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You have a queue of customers at a ticket counter represented by an array tickets, where tickets[i] is the number of tickets that the i-th customer wants to buy. Each customer can buy only one ticket at a time, and they must wait until the current transaction is complete before proceeding to the next customer in a circular fashion from the start of the queue.

Return the number of seconds needed to complete the ticket buying process for the customer located at position k.

Clarifying Questions

1. Is the number of tickets each customer can buy restricted to positive integers?
   • Yes, each customer can buy a positive integer number of tickets.
2. Does the queue form a strictly circular pattern, meaning after the last customer the process continues with the first customer again?
   • Yes, the queue follows a circular pattern. When the last customer has bought a ticket, the process continues with the first customer again.
3. What are the constraints on the length of the tickets array and the values it can contain?
   • Typical constraints for such problems:
     • 1 <= tickets.length <= 1000
     • 1 <= tickets[i] <= 1000
     • 0 <= k < tickets.length

Strategy

1. Start the clock at 0 seconds.
2. Simulate the process where each customer in the queue buys one ticket at a time.
3. Use a loop to iterate over the customers circularly:
   • Decrease the ticket count for the current customer by one.
   • Increase the time counter by one.
   • If the customer at position k gets their last ticket, stop the process and return the current time.
4. Continue until the k-th customer’s ticket count reaches 0.

Time Complexity

• The time complexity of this solution is O(n * m), where n is the length of the tickets array and m is the maximum number of tickets any customer wants to buy. This is because, in the worst case, each customer buys a maximum of m tickets, and we check every customer in each iteration.

Code

def timeRequiredToBuy(tickets, k):
    time = 0
    
    while tickets[k] > 0:
        for i in range(len(tickets)):
            if tickets[i] > 0:  # If the customer still has tickets to buy
                time += 1
                tickets[i] -= 1
            
            if tickets[k] == 0:  # Check if the k-th customer has finished buying
                return time

# Example usage
tickets = [2, 3, 2]
k = 2
print(timeRequiredToBuy(tickets, k))  # Outputs: 6

The above code tracks the total time required for the k-th customer to buy all of their tickets. Each iteration of the loop simulates one customer buying a single ticket in a circular fashion until the process completes for the k-th customer.

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