Brute Force Approach for Take Gifts From the Richest Pile

The naive strategy is to simulate the process exactly as described using a standard array or list.

1. Iterate $k$ times.
2. In each iteration, scan the entire array to find the index of the maximum value.
3. Replace the value at that index with the floor of its square root.
4. After $k$ iterations, sum the array.

Pseudo-code:

text
function bruteForce(gifts, k):
    repeat k times:
        max_val = -1
        max_idx = -1
        for i from 0 to length(gifts):
            if gifts[i] > max_val:
                max_val = gifts[i]
                max_idx = i
        gifts[max_idx] = floor(sqrt(max_val))
    return sum(gifts)

Time Complexity: $O(k \cdot N)$, where $N$ is the number of piles. For every second $k$, we iterate through all $N$ elements. Why it fails: While the constraints for LeetCode 2558 are small enough ($N \le 10^3, k \le 10^3$) that this might pass, it is inefficient. If $N$ and $k$ were larger (e.g., $10^5$), the $O(k \cdot N)$ approach would result in a Time Limit Exceeded (TLE). It fails to utilize the property that we only care about the single maximum element at any given time.

Core Insight for Take Gifts From the Richest Pile

The core inefficiency in the brute force approach is the linear scan to find the maximum. We need a data structure that allows us to:

1. Access the maximum element in $O(1)$ time.
2. Remove the maximum and insert a new value (the square root) in logarithmic time, $O(\log N)$.

A Max-Heap (or Priority Queue) is the ideal data structure for this requirement. The heap invariant guarantees that the root node always contains the maximum value.

Visual Description: Imagine the data organized as a binary tree where every parent node is greater than or equal to its children. The root of this tree is the "richest pile."

1. Extraction: When we select the richest pile, we remove the root.
2. Restoration: To maintain the complete binary tree structure, the last leaf node is moved to the root position.
3. Sifting Down: This new root (likely a small value) violates the heap property. It is swapped with its larger child repeatedly until it finds its correct position. This "bubbling down" or "sifting" ensures the new maximum bubbles up to the root efficiently.

This allows us to perform the simulation in $O(k \log N)$ time, which is significantly faster than repeatedly sorting or scanning the array.

Execution Flow

Let's trace the algorithm with gifts = [25, 64, 9], k = 2.

1. Build Heap:

   • Input: [25, 64, 9]
   • Max-Heap Structure: Root is 64. Children are 25 and 9.

2. Iteration 1 (k=1):

   • Pop Max: Remove 64. Heap becomes [25, 9].
   • Calculate: $\lfloor\sqrt{64}\rfloor = 8$.
   • Push: Insert 8.
   • Rebalance: 25 is now the largest. Heap structure roughly: Root 25, children 8, 9.

3. Iteration 2 (k=2):

   • Pop Max: Remove 25. Heap becomes [9, 8].
   • Calculate: $\lfloor\sqrt{25}\rfloor = 5$.
   • Insert .

4. Final Sum:

   • Remaining elements: 9, 8, 5.
   • Total: $9 + 8 + 5 = 22$.

Proof of Correctness

The problem requires a greedy strategy: at every step, we must choose the pile with the maximum number of gifts. A Max-Heap enforces the invariant that the root is always the global maximum of the current dataset. By extracting the root, modifying it, and re-inserting it, we strictly follow the simulation rules. Since the heap reorders itself after every insertion ($O(\log N)$), the next extraction is guaranteed to be the current maximum, ensuring the correctness of the greedy choice.

Reference Implementation

C++ Solution for LeetCode 2558

cpp
#include <vector>
#include <queue>
#include <cmath>
#include <numeric>

class Solution {
public:
    long long pickGifts(std::vector<int>& gifts, int k) {
        // Use a Max-Heap (priority_queue defaults to max-heap in C++)
        std::priority_queue<int> pq(gifts.begin(), gifts.end());
        
        // Perform the operation k times
        while (k > 0) {
            // Get the richest pile
            int max_gift = pq.top();
            pq.pop();
            
            // Calculate remaining gifts after taking sqrt
            int remaining = std::floor(std::sqrt(max_gift));
            
            // Push the reduced pile back into the heap
            pq.push(remaining);
            
            k--;
        }
        
        // Calculate the total sum of remaining gifts
        long long total_gifts = 0;
        while (!pq.empty()) {
            total_gifts += pq.top();
            pq.pop();
        }
        
        return total_gifts;
    }
};

Java Solution for LeetCode 2558

java
import java.util.PriorityQueue;
import java.util.Collections;

class Solution {
    public long pickGifts(int[] gifts, int k) {
        // Use a PriorityQueue with reverseOrder to simulate a Max-Heap
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        
        // Add all gifts to the heap
        for (int gift : gifts) {
            pq.offer(gift);
        }
        
        // Perform the operation k times
        while (k > 0) {
            if (pq.isEmpty()) break;
            
            int maxGift = pq.poll();
            // Calculate sqrt and cast to integer (floor)
            int remaining = (int) Math.sqrt(maxGift);
            pq.offer(remaining);
            
            k--;
        }
        
        // Calculate sum
        long totalGifts = 0;
        while (!pq.isEmpty()) {
            totalGifts += pq.poll();
        }
        
        return totalGifts;
    }
}

Python Solution for LeetCode 2558

python
import heapq
import math

class Solution:
    def pickGifts(self, gifts: list[int], k: int) -> int:
        # Python's heapq is a Min-Heap. 
        # We negate values to simulate a Max-Heap.
        max_heap = [-g for g in gifts]
        heapq.heapify(max_heap)
        
        for _ in range(k):
            # Pop the smallest value (which is the negated largest)
            max_val_neg = heapq.heappop(max_heap)
            
            # Convert back to positive to calculate sqrt
            max_val = -max_val_neg
            remaining = math.isqrt(max_val) # Integer square root
            
            # Push the negated result back
            heapq.heappush(max_heap, -remaining)
            
        # Sum the values (remember to negate back to positive)
        return -sum(max_heap)