#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.


class Solution:
    def connectSticks(self, sticks: List[int]) -> int:        
        
        ## minheap : time limit exceeded
        out=0
        heapq.heapify(sticks)
        while len(sticks)>1:
            
            a1, a2 = heapq.heappop(sticks), heapq.heappop(sticks)
            heapq.heappush(sticks,a1+a2)
            out+=a1+a2
        return out
        
