Intuition

Each attack poisons Ashe for duration seconds, but a new attack before the effect ends only extends coverage by the gap since the previous attack (capped at duration). The last attack always contributes a full duration seconds.

Approach: One-Pass Interval Merging

1. Start with ans = duration for the poison window after the first attack.
2. For each consecutive pair timeSeries[i - 1] and timeSeries[i], add min(duration, timeSeries[i] - timeSeries[i - 1]) — the extra poison time gained before the timer resets.
3. Return ans.

Complexity

• Time complexity: $$O(n)$$, where n is timeSeries.length — one pass over adjacent pairs.
• Space complexity: $$O(1)$$ extra space.

Code

Go

func findPoisonedDuration(timeSeries []int, duration int) int {
    ans := duration
    for i := 1; i < len(timeSeries); i++ {
        ans += min(duration, timeSeries[i]-timeSeries[i-1])
    }
    return ans
}

Rust

impl Solution {
    pub fn find_poisoned_duration(time_series: Vec<i32>, duration: i32) -> i32 {
        time_series
            .windows(2)
            .fold(duration, |acc, num| acc + duration.min(num[1] - num[0]))
    }
}