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You are given a 0-indexed integer array gifts denoting the number of gifts in several piles. Every second, you pick the pile with the maximum gifts and take half of them (half the gifts are rounded down).

Your job is to return the number of gifts remaining after k seconds.

Clarifying Questions

  1. What should we do if multiple piles have the same number of gifts?
    • Pick any pile with the maximum gifts.
  2. How should we handle rounding when taking half of the gifts?
    • The problem specifies that half the gifts are rounded down.
  3. Are there constraints on the size of gifts and the value of k?
    • Usually, problems have constraints. Assuming typical constraints, the length of gifts and the value of k can be assumed to be reasonably large but manageable within efficient computation time.

Strategy

  1. Use a max-heap to efficiently retrieve and update the pile with the maximum gifts.
    • Python’s heapq library provides a min-heap, so we can store negative values to simulate a max-heap.
  2. For each second up to k:
    • Extract the maximum element from the heap.
    • Calculate the number of gifts to take.
    • Update the heap with the remaining gifts.
  3. Sum the remaining gifts in the heap after k seconds and return the result.

Code

import heapq

def pickGifts(gifts, k):
    max_heap = [-gift for gift in gifts]
    heapq.heapify(max_heap)
    
    for _ in range(k):
        max_gifts = -heapq.heappop(max_heap)
        remaining_gifts = max_gifts - (max_gifts // 2)
        heapq.heappush(max_heap, -remaining_gifts)
    
    remaining_total = -sum(max_heap)
    return remaining_total

# Example usage:
gifts = [5, 4, 9]
k = 3
print(pickGifts(gifts, k))  # Expected output depends on the processing of the function

Time Complexity

  1. Heap Initialization:
    • Building the heap initially takes O(n) time, where n is the number of piles.
  2. Heap Operations:
    • Each of the k operations involves extracting the max (which takes O(log n)) and pushing an updated value back (which also takes O(log n)).
  3. Total Complexity:
    • The overall time complexity is O(n + k log n).

This approach ensures that the solution is efficient and scalable within typical problem constraints.

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