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Leetcode #2558: Take Gifts From the Richest Pile

In this guide, we solve Leetcode #2558 Take Gifts From the Richest Pile in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an integer array gifts denoting the number of gifts in various piles. Every second, you do the following: Choose the pile with the maximum number of gifts.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Array, Simulation, Heap (Priority Queue)

Intuition

We need to repeatedly access the smallest or largest element as the input changes.

A heap provides fast insertions and removals while keeping order.

Approach

Push candidates into the heap as you scan, and pop when you need the best element.

Keep the heap size bounded if the problem requires a top-k structure.

Steps:

  • Push candidates into a heap.
  • Pop the best candidate when needed.
  • Maintain heap size or invariants.

Example

Input: gifts = [25,64,9,4,100], k = 4 Output: 29 Explanation: The gifts are taken in the following way: - In the first second, the last pile is chosen and 10 gifts are left behind. - Then the second pile is chosen and 8 gifts are left behind. - After that the first pile is chosen and 5 gifts are left behind. - Finally, the last pile is chosen again and 3 gifts are left behind. The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.

Python Solution

class Solution: def pickGifts(self, gifts: List[int], k: int) -> int: h = [-v for v in gifts] heapify(h) for _ in range(k): heapreplace(h, -int(sqrt(-h[0]))) return -sum(h)

Complexity

The time complexity is O(n+k×log⁡n)O(n + k \times \log n)O(n+k×logn), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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