Take Gifts From the Richest Pile

Easy

DE Shaw

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Topics:

ArraysGreedy Algorithms

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.
If there is more than one pile with the maximum number of gifts, choose any.
Reduce the number of gifts in the pile to the floor of the square root of the original number of gifts in the pile.

Return the number of gifts remaining after k seconds.

Example 1:

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.

Example 2:

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

Take Gifts From the Richest Pile

Easy

DE Shaw

0 views

Topics:

ArraysGreedy Algorithms

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.
If there is more than one pile with the maximum number of gifts, choose any.
Reduce the number of gifts in the pile to the floor of the square root of the original number of gifts in the pile.

Return the number of gifts remaining after k seconds.

Example 1:

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.

Example 2:

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

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the data type of the pile sizes? Can they be floating-point numbers or only integers?
Can the pile sizes be negative or zero?
What is the expected return value if the input array is null or empty?
If there are multiple piles with the same largest size, which pile should I take the gifts from?
How many times will this gift-taking operation be performed? Is it a fixed number or determined by some other input?

Brute Force Solution

Approach

The brute force approach to taking gifts involves simulating every possible combination of gift taking. We will methodically explore each possibility until we find the best outcome. This method guarantees finding the optimal solution, but it might take a very long time if we have a lot of piles of gifts.

Here's how the algorithm would work step-by-step:

Look at all the piles of gifts.
Find the pile with the most gifts.
Take gifts from that pile according to the given rules.
Repeat the process of finding the pile with the most gifts and taking gifts from it for the specified number of rounds.
Once all rounds are complete, add up the number of gifts remaining in all the piles.
Return the total number of remaining gifts.

Code Implementation

def take_gifts_brute_force(gifts, rounds):
    piles = gifts[:] # Create a copy to avoid modifying the original list

    for _ in range(rounds):
        # Find the index of the pile with the most gifts
        most_gifts_index = 0
        for index in range(1, len(piles)):
            if piles[index] > piles[most_gifts_index]:
                most_gifts_index = index

        # Take gifts from the richest pile
        piles[most_gifts_index] = int(piles[most_gifts_index]**0.5)

    # Calculate the total number of remaining gifts
    total_gifts = sum(piles)
    return total_gifts

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n * k) – The algorithm iterates 'k' times, representing the number of rounds. Within each round, we need to find the pile with the most gifts. Finding the maximum element in an array of 'n' piles takes O(n) time. Since we repeat this process 'k' times, the overall time complexity is O(n * k), where 'n' is the number of piles and 'k' is the number of rounds.

Space Complexity

O(1) – The provided plain English explanation describes an iterative process that finds the maximum value and modifies it in place. No auxiliary data structures like lists, hashmaps, or recursion are used to store intermediate results or track visited piles. The operations are done directly on the input array, and only a few constant-size variables (e.g., for storing the index of the pile with the most gifts) are needed. Therefore, the space complexity is constant, independent of the number of piles.

Optimal Solution

Approach

The core idea is to repeatedly find the pile with the most gifts and reduce its number by its square root. We continue this process for a fixed number of rounds, ensuring that we focus on maximizing the gift reduction at each step.

Here's how the algorithm would work step-by-step:

Find the pile that currently has the most gifts.
Calculate the square root of the number of gifts in that pile.
Replace the original number of gifts in the pile with the calculated square root (rounded down to the nearest whole number).
Repeat the above steps for the given number of rounds.

Code Implementation

def take_gifts_from_richest_pile(gifts, rounds):
    for _ in range(rounds):
        # Find the index of the pile with the maximum gifts.
        richest_pile_index = gifts.index(max(gifts))

        # Calculate the square root and update the pile.
        gifts[richest_pile_index] = int(gifts[richest_pile_index]**0.5)

    # Calculate and return the total number of remaining gifts.
    total_gifts = sum(gifts)
    return total_gifts

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n * k) – The algorithm iterates for k rounds. In each round, it needs to find the maximum element among n piles. Finding the maximum element in an array of size n takes O(n) time. Since this process is repeated k times, the overall time complexity becomes O(n * k), where n is the number of piles and k is the number of rounds.

Space Complexity

O(1) – The algorithm operates directly on the input array (the 'piles' of gifts). It doesn't create any auxiliary data structures like lists, hash maps, or recursion stacks. The only extra space used consists of a few variables to keep track of the index of the richest pile and potentially a temporary variable for calculations. Therefore, the space complexity remains constant, regardless of the number of piles (N).

Edge Cases

Case	How to Handle
Empty gifts array	Return 0 since there are no gifts to take.
gifts array contains negative values	Take the absolute value of each gift before applying the operation, assuming the prompt requires that.
gifts array contains zero values	The square root of zero is zero, handle zeros normally without causing any errors.
k is zero	Return the original gifts array, as no operations are performed.
Large k value exceeding the number of elements in gifts	Iterate through the gifts array up to k times, or stop when all values reach zero.
Integer overflow when calculating square root on a large value in gifts array	Cast to a double for calculation to avoid integer overflow errors, and ensure the result is cast back to an integer.
gifts array contains extremely large positive integer values	Confirm that the square root implementation is resilient to very large numbers without infinite looping or performance degradation.
gifts array is already sorted in non-increasing order	Algorithm works normally as the largest gift will always be processed first.