Minimum Cost to Connect Sticks Python Solution

lcpy gen -n 1167   # by problem number
lcpy gen -s minimum_cost_to_connect_sticks   # by problem name

Input: sticks = [2,4,3]
Output: 14
Explanation: You start with sticks = [2,4,3].
1. Combine sticks 2 and 3 for a cost of 2 + 3 = 5. Now you have sticks = [5,4].
2. Combine sticks 5 and 4 for a cost of 5 + 4 = 9. Now you have sticks = [9].
There is only one stick left, so you are done. The total cost is 5 + 9 = 14.

Input: sticks = [1,8,3,5]
Output: 30
Explanation: You start with sticks = [1,8,3,5].
1. Combine sticks 1 and 3 for a cost of 1 + 3 = 4. Now you have sticks = [4,8,5].
2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have sticks = [9,8].
3. Combine sticks 9 and 8 for a cost of 9 + 8 = 17. Now you have sticks = [17].
There is only one stick left, so you are done. The total cost is 4 + 9 + 17 = 30.

Input: sticks = [5]
Output: 0
Explanation: There is only one stick, so you don't need to do anything. The total cost is 0.

import heapq


class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def connect_sticks(self, sticks: list[int]) -> int:
        heap = list(sticks)
        heapq.heapify(heap)
        cost = 0
        while len(heap) > 1:
            merged = heapq.heappop(heap) + heapq.heappop(heap)
            cost += merged
            heapq.heappush(heap, merged)
        return cost