What is the time complexity of Teemo Attacking?

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

What data structures are used to solve Teemo Attacking?

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

What are the key insights for solving Teemo Attacking?

Insight 1: The key is to iterate through the `timeSeries` and determine whether each attack's duration overlaps with the previous attack. If there's an overlap, we only count the additional poisoned time, not the full duration. Insight 2: The problem leverages the fact that `timeSeries` is sorted. This allows us to process attacks sequentially and easily determine overlap.

Which companies ask LeetCode 495. Teemo Attacking in interviews?

This problem is frequently asked at Riot Games.

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Teemo Attacking - LeetCode 495 Solution

Teemo Attacking - Complete Solution Guide

Teemo Attacking is LeetCode problem 495, 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

Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds. More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1] . If Teemo attacks again before the poison effect ends, the timer for it is reset , and the poison effect will end duration seconds after the new attack. You are given a non-decreasing integer array timeSeries , where timeSeries[i] denotes

Detailed Explanation

The problem asks us to calculate the total time Ashe is poisoned by Teemo's attacks, given a list of attack times (`timeSeries`) and the duration of the poison effect (`duration`). Each attack applies poison for `duration` seconds. If Teemo attacks again before the current poison effect ends, the timer is reset. The goal is to find the sum of all poisoned seconds, considering overlapping poison effects.

Solution Approach

The provided solution uses a straightforward iterative approach. It calculates the total poisoned time by iterating through the `timeSeries` array. For each attack, it checks if the current attack's start time overlaps with the previous attack's poison duration. If there's an overlap, it adds only the non-overlapping portion of the duration. Otherwise, it adds the full duration. It effectively simulates the poison effect and its reset mechanism to get the total time.

Step-by-Step Algorithm

Step 1: Initialize `total_poisoned_time` to 0.
Step 2: Handle the edge case where `timeSeries` is empty. If it is, return 0 because there are no attacks.
Step 3: Initialize `prev_attack_time` with the first attack time in `timeSeries`.
Step 4: Add the full duration of the first attack to `total_poisoned_time`.
Step 5: Iterate through the `timeSeries` from the second attack time onward (index 1).
Step 6: For each attack, calculate the time difference between the current attack and the previous attack.
Step 7: If the time difference is greater than or equal to the `duration`, add the full `duration` to `total_poisoned_time` (no overlap).
Step 8: If the time difference is less than `duration`, add the time difference to `total_poisoned_time` (overlap occurs).
Step 9: Update `prev_attack_time` with the current attack time for the next iteration.
Step 10: After iterating through all attacks, return the final `total_poisoned_time`.

Key Insights

Insight 1: The key is to iterate through the `timeSeries` and determine whether each attack's duration overlaps with the previous attack. If there's an overlap, we only count the additional poisoned time, not the full duration.
Insight 2: The problem leverages the fact that `timeSeries` is sorted. This allows us to process attacks sequentially and easily determine overlap.
Insight 3: The core logic revolves around comparing the current attack's start time with the end time of the previous attack's poison effect.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: Riot Games.

Teemo Attacking | LeetCode 495

Easy

Array Simulation

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

Solution Code

Available in:

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int totalPoisonedTime = 0;
        if (timeSeries.length == 0) {
            return 0;
        }
        totalPoisonedTime += duration;
        for (int i = 1; i < timeSeries.length; i++) {
            if (timeSeries[i] - timeSeries[i - 1] < duration) {
                totalPoisonedTime += timeSeries[i] - timeSeries[i - 1];
            } else {
                totalPoisonedTime += duration;
            }
        }
        return totalPoisonedTime;
    }
}

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Problem Info

Description

Problem 495: Teemo Attacking - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

Riot Games

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:495

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

10 steps

1
Step 1: Initialize `total_poisoned_time` to 0.
2
Step 2: Handle the edge case where `timeSeries` is empty. If it is, return 0 because there are no attacks.
3
Step 3: Initialize `prev_attack_time` with the first attack time in `timeSeries`.
4
Step 4: Add the full duration of the first attack to `total_poisoned_time`.
5
Step 5: Iterate through the `timeSeries` from the second attack time onward (index 1).
6
Step 6: For each attack, calculate the time difference between the current attack and the previous attack.
7
Step 7: If the time difference is greater than or equal to the `duration`, add the full `duration` to `total_poisoned_time` (no overlap).
8
Step 8: If the time difference is less than `duration`, add the time difference to `total_poisoned_time` (overlap occurs).
9
Step 9: Update `prev_attack_time` with the current attack time for the next iteration.
10
Step 10: After iterating through all attacks, return the final `total_poisoned_time`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single loop that iterates through the `timeSeries` array. The space complexity is constant because the memory used does not grow with the input size.

📖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

Using a Set to track poisoned seconds

Iterate through the timeSeries. For each attack, add the poisoned seconds to a set. The set will automatically handle duplicate seconds, and the final size of the set is the total poisoned time.

Time: O(n * duration) in the worst case, where n is the length of the `timeSeries` array. Creating the set of poisoned times for each attack has a complexity of O(duration), and this is performed for n attacks.Space: O(n * duration) in the worst case, since the set could potentially store up to n * duration seconds.

Trade-offs: This approach is less efficient than the main solution in terms of both time and space complexity. It's less desirable due to the large possible space requirement and larger time complexity. This would also be affected by the maximum value of the timeSeries array.

✨

Key Insights

1Insight 1: The key is to iterate through the `timeSeries` and determine whether each attack's duration overlaps with the previous attack. If there's an overlap, we only count the additional poisoned time, not the full duration.
2Insight 2: The problem leverages the fact that `timeSeries` is sorted. This allows us to process attacks sequentially and easily determine overlap.
3Insight 3: The core logic revolves around comparing the current attack's start time with the end time of the previous attack's poison effect.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty `timeSeries`. The code explicitly handles this case by returning 0, as there are no attacks and therefore no poisoned time.
Edge case 2: `duration` is 0. The code should work correctly even if duration is zero because, in that case, total_poisoned_time remains 0 after going through the loop.
Edge case 3: Consecutive attacks with time difference less than duration. The code correctly handles overlapping poison durations by only adding the non-overlapping portion.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the overlap. It's crucial to compare the current attack's start time with the *end* time of the previous attack's poison effect (start + duration).

✗Common mistake 2: Not handling the edge case of an empty `timeSeries`. Failing to do so can lead to errors.

✗Common mistake 3: Assuming that the time series is always strictly increasing. The problem states non-decreasing, meaning duplicates are possible. While duplicates do not change the result, one should keep that in mind.

✅

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: Server request throttling. You can use a similar approach to limit the number of requests a server processes within a certain time window, preventing overload. 'Attack time' represents a request, and duration represents the cool down period.

🏦Real-world scenario 2: Modeling overlapping events. For example, scheduling tasks where each task has a start time and duration, and you need to calculate the total time that any task is running. Time series would be a list of task start times.

💡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.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: October 14, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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 "Teemo Attacking"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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