# 1167 Minimum Cost to Connect Sticks

You have some number of sticks with positive integer lengths. These lengths are given as an array `sticks`, where `sticks[i]` is the length of the `ith` stick.

You can connect any two sticks of lengths `x` and `y` into one stick by paying a cost of `x + y`. You must connect all the sticks until there is only one stick remaining.

Return the minimum cost of connecting all the given sticks into one stick in this way.

Example 1:

``````Input: sticks = [2,4,3]
Output: 14
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 = .
There is only one stick left, so you are done. The total cost is 5 + 9 = 14.
``````

Example 2:

``````Input: sticks = [1,8,3,5]
Output: 30
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 = .
There is only one stick left, so you are done. The total cost is 4 + 9 + 17 = 30.
``````

Example 3:

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

Constraints:

• `1 <= sticks.length <= 104`
• `1 <= sticks[i] <= 104`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 `````` ``````import heapq class Solution: def connectSticks(self, sticks: List[int]) -> int: heap = [] for s in sticks: heapq.heappush(heap, s) cost = 0 while len(heap) > 1: shortest = heapq.heappop(heap) second_shortest = heapq.heappop(heap) length = shortest + second_shortest heapq.heappush(heap, length) cost += length return cost``````