Skip to content
Programming101
Programming101

Learn everything about programming

  • Home
  • CS Subjects
    • IoT – Internet of Things
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programming101
Programming101

Learn everything about programming

Leetcode Teemo Attacking problem solution

YASH PAL, 31 July 2024

In this Leetcode Teemo Attacking problem solution 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.

Leetcode Teemo Attacking problem solution

Problem solution in Python.

class Solution:
    def findPoisonedDuration(self, timeSeries: List[int], duration: int) -> int:
        ret = len(timeSeries) * duration          # first, sum up all
        
        for i in range(len(timeSeries)-1):
            if timeSeries[i+1] - timeSeries[i] < duration:   
                ret -= (duration - (timeSeries[i+1] - timeSeries[i]))  # and then, weep out
        return ret

Problem solution in Java.

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

Problem solution in C++.

int findPoisonedDuration(vector<int>& timeSeries, int duration) {
    int total = 0;
    for (int i = 0; i < timeSeries.size(); i++) 
        total += (i > 0 && timeSeries[i] - timeSeries[i-1] < duration ? timeSeries[i] - timeSeries[i - 1] : duration);
    return total;
}

coding problems

Post navigation

Previous post
Next post
  • HackerRank Separate the Numbers solution
  • How AI Is Revolutionizing Personalized Learning in Schools
  • GTA 5 is the Game of the Year for 2024 and 2025
  • Hackerrank Day 5 loops 30 days of code solution
  • Hackerrank Day 6 Lets Review 30 days of code solution
How to download udemy paid courses for free

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
©2025 Programming101 | WordPress Theme by SuperbThemes