Teemo Attacking

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.

Examples

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.

Solution for Teemo Attacking

Approach 1: One pass

Intuition

The problem is an example of merge interval questions which are now quite popular in Google.

Typically such problems could be solved in a linear time in the case of sorted input, and in O(Nlog⁡N) time otherwise.

Here one deals with a sorted input, and the problem could be solved in one pass with a constant space. The idea is straightforward: consider only the interval between two attacks. Ashe spends in a poisoned condition the whole time interval if this interval is shorter than the poisoning time duration duration, and duration otherwise.

Algorithm

• Initiate total time in poisoned condition total = 0.

• Iterate over timeSeries list. At each step add to the total time the minimum between interval length and the poisoning time duration duration.

• Return total + duration to take the last attack into account.

Code in Different Languages

C++

Written by @Shreyash3087

class Solution {
public:
    int findPoisonedDuration(std::vector<int>& timeSeries, int duration) {
        int n = timeSeries.size();
        if (n == 0) return 0;

        int total = 0;
        for(int i = 0; i < n - 1; ++i)
            total += std::min(timeSeries[i + 1] - timeSeries[i], duration);
        return total + duration;
    }
};

Java

Written by @Shreyash3087

class Solution {
  public int findPoisonedDuration(int[] timeSeries, int duration) {
    int n = timeSeries.length;
    if (n == 0) return 0;

    int total = 0;
    for(int i = 0; i < n - 1; ++i)
      total += Math.min(timeSeries[i + 1] - timeSeries[i], duration);
    return total + duration;
  }
}

Python

Written by @Shreyash3087

class Solution:
    def findPoisonedDuration(self, timeSeries: List[int], duration: int) -> int:
        n = len(timeSeries)
        if n == 0:
            return 0
        
        total = 0
        for i in range(n - 1):
            total += min(timeSeries[i + 1] - timeSeries[i], duration)
        return total + duration     

Complexity Analysis

Time Complexity: $O(N)$

Reason: where N is the length of the input list since we iterate the entire list.

Space Complexity: $O(1)$

Reason: it's a constant space solution.

References

• LeetCode Problem: Teemo Attacking

• Solution Link: Teemo Attacking

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