Leetcode Teemo Attacking problem solution

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

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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;
}

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