Teemo Attacking

LeetCode 495 | Difficulty: Easy

Easy

Problem Description

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 <= 10^4`

- `0 <= timeSeries[i], duration <= 10^7`

- `timeSeries` is sorted in **non-decreasing** order.

Topics: Array, Simulation

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Approach

Direct Approach

This problem can typically be solved with straightforward iteration or simple data structure usage. Focus on correctness first, then optimize.

When to use

Basic problems that test fundamental programming skills.

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Solutions

Solution 1: C# (Best: 168 ms)

Metric	Value
Runtime	168 ms
Memory	43 MB
Date	2022-01-11

Solution

public class Solution {
    public int FindPoisonedDuration(int[] timeSeries, int duration) {
        int poisoned = duration;
    for (int i = 1; i < timeSeries.Length; i++)
    {
        int prev = timeSeries[i-1];
        int current = timeSeries[i];
        
        poisoned += (prev+duration <= current) ? duration : current-prev;
    }
    return poisoned;
    }
}

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.