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Home Code Solutions

Teemo Attacking – LeetCode Solution

Teemo Attacking - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

Leetcode All Problems Solutions

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Table of Contents

  • Teemo Attacking – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Teemo Attacking LeetCode Solution
  • Java Teemo Attacking LeetCode Solution
  • Python 3 Teemo Attacking LeetCode Solution
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Teemo Attacking – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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 that Teemo attacks Ashe at second timeSeries[i], and an integer duration.

Return the total number of seconds that Ashe is poisoned.

 

Example 1:

Input: timeSeries = [1,4], duration = 2
Output: 4
Explanation: Teemo's attacks on Ashe go as follows:
- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
- At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.

Example 2:

Input: timeSeries = [1,2], duration = 2
Output: 3
Explanation: Teemo's attacks on Ashe go as follows:
- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
- At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.

 

Constraints:

  • 1 <= timeSeries.length <= 104
  • 0 <= timeSeries[i], duration <= 107
  • timeSeries is sorted in non-decreasing order.

C++ Teemo Attacking LeetCode Solution


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class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
        int len = timeSeries.size();
        if (!len) return 0;
        int result = duration;
        for (int i = 1; i < len; i++) {
            result += min(duration, timeSeries[i] - timeSeries[i - 1]);
        }
        return result;
    }
};

Java Teemo Attacking LeetCode Solution


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public class Solution {
    public int findPosisonedDuration(int[] timeSeries, int duration) {
        if (timeSeries == null || timeSeries.length == 0 || duration == 0) return 0;
        
        int result = 0, start = timeSeries[0], end = timeSeries[0] + duration;
        for (int i = 1; i < timeSeries.length; i++) {
            if (timeSeries[i] > end) {
                result += end - start;
                start = timeSeries[i];
            }
            end = timeSeries[i] + duration;
        }
        result += end - start;
        
        return result;
    }
}

 



Python 3 Teemo Attacking LeetCode Solution


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class Solution(object):
    def findPoisonedDuration(self, timeSeries, duration):
        ans = duration * len(timeSeries)
        for i in range(1,len(timeSeries)):
            ans -= max(0, duration - (timeSeries[i] - timeSeries[i-1]))
        return ans



 

Array-1180
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Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolutionTeemo Attacking
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