# Teemo Attacking Problem

## Description

LeetCode Problem 495.

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:

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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:

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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 <= 10^4
- 0 <= timeSeries[i], duration <= 10^7
- timeSeries is sorted in non-decreasing order.

## Sample C++ Code

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