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

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Time Needed to Buy Tickets?

This problem uses Array, Queue, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Time Needed to Buy Tickets?

Insight 1: The key is to realize that the order of people matters, and people who arrive *after* person `k` in the line, but *before* `k` finishes, will need to be accounted for separately, since they are not blocking the `k`th person. Insight 2: We don't need a queue data structure. A simple array is sufficient, as we can simulate the line's behavior by iterating through it.

Which companies ask LeetCode 2073. Time Needed to Buy Tickets in interviews?

This problem is frequently asked at Komprise, X.

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Time Needed to Buy Tickets - LeetCode 2073 Solution

Time Needed to Buy Tickets - Complete Solution Guide

Time Needed to Buy Tickets is LeetCode problem 2073, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n people in a line queuing to buy tickets, where the 0 th person is at the front of the line and the (n - 1) th person is at the back of the line. You are given a 0-indexed integer array tickets of length n where the number of tickets that the i th person would like to buy is tickets[i] . Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously ) in order to buy more tickets. If

Detailed Explanation

The problem simulates a ticket purchasing line. We have `n` people, each with a certain number of tickets to buy (represented by the `tickets` array). Each person takes 1 second to buy a single ticket. After buying a ticket, they go to the back of the line. The goal is to determine the time it takes for the person initially at position `k` to finish buying all their tickets.

Solution Approach

The provided solutions use a simulation approach. They iterate through the line, decrementing each person's ticket count by one. After each person's turn, if that person has tickets remaining, they are moved to the end of the line implicitly (by continuing the iteration). The time is incremented with each ticket purchased. The loop continues until the person at position `k` has bought all their tickets.

Step-by-Step Algorithm

Step 1: Initialize `time` to 0.
Step 2: Iterate through the `tickets` array. For each person:
Step 3: Decrement the person's ticket count by 1 if they have tickets remaining.
Step 4: Increment `time` by 1 (representing the time spent buying a ticket).
Step 5: Check if the person at position `k` has finished buying tickets. If so, return `time`.
Step 6: Continue until the person at position `k` has finished.

Key Insights

Insight 1: The key is to realize that the order of people matters, and people who arrive *after* person `k` in the line, but *before* `k` finishes, will need to be accounted for separately, since they are not blocking the `k`th person.
Insight 2: We don't need a queue data structure. A simple array is sufficient, as we can simulate the line's behavior by iterating through it.
Insight 3: Optimizations are possible by considering that the person at position `k`'s ticket count is the upper bound for some of the calculations. If someone else has more tickets than the `k`th person, the `k`th person will certainly finish before they exhaust their tickets.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array, Queue, Simulation.

Companies

Asked at: Komprise, X.

Time Needed to Buy Tickets | LeetCode 2073

Easy

Array Queue Simulation

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int n = tickets.length;
        int time = 0;
        for (int i = 0; i < n; i++) {
            if (tickets[i] <= tickets[k]) {
                if (i <= k) {
                    time += tickets[i];
                } else {
                    time += tickets[i];
                }
            } else {
                if (i <= k) {
                    time += tickets[k];
                } else {
                    time += tickets[k];
                }
            }
        }
        return time;
    }
}

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Problem Info

Description

Problem 2073: Time Needed to Buy Tickets - Detailed explanation and solution approaches.

Topics

Array Queue Simulation

Companies

KompriseX

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2073

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize `time` to 0.
2
Step 2: Iterate through the `tickets` array. For each person:
3
Step 3: Decrement the person's ticket count by 1 if they have tickets remaining.
4
Step 4: Increment `time` by 1 (representing the time spent buying a ticket).
5
Step 5: Check if the person at position `k` has finished buying tickets. If so, return `time`.
6
Step 6: Continue until the person at position `k` has finished.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n*m) because in the worst case, we might need to iterate through the entire array multiple times (at most 'm' times where m is max(tickets)). The space complexity is O(1) because we are using only a few integer variables. We are modifying the tickets array in place.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Optimized Simulation

Instead of iterating through the entire line each time, we can iterate till the person at k is done. We can calculate the total time taken till person k's turn comes first. After that, we just need to check how many more times the loop goes around till person k finishes.

Time: O(n + m)Space: O(1)

Trade-offs: This approach is significantly faster, especially when k is large or ticket counts are high. The initial simulation approach has a worst-case runtime that's the product of the number of people and the max tickets; the optimized approach's worst-case time is the sum, making it asymptotically superior.

✨

Key Insights

1Insight 1: The key is to realize that the order of people matters, and people who arrive *after* person `k` in the line, but *before* `k` finishes, will need to be accounted for separately, since they are not blocking the `k`th person.
2Insight 2: We don't need a queue data structure. A simple array is sufficient, as we can simulate the line's behavior by iterating through it.
3Insight 3: Optimizations are possible by considering that the person at position `k`'s ticket count is the upper bound for some of the calculations. If someone else has more tickets than the `k`th person, the `k`th person will certainly finish before they exhaust their tickets.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: k = 0 (the person at the front of the line). The solution correctly handles this by immediately processing the person at position 0.
Edge case 2: `tickets` array containing all zeros. This will result in immediate return of time = 0, which is handled correctly.
Edge case 3: All people (except person k) have fewer tickets than person k. The solution handles this correctly by iterating till person k completes

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the movement of people to the end of the line. Failing to simulate the queue properly leads to inaccurate time calculations.

✗Common mistake 2: Not considering edge cases where the person at index k has a smaller number of tickets than others. This can lead to overestimating the total time.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Simulating resource allocation in systems with priority queues (e.g., job scheduling in an operating system).

🏦Real-world scenario 2: Modeling customer service lines where customers with higher priority (e.g., more urgent issues) get served first.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Time Needed to Buy Tickets"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether queue techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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