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1167. Minimum Cost to Connect Sticks 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int connectSticks(vector<int>& sticks) {
    int ans = 0;
    priority_queue<int, vector<int>, greater<>> minHeap;

    for (const int stick : sticks)
      minHeap.push(stick);

    while (minHeap.size() > 1) {
      const int x = minHeap.top();
      minHeap.pop();
      const int y = minHeap.top();
      minHeap.pop();
      ans += x + y;
      minHeap.push(x + y);
    }

    return ans;
  }
};

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class Solution {
  public int connectSticks(int[] sticks) {
    int ans = 0;
    Queue<Integer> minHeap = new PriorityQueue<>();

    for (final int stick : sticks)
      minHeap.offer(stick);

    while (minHeap.size() > 1) {
      final int x = minHeap.poll();
      final int y = minHeap.poll();
      ans += x + y;
      minHeap.offer(x + y);
    }

    return ans;
  }
}

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class Solution:
  def connectSticks(self, sticks: list[int]) -> int:
    ans = 0
    heapq.heapify(sticks)

    while len(sticks) > 1:
      x = heapq.heappop(sticks)
      y = heapq.heappop(sticks)
      ans += x + y
      heapq.heappush(sticks, x + y)

    return ans