programmers lv3 : Connecting islands commenting

def solution(n, costs):
    answer = 0
    V = set()
    for v1, v2, cost in costs:
        V.add(v1)
        V.add(v2)
    # collectiong total index and number of islands without duplication    # used as unvisited set
    sortedCosts = sorted(costs, key = lambda x: x[2])
    # make a sorted list, key : node connection, value : cost    # it keeps the sequence of each elem in inner list,    # but each inner list is sorted by lower cost    # means this sort meet the minial cost condition    # x:x[2] means, key value to sort by is x[2] item in inner list    visited = set()

    visited.add(V.pop())
    # put the first node index and thats it    while V:
        for i in range(len(sortedCosts)):
            v1, v2, cost = sortedCosts[i]
            # how to spead and assign multiple items in one line            if v1 in visited and v2 in visited:
                # means same connection with lower cost has already been confirmed                # if the connection with lower cost has already confirmed                sortedCosts.pop(i)
                break            elif v1 in visited or v2 in visited:
                # add a new island, unvisited island into the graph                #   remove the one unvisited island from unvisited set, V                if v1 in V:
                    V.remove(v1)
                if v2 in V:
                    V.remove(v2)

                visited.add(v1)
                visited.add(v2)
                # visited is a set, we dont know which one was unvisited                # so we simply add two and only unvisited node will be added safely
                answer += cost
                # this tuple is counted as confirmed cost, add it                sortedCosts.pop(i)
                # this tuple has processed, remove it                # since this is whiled until there will be no node unvisited (V)                # and start again from new index and length of sortedCosts                break
        # I think reduced index of enumerate object(popping sortedCosts) might cause indexing error later        # that is why each loop action does break and start again        # until all islands are connected    return answer

costs = [[0,1,1],[0,2,2],[1,2,5],[1,3,1],[2,3,8]]
n = 4solution(4, costs)