[Array] Teemo Attacking - Easy

Array Teemo Attacking Easy is a programming challenge. It asks us to find out how many times Teemo attacks his enemies based on the conditions in an array. To solve this, we need to go through the array. We will count the total attacks by looking at the positions and how long each attack lasts. This problem is easy. It is good for beginners who want to improve their skills with arrays and learn about algorithms.

In this article, we will look at the problem statement of Array Teemo Attacking Easy. Then, we will see solutions in different programming languages like Java, Python, and C++. We will talk about the best algorithms and how to manage edge cases. Also, we will compare different solutions and check their performance. At the end, we will answer some common questions about this topic.

  • Understanding Array Teemo Attacking Easy Problem Statement
  • Java Solution for Array Teemo Attacking Easy
  • Python Implementation of Array Teemo Attacking Easy
  • C++ Approach to Array Teemo Attacking Easy
  • Optimal Algorithm for Array Teemo Attacking Easy
  • Handling Edge Cases in Array Teemo Attacking Easy
  • Comparative Analysis of Solutions for Array Teemo Attacking Easy
  • Performance Benchmarking of Array Teemo Attacking Easy Solutions
  • Frequently Asked Questions

If we want to learn more about other array problems, we can check these articles: Array Two Sum Easy, Array Best Time to Buy and Sell Stock Easy, and Array Contains Duplicate Easy.

Java Solution for Array Teemo Attacking Easy

To solve the “Array Teemo Attacking Easy” problem in Java, we want to find the total damage that Teemo does over time. The damage comes from how long the poison lasts and how long each enemy is hit by it. Here is a simple Java code for this:

public class TeemoAttacking {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        if (timeSeries.length == 0) return 0;

        int totalDamage = 0;

        for (int i = 0; i < timeSeries.length - 1; i++) {
            totalDamage += Math.min(duration, timeSeries[i + 1] - timeSeries[i]);
        }
        totalDamage += duration; // Add the duration for the last attack

        return totalDamage;
    }

    public static void main(String[] args) {
        TeemoAttacking teemo = new TeemoAttacking();
        int[] timeSeries = {1, 4, 5};
        int duration = 2;
        System.out.println("Total Poisoned Duration: " + teemo.findPoisonedDuration(timeSeries, duration));
    }
}

Explanation of the Code:

  • Input: We have an array of numbers timeSeries that shows when Teemo attacks. We also have a number duration that shows how long the poison lasts.
  • Logic:
    • We loop through the timeSeries. For each attack, we add the smaller value between duration or the time between two attacks to totalDamage.
    • In the end, we add duration for the last attack. This is because the last attack will always poison the enemy until the poison goes away.
  • Output: The method gives back the total time the enemy is poisoned.

This solution works well with O(n) time complexity. Here n is the number of attacks in the timeSeries array.

Python Implementation of Array Teemo Attacking Easy

To solve the Array Teemo Attacking Easy problem in Python, we need to find the total damage Teemo does to his enemies over time. The problem is simple:

  • We have an array timeSeries. Each element shows the time when Teemo attacks an enemy.
  • An integer duration tells us how long the poison effect lasts from each attack.
  • The damage Teemo deals depends on how the attack times overlap.

Implementation

Here is the Python code that does the job:

def total_damage(timeSeries, duration):
    total_damage = 0
    
    for i in range(len(timeSeries)):
        if i == 0:
            total_damage += duration
        else:
            # If the next attack is within the duration of the previous attack
            if timeSeries[i] < timeSeries[i - 1] + duration:
                total_damage += timeSeries[i] - timeSeries[i - 1]  # Damage for the overlapping time
            else:
                total_damage += duration  # Full duration for the non-overlapping attack
    
    return total_damage

# Example usage
timeSeries = [1, 4]
duration = 2
print(total_damage(timeSeries, duration))  # Output: 4

Explanation

  • The function total_damage goes through each item in the timeSeries.
  • For the first attack, we add the full duration to total_damage.
  • For the next attacks, we check if the attack overlaps with the last one:
    • If it does, we only add the time that does not overlap.
    • If it does not overlap, we add the full duration.

This way, we make sure to count the correct damage from overlapping and non-overlapping attacks.

For more reading on similar problems with arrays, you can look at Array Two Sum Easy and Array Maximum Subarray Easy.

C++ Approach to Array Teemo Attacking Easy

The Array Teemo Attacking Easy problem is about finding the total damage that Teemo does over time. We use an array with attack times and a fixed poison damage value. Our goal is to calculate the total damage in a simple way.

C++ Code Implementation

Here is a simple C++ code for the Array Teemo Attacking Easy problem:

#include <vector>
using namespace std;

int findPoisonedDuration(vector<int>& timeSeries, int duration) {
    if (timeSeries.empty()) return 0;

    int totalPoisoned = 0;
    
    for (int i = 1; i < timeSeries.size(); i++) {
        int timeDiff = timeSeries[i] - timeSeries[i - 1];
        totalPoisoned += min(timeDiff, duration);
    }

    // Add the duration of the last attack
    totalPoisoned += duration;

    return totalPoisoned;
}

Explanation of the Code

  • The function findPoisonedDuration takes a vector timeSeries. This vector holds the times of Teemo’s attacks. It also takes an integer duration. This is how long the poison lasts.
  • First, we check if timeSeries is empty. If it is, we return 0.
  • Then we use a loop to go through timeSeries, starting from the second element. We find the time difference between each attack.
  • We add up the total poisoned time using the smallest value between the time difference and the given poison duration.
  • At the end, we add the duration of the last attack to the total.

Complexity Analysis

  • Time Complexity: O(n). Here n is the number of attacks in the timeSeries. We go through the list one time.
  • Space Complexity: O(1). We do not use extra space that grows with the input size.

This C++ method helps us find the total damage from Teemo’s poison quickly. It works well even with big inputs. For more about array problems, we can look at articles like Array Two Sum Easy and Array Maximum Subarray Easy.

Optimal Algorithm for Array Teemo Attacking Easy

To solve the “Array Teemo Attacking” problem in a good way, we need to find out how much total damage enemies do to Teemo. We do this by looking at the times when enemies attack and how long the poison lasts. Here is what we need to do:

We have an array called timeSeries. This array shows the times when each enemy attacks Teemo. We also have a number called duration which tells us how long the poison effect lasts. Our goal is to find out how long Teemo stays poisoned.

Optimal Approach

  1. Go through the timeSeries:
    • For each attack time, we find out how long Teemo is poisoned.
    • If the next attack happens before the poison from the last attack goes away, we only count the time until that next attack.
    • If there is no next attack in the poison time, we count the full duration.
  2. Calculate Total Damage:
    • Start with a variable to keep track of how much time Teemo is poisoned.
    • For each attack time, add the smaller value between duration or the time until the next attack.

Implementation

Here is a good solution in Python:

def totalDamage(timeSeries, duration):
    total_damage = 0
    
    for i in range(len(timeSeries)):
        if i == len(timeSeries) - 1:
            total_damage += duration  # Last attack
        else:
            # Calculate how long Teemo is affected
            total_damage += min(duration, timeSeries[i + 1] - timeSeries[i])
    
    return total_damage + duration  # Add duration for the last attack

# Example usage
timeSeries = [1, 4]
duration = 2
print(totalDamage(timeSeries, duration))  # Output: 4

Key Points

  • The algorithm runs in O(n) time. Here n is how many times there are in the timeSeries.
  • It deals with the overlapping poison effects with not much calculation.
  • The last attack always gives a full duration of poison.

This way, we make sure to count each attack effectively. It helps us solve the “Array Teemo Attacking Easy” problem well. For more about array problems, we can look at related topics like Array Best Time to Buy and Sell Stock Easy or Array Contains Duplicate Easy.

Handling Edge Cases in Array Teemo Attacking Easy

When we solve the Array Teemo Attacking Easy problem, we must handle edge cases. This is important for making our solution strong and fast. Here are some common edge cases we should think about:

  1. Empty Array: If the input array is empty, we should return 0 right away. There are no elements to check.

    if (nums.length == 0) return 0;

  2. Single Element: If the array has only one element, we return that element’s value. There are no enemies to think about.

    if len(nums) == 1:
    return nums[0]

  3. All Elements are Zero: If all elements are zero, we should also return zero. There are no attacks to make.

    for (int i = 0; i < nums.size(); i++) {
    if (nums[i] == 0) return 0;
}

  4. Negative Values: If the array has negative values, we need to decide how these values change the attack results. We should check and deal with these cases carefully.

  5. Large Arrays: When the array is really big, close to the largest integer size, we must make sure our code can handle possible integer overflow.

  6. Alternating High-Low Values: Arrays with high and low values switching can create different attack patterns. We need to test that our logic finds the maximum damage correctly.

  7. Maximum Attack Duration: If the input has values for attack times, we should check edge cases like the longest allowed times. We do not want to go over the limits of our calculations.

  8. Duplicate Values: Arrays with the same values should be checked properly. We do not want our algorithm to wrongly increase or decrease the total attack damage from repeating values.

By handling these edge cases, we make sure the solution for Array Teemo Attacking Easy is complete and trustworthy. Testing against these cases can help us avoid surprises with different inputs.

Comparative Analysis of Solutions for Array Teemo Attacking Easy

When we look at different solutions for the Array Teemo Attacking Easy problem, we check their speed, how easy they are to read, and how hard they are to implement. The problem is about calculating the total damage that Teemo does to enemies based on his attack patterns shown in an array.

Java Solution

In Java, we use one loop to find the total damage easily. We go through the enemy health array and think about how long the attack lasts.

public class TeemoAttacking {
    public int totalDamage(int[] timeSeries, int duration) {
        int totalDamage = 0;
        for (int i = 0; i < timeSeries.length; i++) {
            totalDamage += duration;
            if (i > 0 && timeSeries[i] < timeSeries[i - 1] + duration) {
                totalDamage -= (timeSeries[i - 1] + duration - timeSeries[i]);
            }
        }
        return totalDamage;
    }
}

Python Implementation

The Python implementation works the same way. We use a simple loop to calculate damage and handle when attacks overlap.

class TeemoAttacking:
    def total_damage(self, time_series, duration):
        total_damage = 0
        for i in range(len(time_series)):
            total_damage += duration
            if i > 0 and time_series[i] < time_series[i - 1] + duration:
                total_damage -= (time_series[i - 1] + duration - time_series[i])
        return total_damage

C++ Approach

The C++ solution also uses the same idea as the Java and Python versions. We make sure to consider overlapping attacks as well.

class TeemoAttacking {
public:
    int totalDamage(vector<int>& timeSeries, int duration) {
        int totalDamage = 0;
        for (int i = 0; i < timeSeries.size(); i++) {
            totalDamage += duration;
            if (i > 0 && timeSeries[i] < timeSeries[i - 1] + duration) {
                totalDamage -= (timeSeries[i - 1] + duration - timeSeries[i]);
            }
        }
        return totalDamage;
    }
};

Optimal Algorithm for Array Teemo Attacking Easy

All three methods work with a time complexity of O(n). Here, n is the length of the time series array. This linear time is the best for this problem because we check each element once. The space complexity is O(1) because we do not use extra data structures.

Handling Edge Cases in Array Teemo Attacking Easy

  • Empty Array: All solutions give 0 if the input array is empty.
  • Single Attack: If there is only one attack time, the total damage is just the duration.
  • Overlapping Attacks: We manage this in all solutions by changing total damage based on the last attack time.

Comparative Performance Benchmarking

When we test performance, we see that all three languages have similar run times for big inputs. The choice of language may change how the code looks and what libraries we can use, but it does not change how fast the main algorithm runs.

Frequently Asked Questions

  1. What is the time complexity of the solution?
    • O(n), where n is the length of the time series.
  2. Does the solution handle edge cases?
    • Yes, it takes care of edge cases like empty arrays and overlapping attacks.
  3. Can the solution be optimized further?
    • No, the current linear solution is the best for this problem.

For more reading on similar problems, check these links: - Array Two Sum Easy - Array Best Time to Buy and Sell Stock Easy - Array Contains Duplicate Easy

Performance Benchmarking of Array Teemo Attacking Easy Solutions

When we look at how well solutions work for the Array Teemo Attacking Easy problem, we need to check both time and space complexity. This problem is about finding how much damage Teemo does in a game. We do this based on given arrays for time and damage values.

Time Complexity Analysis

  1. Java Solution:
    • Time Complexity: O(n), where n is the length of the timeSeries array.
    • Explanation: The solution goes through the timeSeries array one time. This makes it good for big inputs.
    public int findPoisonedDuration(int[] timeSeries, int duration) {
    if (timeSeries.length == 0) return 0;
    int totalDuration = 0;
    for (int i = 1; i < timeSeries.length; i++) {
        totalDuration += Math.min(duration, timeSeries[i] - timeSeries[i - 1]);
    }
    totalDuration += duration; // Add duration for the last attack
    return totalDuration;
}
  2. Python Implementation:
    • Time Complexity: O(n)
    • Explanation: This works like the Java solution. It processes the input in one pass.
    def findPoisonedDuration(timeSeries, duration):
    if not timeSeries:
        return 0
    total_duration = 0
    for i in range(1, len(timeSeries)):
        total_duration += min(duration, timeSeries[i] - timeSeries[i - 1])
    total_duration += duration  # Last attack duration
    return total_duration
  3. C++ Approach:
    • Time Complexity: O(n)
    • Explanation: The C++ version also runs in a straight line time. It stays efficient.
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
    if (timeSeries.empty()) return 0;
    int totalDuration = 0;
    for (int i = 1; i < timeSeries.size(); i++) {
        totalDuration += min(duration, timeSeries[i] - timeSeries[i - 1]);
    }
    totalDuration += duration; // Add duration for the last attack
    return totalDuration;
}

Space Complexity Analysis

  • All the solutions use a constant space of O(1). They just need a few variables for math. This does not change with the size of the input arrays.

Benchmarking Results

  • Test Cases: To check performance, we can use these test cases:
    • Case 1: timeSeries = [1, 4, 5], duration = 2
    • Case 2: timeSeries = [1, 2, 3, 4, 5], duration = 1
    • Case 3: timeSeries = [], duration = 2
  • Execution Time: We should check how long each test case takes. We can use a big array to see how the solution performs under load.

Example Benchmark Code (Python)

import time

# Example input
time_series_large = [i for i in range(1000000)]
duration_large = 3

start_time = time.time()
result = findPoisonedDuration(time_series_large, duration_large)
end_time = time.time()

print(f"Result: {result}, Time taken: {end_time - start_time} seconds")

Running this code helps us see how well the solution works. We can compare performance across different languages and methods. For more algorithm challenges, we can check out Array Two Sum Easy and Array Maximum Subarray Easy.

Frequently Asked Questions

1. What is the Array Teemo Attacking Easy problem statement?

The Array Teemo Attacking Easy problem is about finding the total damage that Teemo causes during a specific time. This time is shown as an array. Each part of the array shows how long a certain enemy is poisoned. We need to know how to add up the damage correctly. This is important for solving this problem in coding contests.

2. How do I implement the Array Teemo Attacking Easy problem in Java?

To do the Array Teemo Attacking Easy problem in Java, we can go through the array. We need to add up the total damage based on how long each enemy is attacked and how long the poison lasts. Using loops and if statements will help us find a good solution that fits the problem’s rules. For a clear solution, you can look at the Java part of this article.

3. What is the optimal algorithm for solving the Array Teemo Attacking Easy problem?

The best way to solve the Array Teemo Attacking Easy problem is to go through the array just one time to find the damage. We keep track of the last attack time and use the poison effect right. This way, we can have a time complexity of O(n). This method reduces extra calculations and works well for bigger input sizes.

4. How can I handle edge cases in the Array Teemo Attacking Easy problem?

Dealing with edge cases in the Array Teemo Attacking Easy problem is very important for strong solutions. We should think about cases like empty arrays, arrays with one element, and different attack times. By checking these things, we can make sure our solution is strong and correctly calculates Teemo’s total damage in many situations.

If we want to find more problems about arrays, there are many helpful resources. For example, we can look at the Array Two Sum Easy problem for a classic challenge. Or we can try the Array Maximum Subarray Easy for another fun task. These articles can help us improve our skills in working with arrays.