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 `i`

stick.^{th}

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:14Explanation: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.

**Example 2:**

Input:sticks = [1,8,3,5]Output:30Explanation: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.

**Example 3:**

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

**Constraints:**

`1 <= sticks.length <= 10`

^{4}`1 <= sticks[i] <= 10`

^{4}

Solution1:

```
import heapq
class Solution(object):
def connectSticks(self, sticks):
"""
:type sticks: List[int]
:rtype: int
"""
heapq.heapify(sticks)
total = 0
while len(sticks) > 1:
stick0 = heapq.heappop(sticks)
stick1 = heapq.heappop(sticks)
cost = stick0 + stick1
total += cost
heapq.heappush(sticks, cost)
return total
```

Solution2:

### Explanation

- We need to combine everything anyway, thus always sticks currently 2 smallest ones
- Best data structure for this purpose is
`heap`

### Implementation

```
class Solution:
def connectSticks(self, sticks: List[int]) -> int:
n, heap = len(sticks), sticks
heapq.heapify(heap)
ans = 0
for i in range(n-1):
s = heapq.heappop(heap) + heapq.heappop(heap)
heapq.heappush(heap, s)
ans += s
return ans
```

A really cool feature of python heap is `heapq.heapreplace`

. It’s more efficient than `heapq.heappop + heapq.heappush`

, below is an example