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3066. Minimum Operations to Exceed Threshold Value II 👍

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

    for (const int num : nums)
      minHeap.push(num);

    while (minHeap.size() > 1 && minHeap.top() < k) {
      const int x = minHeap.top();
      minHeap.pop();
      const int y = minHeap.top();
      minHeap.pop();
      minHeap.push(min(x, y) * 2L + max(x, y));
      ++ans;
    }

    return ans;
  }
};

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class Solution {
  public int minOperations(int[] nums, int k) {
    int ans = 0;
    Queue<Long> minHeap = new PriorityQueue<>();

    for (final int num : nums)
      minHeap.add((long) num);

    while (minHeap.size() > 1 && minHeap.peek() < k) {
      final long x = minHeap.poll();
      final long y = minHeap.poll();
      minHeap.add(Math.min(x, y) * 2 + Math.max(x, y));
      ++ans;
    }

    return ans;
  }
}

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class Solution:
  def minOperations(self, nums: list[int], k: int) -> int:
    ans = 0
    minHeap = nums.copy()
    heapq.heapify(minHeap)

    while len(minHeap) > 1 and minHeap[0] < k:
      x = heapq.heappop(minHeap)
      y = heapq.heappop(minHeap)
      heapq.heappush(minHeap, min(x, y) * 2 + max(x, y))
      ans += 1

    return ans