1962. Remove Stones to Minimize the Total LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Remove Stones to Minimize the Total solution in C++
4. Remove Stones to Minimize the Total solution in Java
5. Remove Stones to Minimize the Total solution in Python
6. Additional Resources

Problem Statement of Remove Stones to Minimize the Total

You are given a 0-indexed integer array piles, where piles[i] represents the number of stones in the ith pile, and an integer k. You should apply the following operation exactly k times:

Choose any piles[i] and remove floor(piles[i] / 2) stones from it.

Notice that you can apply the operation on the same pile more than once.
Return the minimum possible total number of stones remaining after applying the k operations.
floor(x) is the greatest integer that is smaller than or equal to x (i.e., rounds x down).

Example 1:

Input: piles = [5,4,9], k = 2
Output: 12
Explanation: Steps of a possible scenario are:
– Apply the operation on pile 2. The resulting piles are [5,4,5].
– Apply the operation on pile 0. The resulting piles are [3,4,5].
The total number of stones in [3,4,5] is 12.

Example 2:

Input: piles = [4,3,6,7], k = 3
Output: 12
Explanation: Steps of a possible scenario are:
– Apply the operation on pile 2. The resulting piles are [4,3,3,7].
– Apply the operation on pile 3. The resulting piles are [4,3,3,4].
– Apply the operation on pile 0. The resulting piles are [2,3,3,4].
The total number of stones in [2,3,3,4] is 12.

Constraints:

1 <= piles.length <= 105
1 <= piles[i] <= 104
1 <= k <= 105

Complexity Analysis

• Time Complexity: O(k\log n)
• Space Complexity: O(n)

1962. Remove Stones to Minimize the Total LeetCode Solution in C++

class Solution {
 public:
  int minStoneSum(vector<int>& piles, int k) {
    int ans = accumulate(piles.begin(), piles.end(), 0);
    priority_queue<int> maxHeap;

    for (const int pile : piles)
      maxHeap.push(pile);

    for (int i = 0; i < k; ++i) {
      const int maxPile = maxHeap.top();
      maxHeap.pop();
      maxHeap.push(maxPile - maxPile / 2);
      ans -= maxPile / 2;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1962. Remove Stones to Minimize the Total LeetCode Solution in Java

class Solution {
  public int minStoneSum(int[] piles, int k) {
    int ans = Arrays.stream(piles).sum();
    Queue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());

    for (final int pile : piles)
      maxHeap.offer(pile);

    for (int i = 0; i < k; ++i) {
      final int maxPile = maxHeap.poll();
      maxHeap.offer(maxPile - maxPile / 2);
      ans -= maxPile / 2;
    }

    return ans;
  }
}
// code provided by PROGIEZ

1962. Remove Stones to Minimize the Total LeetCode Solution in Python

class Solution:
  def minStoneSum(self, piles: list[int], k: int) -> int:
    maxHeap = [-pile for pile in piles]
    heapq.heapify(maxHeap)

    for _ in range(k):
      heapq.heapreplace(maxHeap, maxHeap[0] // 2)

    return -sum(maxHeap)
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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