README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2582.Pass%20the%20Pillow/README_EN.md	1278	Weekly Contest 335 Q1	Math Simulation

2582. Pass the Pillow

中文文档

Description

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.

 

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 2nd 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 3rd 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.

Solutions

Solution 1: Simulation

We can simulate the process of passing the pillow, and each time the pillow is passed, if the pillow reaches the front or the end of the queue, the direction of the pillow will change, and the queue will continue to pass the pillow along the opposite direction.

The time complexity is $O(time)$ and the space complexity is $O(1)$, where $time$ is the given time.

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    int passThePillow(int n, int time) {
        int ans = 1, k = 1;
        while (time--) {
            ans += k;
            if (ans == 1 || ans == n) {
                k *= -1;
            }
        }
        return ans;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

Solution 2: Math

We notice that there are $n - 1$ passes in each round. Therefore, we can divide $time$ by $n - 1$ to get the number of rounds $k$ that the pillow is passed, and then take the remainder of $time$ modulo $n - 1$ to get the remaining passes $mod$ in the current round.

Then we judge the current round $k$:

• If $k$ is odd, then the current direction of the pillow is from the end of the queue to the front, so the pillow will be passed to the person with the number $n - mod$.
• If $k$ is even, then the current direction of the pillow is from the front of the queue to the back, so the pillow will be passed to the person with the number $mod + 1$.

The time complexity is $O(1)$ and the space complexity is $O(1)$.

Python3

class Solution:
    def passThePillow(self, n: int, time: int) -> int:
        k, mod = divmod(time, n - 1)
        return n - mod if k & 1 else mod + 1

Java

class Solution {
    public int passThePillow(int n, int time) {
        int k = time / (n - 1);
        int mod = time % (n - 1);
        return (k & 1) == 1 ? n - mod : mod + 1;
    }
}

C++

class Solution {
public:
    int passThePillow(int n, int time) {
        int k = time / (n - 1);
        int mod = time % (n - 1);
        return k & 1 ? n - mod : mod + 1;
    }
};

Go

func passThePillow(n int, time int) int {
	k, mod := time/(n-1), time%(n-1)
	if k&1 == 1 {
		return n - mod
	}
	return mod + 1
}

TypeScript

function passThePillow(n: number, time: number): number {
    const k = time / (n - 1);
    const mod = time % (n - 1);
    return (k & 1) == 1 ? n - mod : mod + 1;
}

Rust

impl Solution {
    pub fn pass_the_pillow(n: i32, time: i32) -> i32 {
        let mut k = time / (n - 1);
        let mut _mod = time % (n - 1);

        if (k & 1) == 1 {
            return n - _mod;
        }

        _mod + 1
    }
}