Problem Description

You have an integer array gifts where each element represents the number of gifts in different piles. You need to perform a specific operation for k seconds.

Every second, you must:

1. Find the pile with the maximum number of gifts
2. If multiple piles have the same maximum value, you can choose any one of them
3. Replace the number of gifts in that pile with the floor of its square root

For example, if a pile has 25 gifts, after the operation it will have floor(√25) = 5 gifts. If a pile has 10 gifts, it becomes floor(√10) = 3 gifts.

After performing this operation exactly k times (one per second), return the total number of gifts remaining across all piles.

The solution uses a max heap to efficiently find and update the pile with the most gifts. By storing negative values in Python's min heap, we simulate a max heap. In each of the k iterations, we extract the maximum value, compute its square root, floor it, and push it back into the heap. After k operations, we sum all remaining values to get the final answer.

Intuition

The key insight is that we need to repeatedly find and modify the maximum element in the array. This immediately suggests using a data structure that efficiently supports finding the maximum element - a max heap is perfect for this.

Why a heap? Consider the alternatives:

• If we use a simple array and search for the maximum each time, we'd spend O(n) time per operation just to find the max, resulting in O(k*n) total time.
• If we sort the array initially, the first maximum is easy to find, but after taking its square root and reinserting it, we'd need to maintain the sorted order, which is inefficient.

A max heap gives us O(log n) for both extracting the maximum and inserting the modified value back. Since we're always operating on the largest value and reducing it (square root always reduces values greater than 1), we're essentially "flattening" the distribution of gifts across piles.

The square root operation has an interesting property - it reduces larger numbers more aggressively than smaller ones. For instance, √100 = 10 (reduced by 90), while √9 = 3 (reduced by only 6). This means after several operations, the values tend to converge toward smaller numbers, and the differences between piles become less pronounced.

In Python, since the built-in heapq module only provides a min heap, we use a clever trick: store negative values. This way, the smallest (most negative) value in our min heap corresponds to the largest positive value in the original array. The heapreplace operation is particularly efficient as it combines pop and push in a single operation, maintaining the heap property with just one percolation.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The solution implements a max heap approach to efficiently handle the repeated operations of finding and modifying the maximum element.

Step 1: Initialize the Max Heap

h = [-v for v in gifts]
heapify(h)

Since Python's heapq module only provides a min heap, we store negative values to simulate a max heap. We negate all values in the gifts array and then call heapify() to build the heap in O(n) time. Now the smallest (most negative) element in our heap represents the largest positive value.

Step 2: Perform k Operations

for _ in range(k):
    heapreplace(h, -int(sqrt(-h[0])))

For each of the k seconds:

• h[0] gives us the minimum element (most negative), which corresponds to the maximum positive value
• -h[0] converts it back to the positive value
• sqrt(-h[0]) computes the square root of this maximum value
• int(sqrt(-h[0])) applies the floor operation (since int() truncates decimals for positive numbers)
• We negate the result again with -int(sqrt(-h[0])) to maintain our negative value convention
• heapreplace() efficiently replaces the top element with the new value and maintains the heap property in O(log n) time

The heapreplace() function is particularly efficient because it combines extracting the minimum and inserting a new value in a single operation with only one percolation through the heap.

Step 3: Calculate the Final Sum

return -sum(h)

After k operations, we sum all elements in the heap. Since our heap contains negative values, we negate the sum to get the positive total of remaining gifts.

Time Complexity: O(n + k log n) where n is the length of the gifts array. Building the heap takes O(n), and we perform k operations, each taking O(log n).

Space Complexity: O(n) for storing the heap.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

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

Step 1: Initialize the Max Heap

First, we negate all values and build a heap:

• Original: [25, 64, 9, 4, 100]
• Negated: [-25, -64, -9, -4, -100]
• After heapify: [-100, -64, -25, -4, -9] (min heap structure)

The heap visually looks like:

       -100
      /     \
   -64      -25
   /  \
 -4   -9

Step 2: Perform k=4 Operations

Operation 1:

• Current max (top of heap): -100 → actual value: 100
• Calculate: floor(√100) = 10
• Replace with: -10
• Heap after heapreplace: [-64, -10, -25, -4, -9]
• Actual values: [64, 10, 25, 4, 9]

Operation 2:

• Current max: -64 → actual value: 64
• Calculate: floor(√64) = 8
• Replace with: -8
• Heap after heapreplace: [-25, -10, -8, -4, -9]
• Actual values: [25, 10, 8, 4, 9]

Operation 3:

• Current max: -25 → actual value: 25
• Calculate: floor(√25) = 5
• Replace with: -5
• Heap after heapreplace: [-10, -9, -8, -4, -5]
• Actual values: [10, 9, 8, 4, 5]

Operation 4:

• Current max: -10 → actual value: 10
• Calculate: floor(√10) = 3
• Replace with: -3
• Heap after heapreplace: [-9, -5, -8, -4, -3]
• Actual values: [9, 5, 8, 4, 3]

Step 3: Calculate Final Sum

• Heap contains: [-9, -5, -8, -4, -3]
• Sum of heap: -29
• Return: -(-29) = 29

Final Answer: 29

Notice how the max heap efficiently maintains access to the largest element at each step, and how the square root operation progressively "flattens" the distribution - the initial range was 4 to 100 (difference of 96), but after 4 operations, the range is 3 to 9 (difference of only 6).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + k × log n)

• Converting the list to a negated heap: O(n) - The list comprehension takes O(n) time to negate all values
• Building the heap with heapify(): O(n) - This operation transforms the list into a heap structure in linear time
• Loop iterations: The loop runs k times, and each iteration performs:
  • heapreplace(): O(log n) - This operation pops the minimum element and pushes a new element in one operation, maintaining the heap property
  • Square root calculation: O(1) - Computing sqrt() is constant time
  • Total for loop: k × O(log n) = O(k × log n)
• Final sum calculation: O(n) - Summing all elements in the heap

Overall: O(n) + O(n) + O(k × log n) + O(n) = O(n + k × log n)

Space Complexity: O(n)

• The heap h stores n elements (negated values from the original gifts list): O(n)
• No additional data structures are created that scale with input size
• The heapify() operation modifies the list in-place and doesn't require extra space beyond the list itself

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Early Termination When All Piles Become 1

A common pitfall is not recognizing that once a pile has 1 gift, applying the square root operation keeps it at 1 (since floor(√1) = 1). If all piles reach 1 before k operations are complete, continuing the operations is wasteful.

Problem Example:

gifts = [1, 2, 3], k = 10

After just 2 operations, all piles become 1, but the naive solution still performs 8 more unnecessary operations.

Optimized Solution:

def pickGifts(self, gifts: List[int], k: int) -> int:
    max_heap = [-gift for gift in gifts]
    heapify(max_heap)
  
    for _ in range(k):
        # Early termination: if max value is 1, no point continuing
        if -max_heap[0] == 1:
            break
        max_value = -max_heap[0]
        new_value = int(sqrt(max_value))
        heapreplace(max_heap, -new_value)
  
    return -sum(max_heap)

2. Integer Overflow in Other Languages

While Python handles arbitrary precision integers automatically, implementing this in languages like Java or C++ requires careful consideration of integer overflow when dealing with large gift values.

Problem: If gifts[i] can be up to 10^9, you need to ensure your data types can handle these values and their transformations.

Solution for Static-Typed Languages:

• Use appropriate data types (e.g., long long in C++, long in Java)
• Be mindful of the sum calculation at the end which could overflow

3. Incorrect Floor Implementation

Using int() for flooring only works correctly for positive numbers. If the problem were modified to allow negative values, int() would round toward zero, not floor.

Example of the Issue:

import math
# For positive numbers, int() and floor() give same result
print(int(sqrt(10)))    # 3
print(math.floor(sqrt(10)))  # 3

# But for negative numbers (if problem allowed):
print(int(-3.7))        # -3 (rounds toward zero)
print(math.floor(-3.7)) # -4 (actual floor)

Robust Solution:

from math import floor, sqrt

# Use floor() explicitly for clarity and correctness
new_value = floor(sqrt(max_value))

4. Unnecessary Heap Operations When k > Useful Operations

When k is very large relative to the gift values, many operations become redundant once most piles stabilize at small values.

Optimization with Operation Counting:

def pickGifts(self, gifts: List[int], k: int) -> int:
    max_heap = [-gift for gift in gifts]
    heapify(max_heap)
  
    operations = 0
    for _ in range(k):
        max_value = -max_heap[0]
        new_value = int(sqrt(max_value))
      
        # If the value doesn't change, we've reached a stable state
        if new_value == max_value:
            break
          
        heapreplace(max_heap, -new_value)
        operations += 1
  
    return -sum(max_heap)