2558. Take Gifts From the Richest Pile - Detailed Explanation

Problem Statement

You are given an array of integers, where each element represents a pile of gifts. You are allowed to perform exactly k operations. In each operation, you must choose the pile with the maximum number of gifts and then replace that pile’s count with the floor value of half of its current count (i.e., update that pile to ⌊gifts[i] / 2⌋). Your goal is to determine the total number of gifts remaining after performing the k operations.

Examples

Example 1

• Input: gifts = [8, 4, 6, 6], k = 3
• Explanation:
  • Operation 1: The richest pile is 8. Replace 8 with ⌊8/2⌋ = 4. New gifts array: [4, 4, 6, 6].
  • Operation 2: The richest pile is 6. Replace one of the 6’s with ⌊6/2⌋ = 3. New gifts array: [4, 4, 3, 6].
  • Operation 3: The richest pile is now 6 again. Replace it with ⌊6/2⌋ = 3. New gifts array: [4, 4, 3, 3].
• Output: The sum of the gifts is 4 + 4 + 3 + 3 = 14.

Example 2

• Input: gifts = [10, 5, 8], k = 4
• Explanation:
  Follow four operations, always choosing the pile with the largest count, reducing it to half (using floor division) each time. The order of operations will determine the intermediate states, and after 4 operations the sum of the remaining gifts is returned.
• Output: (The final sum after processing 4 operations.)

Hints

1. Use a Max Heap:
   Since you always need to choose the richest (largest) pile, a max-heap (or priority queue) is the ideal data structure for efficiently retrieving and updating the maximum element.

2. Simulation Through k Operations:
   Instead of trying to find a shortcut by mathematical manipulation, simulate each operation:

   • Pop the maximum element from the heap.
   • Calculate the new value as the floor division of that number by 2.
   • Push the new value back into the heap.
   • Repeat for k iterations.

Approaches

Approach 1: Greedy Simulation with a Max Heap

Explanation

1. Initialize a Max Heap:
   Build a max heap containing all the piles from the input array. This structure allows for efficient retrieval of the largest element.

2. Perform k Operations:
   For each operation:

   • Extract the maximum element from the heap.
   • Compute the new value using floor division by 2.
   • Insert this updated value back into the heap. This guarantees that in each step you are reducing the richest pile.

3. Compute the Final Sum:
   After k operations, sum up all elements remaining in the heap to obtain the final number of gifts.

Complexity Analysis

• Time Complexity:
  Building the initial heap takes O(n). Each operation (pop and push) takes O(log n), leading to a total of O(k log n) for k operations. Thus, the overall time complexity is O(n + k log n).

• Space Complexity:
  The max heap uses O(n) space.

Approach 2: (If k Is Small) Direct Iteration Without Advanced Data Structures

While using a max heap is optimal for large k and n, if the constraints were much smaller, one might consider simply iterating and scanning for the maximum element each time. However, this approach would lead to a time complexity of O(k · n) and is generally less efficient.

Code Solutions

Python Implementation

Java Implementation