Efficient waiting time computation for tickets challenge

waiting_time.rb

def waitingTime(tickets, p) 
  
    # if you only want 1 ticket just count number of people in front, 
    return p + 1 if tickets[p] == 1 
      
    wait = 0
    multiplier = 1
    for i in 0...tickets.length do
      next if i == p
      
      # if the person eventually drops off, they will only add the
      # number of tickets they are buying
      if tickets[i] < tickets[p]
        wait += tickets[i]
      elsif tickets[i] >= tickets[p]
        # person who wants more tickets is in back, so on the final
        # run he won't add any time
        if i > p
          wait -= 1
        end
        # Since the person is always in the queue, he adds 1 for each
        # time through the queue, in this case tickets[p] times
        multiplier += 1
      end
    end
    
    # total wait
    wait + tickets[p] * multiplier
end

# Me figuring out the algorithm...
puts waitingTime([1, 2, 3, 7, 5], 4)  == 16  # 5*2 + 6
puts waitingTime([5, 1, 2, 3, 7], 0)  == 15  # 5*2 - 1 + 6
puts waitingTime([5, 1, 2, 3, 7], 1)  == 2   # 2
puts waitingTime([5, 10, 2, 3, 7], 1) == 27 # 10 + 5 + 5 + 7 = 27
puts waitingTime([5, 10, 2, 3, 7], 3) == 13 # 3*4 - 1 + 2
puts waitingTime([5, 10, 2, 3, 7], 4) == 24 # 7*2 + 10
puts waitingTime([5, 10, 2, 3, 10], 4) == 30  # 7*2 + 10

# testing different cases...
puts waitingTime([10], 0) == 10 
puts waitingTime([1], 0) == 1
puts waitingTime([1, 2, 3, 4, 10], 4) == 20
puts waitingTime([1, 2, 3, 4, 1], 4) == 5
puts waitingTime([2, 2, 2, 2, 2], 4) == 10
puts waitingTime([3, 3, 3, 3, 2], 4) == 10
puts waitingTime([3, 3, 2, 3, 3], 2) == 8