LeetCode #2582 — EASY

Pass the Pillow

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

The Problem

Problem Statement

There are n people standing in a line labeled from 1 to n. The first person in the line is holding a pillow initially. Every second, the person holding the pillow passes it to the next person standing in the line. Once the pillow reaches the end of the line, the direction changes, and people continue passing the pillow in the opposite direction.

For example, once the pillow reaches the n^th person they pass it to the n - 1^th person, then to the n - 2^th person and so on.

Given the two positive integers n and time, return the index of the person holding the pillow after time seconds.

Example 1:

Input: n = 4, time = 5
Output: 2
Explanation: People pass the pillow in the following way: 1 -> 2 -> 3 -> 4 -> 3 -> 2.
After five seconds, the 2^nd person is holding the pillow.

Example 2:

Input: n = 3, time = 2
Output: 3
Explanation: People pass the pillow in the following way: 1 -> 2 -> 3.
After two seconds, the 3^r^d person is holding the pillow.

Constraints:

2 <= n <= 1000
1 <= time <= 1000

Note: This question is the same as 3178: Find the Child Who Has the Ball After K Seconds.

Step 01

Brute Force Baseline

Problem summary: There are n people standing in a line labeled from 1 to n. The first person in the line is holding a pillow initially. Every second, the person holding the pillow passes it to the next person standing in the line. Once the pillow reaches the end of the line, the direction changes, and people continue passing the pillow in the opposite direction. For example, once the pillow reaches the nth person they pass it to the n - 1th person, then to the n - 2th person and so on. Given the two positive integers n and time, return the index of the person holding the pillow after time seconds.

Baseline thinking

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

Pattern signal: Math

Example 1

4
5

Example 2

3
2

Core Insight

What unlocks the optimal approach

Maintain two integer variables, direction and i, where direction denotes the current direction in which the pillow should pass, and i denotes an index of the person holding the pillow.
While time is positive, update the current index with the current direction. If the index reaches the end of the line, multiply direction by - 1.

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 #2582: Pass the Pillow
class Solution {
    public int passThePillow(int n, int time) {
        int ans = 1, k = 1;
        while (time-- > 0) {
            ans += k;
            if (ans == 1 || ans == n) {
                k *= -1;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2582: Pass the Pillow
func passThePillow(n int, time int) int {
	ans, k := 1, 1
	for ; time > 0; time-- {
		ans += k
		if ans == 1 || ans == n {
			k *= -1
		}
	}
	return ans
}

# Accepted solution for LeetCode #2582: Pass the Pillow
class Solution:
    def passThePillow(self, n: int, time: int) -> int:
        ans = k = 1
        for _ in range(time):
            ans += k
            if ans == 1 or ans == n:
                k *= -1
        return ans

// Accepted solution for LeetCode #2582: Pass the Pillow
impl Solution {
    pub fn pass_the_pillow(n: i32, time: i32) -> i32 {
        let mut ans = 1;
        let mut k = 1;

        for i in 1..=time {
            ans += k;

            if ans == 1 || ans == n {
                k *= -1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2582: Pass the Pillow
function passThePillow(n: number, time: number): number {
    let ans = 1,
        k = 1;
    while (time-- > 0) {
        ans += k;
        if (ans === 1 || ans === n) {
            k *= -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(time)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.