You managed to send your friend to queue for tickets in your stead, but there is a catch: he will get there only if you tell him how much that is going to take. And everybody can only take one ticket at a time, then they go back in the last position of the queue if they need more (or go home if they are fine).

Each ticket takes one minutes to emit, the queue is well disciplined, [Brit-style](https://www.codewars.com/kata/english-beggars), and so it moves smoothly, with no waste of time.

You will be given an array/list/vector with all the people queuing and the *initial* position of your buddy, so for example, knowing that your friend is in the third position (that we will consider equal to the index, `2`: he is the guy that wants 3 tickets!) and the initial queue is `[2, 5, 3, 4, 6]`.

The first dude gets his ticket and the queue goes now like this `[5, 3, 4, 6, 1]`, then `[3, 4, 6, 1, 4]` and so on. In the end, our buddy will be queuing for 12 minutes, true story!

Build a function to compute it, resting assured that only positive integers are going to be there and you will be always given a valid index; but we also want to go to pretty popular events, so be ready for big queues with people getting plenty of tickets.

[[hard core version](https://www.codewars.com/kata/queue-time-counter-hard-core-version/solutions/javascript) now available if you don't want the "easy" kata!]

Difficulty: introductory

Starter code:

def queue(queuers,pos):
	

Solutions:
def queue(queuers,pos):
    return sum(min(queuer, queuers[pos] - (place > pos)) for place, queuer in enumerate(queuers))

def queue(queuers,pos):
    friendWait = queuers[pos]
    # Divide the line into the front of the line (up to the friend)
    # and back of the line (behind the friend):
    frontOfLine = queuers[:pos+1]
    backOfLine = queuers[pos+1:]
    # Convert the frontOfLine to the min of friendWait:
    frontOfLine = [min(x, friendWait) for x in frontOfLine]
    # Convert the backOfLine to the min of friendWait-1:
    backOfLine = [min(x, friendWait-1) for x in backOfLine]
    # Return the result, which is the sum of both line parts:
    return sum(frontOfLine) + sum(backOfLine)


def queue(queuers,pos):
    return sum(min(q, queuers[pos]) for q in queuers) - sum(1 for q in queuers[pos + 1:] if q >= queuers[pos])


def queue(q,pos):
    time = 0
    while True:
        time += 1
        if q[0] == 1:
            if pos: q.pop(0)
            else:   return time
        else:
            q.append(q.pop(0) - 1)
        pos = pos - 1 if pos  else len(q) - 1


def queue(q, pos):
    return (
        sum(min(x, q[pos]) for x in q[:pos+1]) + 
        sum(min(x, q[pos]-1) for x in q[pos+1:])
    )

def queue(queuers, pos, time = 0):
    while True:
        for i in range(len(queuers)):
            if queuers[pos] == 0:
                return time
            elif queuers[i]:
                queuers[i] -= 1
                time += 1

def queue(q, pos, r = 0):
    while True:
        r += 1
        q[0] -= 1
        if q[pos] == 0: return r
        q = q[1:] + [q[0]] if q[0] > 0 else q[1:]

        pos -= 1
        if pos<0: pos = len(q) - 1

def queue(queuers,pos):
    count = 0
    while len(queuers) != 0:    #배열이 빌 때까지 반복
        k = queuers.pop(0)    #first-in을 pop
        count += 1    #걸리는 시간을 체크
        if k-1 != 0:    #하나 줄여도 0이 아니라면
            queuers.append(k-1)    #다시 배열의 끝에 첨가
            if pos == 0:    #이 때, 우리가 체크하는 index가 0이었다면
                pos = len(queuers) - 1    #배열의 끝 index로 수정
            else:
                pos -= 1
        elif k-1 == 0 and pos == 0:    #우리의 원소가 0되면 반복문 나감
            break
        else:
            pos -= 1    #다른 원소가 0되면 그냥 배열에서 삭제
    return count

def queue(queuers,pos):
    count = 0
    while queuers[pos] > 0:
        if pos == 0:
            pos = len(queuers) - 1
        else:
            pos -= 1
        queuers[0] -= 1
        queuers.append(queuers[0])
        queuers.pop(0)
        count += 1
    
    return count + sum([i for i in queuers if i < 0])

def queue(queuers,pos):
    min = 0
    while queuers[pos] > 0:
        for i in range(0,len(queuers)):
            if queuers[i] > 0:
                queuers[i]-=1
                min += 1
            if queuers[pos] == 0:
                break
                
    return min