# 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
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.

Example 2:

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.

Example 3:

Input: sticks = [5]
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

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

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