#2558

Easy

auto_awesomeArray plus Heap (Priority Queue)

LeetCode Problem Workspace

Take Gifts From the Richest Pile

Take Gifts From the Richest Pile uses a heap to simulate the process of taking gifts from the richest pile over a number of seconds.

array heap priority queue simulation

Problem Statement

You are given an array of integers where each element represents the number of gifts in a pile. Every second, you must take gifts from the richest pile, removing half of the gifts from it, and placing the remaining half back into the pile. Repeat this process for k seconds. At the end of k seconds, you need to return the total number of gifts left in all piles.

For example, if you are given the array gifts = [25,64,9,4,100] and k = 4, after taking gifts from the piles over 4 seconds, you will return the total number of remaining gifts.

Examples

Example 1

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

Output: 29

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.

Example 2

Input: gifts = [1,1,1,1], k = 4

Output: 4

In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. That is, you can't take any pile with you. So, the total gifts remaining are 4.

Constraints

1 <= gifts.length <= 103
1 <= gifts[i] <= 109
1 <= k <= 103

Solution Approach

Use a Max-Heap

To efficiently track and modify the richest pile, we use a max-heap (priority queue). This allows us to access the pile with the most gifts in O(log n) time. After each second, the largest pile is selected, and its gifts are halved before being pushed back into the heap.

Simulation for k Seconds

For each second, we extract the maximum pile from the heap, halve its gift count, and push the remaining gifts back into the heap. This process is repeated for k seconds, simulating the taking of gifts over time.

Sum Remaining Gifts

After simulating for k seconds, simply sum up all the remaining gifts in the heap to get the total number of gifts left in all piles.

Complexity Analysis

Metric	Value
Time	O(n + k \times \log n)
Space	O(n)

The time complexity is O(n + k \times \log n) because we need to initially build the heap (O(n)) and then for each of the k seconds, we perform a heap operation (O(log n)). The space complexity is O(n) due to storing the heap structure that holds the pile values.

What Interviewers Usually Probe

The candidate demonstrates an understanding of heaps and their operations.
The candidate can effectively simulate the problem using heap operations.
The candidate handles edge cases, such as all piles having the same number of gifts.

Common Pitfalls or Variants

Common pitfalls

Not using a max-heap to efficiently access and update the richest pile.
Confusing the heap structure with a min-heap or failing to properly simulate the gift-taking process.
Incorrectly summing the remaining gifts at the end, missing the final result.

Follow-up variants

What if you are allowed to modify the gift-taking process, such as taking only a quarter or a third of the gifts instead of half?
How would the problem change if the number of seconds were unlimited?
What if we had multiple rounds of taking gifts and the number of seconds varied for each round?

FAQ

What is the time complexity of solving Take Gifts From the Richest Pile?

The time complexity is O(n + k \times \log n) because we build the heap once (O(n)) and perform k heap operations (O(log n)) for each second.

What data structure is used to solve this problem?

A max-heap (priority queue) is used to efficiently track the richest pile during the simulation process.

How does GhostInterview help with heap-related problems?

GhostInterview provides practice and detailed explanations for heap-related problems, guiding you through essential operations like heapify and extraction.

How do I handle edge cases in this problem?

Edge cases can include piles with the same number of gifts, or very small values for k. GhostInterview provides feedback on handling these situations efficiently.

Can I modify the gift-taking process in Take Gifts From the Richest Pile?

Yes, you could modify the amount taken each second, such as taking a quarter or a third of the gifts instead of half. This variant changes the heap update strategy.

terminal

Solution

Solution 1: Priority Queue (Max Heap)

We can store the array $gifts$ in a max heap, and then loop $k$ times, each time taking out the top element of the heap, taking the square root of it, and putting the result back into the heap.

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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)

Continue Practicing

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