How do you solve Time Needed to Buy Tickets on LeetCode?

Time Needed to Buy Tickets (LeetCode #2073) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The number of tickets a person has affects the total time taken.

What is the time complexity of Time Needed to Buy Tickets?

The optimal solution for Time Needed to Buy Tickets runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only need to iterate through the list of tickets once to calculate the total time.

What is the brute force solution for Time Needed to Buy Tickets?

The brute force approach for Time Needed to Buy Tickets is Brute Force, with O(n²) time complexity and O(1) space complexity. We simulate the ticket buying process by iterating through the queue repeatedly until the kth person has bought all their tickets. This approach is straightforward but inefficient as it may require multiple passes through the queue. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Time Needed to Buy Tickets?

Time Needed to Buy Tickets uses the following concepts: Array, Queue, Simulation. The recommended patterns to study are: Simulation, Queue.

What are the interview tips for Time Needed to Buy Tickets?

Start with a brute force approach to demonstrate understanding. Optimize your solution by identifying patterns in the problem. Practice explaining your thought process clearly during the interview.

What are common mistakes when solving Time Needed to Buy Tickets?

Not considering the people in front of the kth person. Overcomplicating the simulation process.

#2073

Time Needed to Buy Tickets

Easy

ArrayQueueSimulationSimulationQueue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of simulating every ticket purchase, we can calculate the time directly based on the number of tickets the kth person wants and the number of people in front of them. This approach is more efficient and avoids unnecessary iterations.

⚙️

Algorithm

3 steps

1Step 1: Count the total time taken by the people in front of the kth person who still have tickets.
2Step 2: Add the tickets the kth person wants to the total time.
3Step 3: Return the total time.

solution.py6 lines

1def timeNeededToBuy(tickets, k):
2    time = 0
3    for i in range(len(tickets)):
4        if tickets[i] > 0:
5            time += min(tickets[i], tickets[k])
6    return time + tickets[k]

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the list of tickets once to calculate the total time.

1The number of tickets a person has affects the total time taken.
2People in front of the kth person can delay their ticket buying.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.