Solution: Connecting n Pipes With Minimum Cost

import heapq
def min_cost(pipes):
    """
    Calculates the minimum cost of connecting pipes
    :param pipes: A list where its length is the number of pipes and indexes are the specific lengths of the pipes.
    :return: The minimum cost
    """
    
    # Create a priority queue out of the
    # given list
    heapq.heapify(pipes)
     
    # Initializ result
    cost = 0
     
    # While size of priority queue
    # is more than 1
    while(len(pipes) > 1):
         
        # Extract shortest two pipes from the priority queue
        first = heapq.heappop(pipes)
        second = heapq.heappop(pipes)
         
        # Connect the pipe: update cost
        # and insert the new pipe to pipes queue
        cost += first + second
        heapq.heappush(pipes, first + second)
         
    return cost
# Main program to test above function
if __name__ == "__main__":
    pipes = [4, 3, 2, 6]
    print(min_cost(pipes))