Solution 1 - Linear Scan

Sum min(gap, duration) for each consecutive pair, plus duration for the last attack.

TimeComplexity: O(n)

SpaceComplexity: O(1)

""" 0495.1 - Teemo Attacking - Solution 1 - Linear Scan """

from typing import List

class Solution:
    def findPoisonedDuration(self, timeSeries: List[int], duration: int) -> int:
        if not timeSeries:
            return 0
        total = duration
        for i in range(1, len(timeSeries)):
            total += min(duration, timeSeries[i] - timeSeries[i - 1])
        return total

if __name__ == "__main__":
    print(Solution().findPoisonedDuration([1,4], 2))  # 4

function findPoisonedDuration(t:number[], duration:number):number{
 if(t.length===0) return 0;
 let total=duration;
 for(let i=1;i<t.length;i++){
  total += Math.min(duration, t[i]-t[i-1]);
 }
 return total;
}

package main

func findPoisonedDuration(timeSeries []int, duration int) int {
	if len(timeSeries) == 0 {
		return 0
	}
	total := duration
	for i := 1; i < len(timeSeries); i++ {
		gap := timeSeries[i] - timeSeries[i-1]
		if gap < duration {
			total += gap
		} else {
			total += duration
		}
	}
	return total
}

func main() {}

fn find_poisoned_duration(t: &[i32], duration: i32) -> i32 {
	if t.is_empty() { return 0; }
	let mut total = duration;
	for i in 1..t.len() {
		total += duration.min(t[i] - t[i-1]);
	}
	total
}
fn main() {}

0495 - Teemo Attacking

Solution 1 - Linear Scan