LeetCode #495 — EASY

Teemo Attacking

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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 <= 10⁴
0 <= timeSeries[i], duration <= 10⁷
timeSeries is sorted in non-decreasing order.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,4]
2

Example 2

[1,2]
2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #495: Teemo Attacking
class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int n = timeSeries.length;
        int ans = duration;
        for (int i = 1; i < n; ++i) {
            ans += Math.min(duration, timeSeries[i] - timeSeries[i - 1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #495: Teemo Attacking
func findPoisonedDuration(timeSeries []int, duration int) (ans int) {
	ans = duration
	for i, x := range timeSeries[1:] {
		ans += min(duration, x-timeSeries[i])
	}
	return
}

# Accepted solution for LeetCode #495: Teemo Attacking
class Solution:
    def findPoisonedDuration(self, timeSeries: List[int], duration: int) -> int:
        ans = duration
        for a, b in pairwise(timeSeries):
            ans += min(duration, b - a)
        return ans

// Accepted solution for LeetCode #495: Teemo Attacking
struct Solution;

impl Solution {
    fn find_poisoned_duration(time_series: Vec<i32>, duration: i32) -> i32 {
        let n = time_series.len();
        if n == 0 {
            return 0;
        }
        let mut start = time_series[0];
        let mut res = 0;
        for i in 1..n {
            let end = time_series[i];
            if start + duration > end {
                res += end - start;
            } else {
                res += duration;
            }
            start = end;
        }
        res += duration;
        res
    }
}

#[test]
fn test() {
    let time_series = vec![1, 4];
    let duration = 2;
    let res = 4;
    assert_eq!(Solution::find_poisoned_duration(time_series, duration), res);
    let time_series = vec![1, 2];
    let duration = 2;
    let res = 3;
    assert_eq!(Solution::find_poisoned_duration(time_series, duration), res);
}

// Accepted solution for LeetCode #495: Teemo Attacking
function findPoisonedDuration(timeSeries: number[], duration: number): number {
    const n = timeSeries.length;
    let ans = duration;
    for (let i = 1; i < n; ++i) {
        ans += Math.min(duration, timeSeries[i] - timeSeries[i - 1]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.