the maximum gifts).

  4. Take the floor of its square root.

  5. Push the new reduced count back into the heap.

  6. After completing k operations, sum all elements in the heap (convert negatives back to positive) to get the total remaining gifts.

Why it works: At each iteration, the heap guarantees that we always operate on the current maximum pile. The invariants are that the heap always contains the current state of all piles, and the reduction operation is applied exactly k times. This guarantees that the sum of the heap elements at the end correctly reflects the remaining gifts.

Python Solution

from typing import List
import heapq
import math

class Solution:
    def pickGifts(self, gifts: List[int], k: int) -> int:
        # Initialize a max heap using negative values
        max_heap = [-gift for gift in gifts]
        heapq.heapify(max_heap)
        
        # Perform the operation k times
        for _ in range(k):
            largest = -heapq.heappop(max_heap)  # Get the largest gift pile
            reduced = math.isqrt(largest)       # Floor of square root
            heapq.heappush(max_heap, -reduced)  # Push back into heap
        
        # Sum remaining gifts
        return -sum(max_heap)

In this implementation, we first convert all gift counts to negative values to use Python's min heap as a max heap. Each iteration pops the largest pile, reduces it, and pushes the updated count back. Finally, we convert the heap values back to positive and sum them.

Go Solution

import (
    "container/heap"
    "math"
)

type MaxHeap []int

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

func pickGifts(gifts []int, k int) int64 {
    h := &MaxHeap{}
    for _, gift := range gifts {
        *h = append(*h, gift)
    }
    heap.Init(h)
    
    for i := 0; i < k; i++ {
        largest := heap.Pop(h).(int)
        reduced := int(math.Sqrt(float64(largest)))
        heap.Push(h, reduced)
    }
    
    var total int64 = 0
    for h.Len() > 0 {
        total += int64(heap.Pop(h).(int))
    }
    return total
}

In Go, we define a custom MaxHeap type and implement the heap.Interface methods. The rest of the logic mirrors Python, with integer square root conversion handled via math.Sqrt. The heap ensures efficient maximum selection and update.

Worked Examples

Example 1: gifts = [25,64,9,4,100], k = 4

Iteration Max Heap (as list) Pile chosen Reduced value Heap after push
1 [100,64,25,4,9] 100 10 [64,25,10,4,9]
2 [64,25,10,4,9] 64 8 [25,10,9,4,8]
3 [25,10,9,4,8] 25 5 [10,9,5,4,8]
4 [10,9,5,4,8] 10 3 [9,8,5,4,3]

Sum = 9 + 8 + 5 + 4 + 3 = 29

Example 2: gifts = [1,1,1,1], k = 4

All piles are 1, reducing any gives 1. After 4 iterations, sum = 1 + 1 + 1 + 1 = 4

Complexity Analysis

Measure Complexity Explanation
Time O(k * log n) Each of the k operations pops and pushes from the heap in O(log n)
Space O(n) Heap stores all n gift piles

The time complexity improves over the brute-force approach by using a heap to avoid scanning the array for the maximum each time. Space complexity is dominated by the heap.

Test Cases

# Provided examples
assert Solution().pickGifts([25,64,9,4,100], 4) == 29  # standard case
assert Solution().pickGifts([1,1,1,1], 4) == 4        # all ones

# Edge cases
assert Solution().pickGifts([1], 1) == 1              # single pile
assert Solution().pickGifts([1000000000], 1) == 31622 # large single value
assert Solution().pickGifts([2,3,5,7,11], 5) == 7    # small primes
assert Solution().pickGifts([4,4,4,4], 2) == 8       # same values
assert Solution().pickGifts([100,64,36,25,16], 0) == 241 # zero k, sum unchanged
Test Why
[25,64,9,4,100], k=4 Normal multi-pile scenario
[1,1,1,1], k=4 Piles with minimum value, no reduction
[1], k=1 Single pile edge case
[1000000000], k=1 Test large numbers and sqrt
[2,3,5,7,11], k=5 Small primes to test multiple operations
[4,4,4,4], k=2 Equal piles scenario
[100,64,36,25,16], k=0 k=0, sum should remain unchanged

Edge Cases

First, the case where all piles contain `