LeetCode 2582. Pass the pillow: explain the O(1) time algorithm in clear words

【LetMeFly】2582. Pass the pillow: explain the O(1) time algorithm in clear words

Leetcode question link: https://leetcode.cn/problems/pass-the-pillow/

nIndividuals stand in a row and are numbered from 1to n.

Initially, the first person at the head of the line holds a pillow. Every second, the person holding the pillow passes the pillow to the next person in line. Once the pillow reaches the beginning or end of the line, the direction of the delivery changes and the team continues passing pillows in the opposite direction.

• For example, when the pillow reaches the -th nperson, TA will pass the pillow to the -th n - 1person, then to the -th n - 2person, and so on.

Given two positive integers nand time, return timethe number of the person holding the pillow seconds later.

 

Example 1:

Input: n = 4, time = 5
 Output: 2
 Explanation: The pillow delivery situation in the team is: 1 -> 2 -> 3 -> 4 -> 3 -> 2. 
After 5 seconds, the pillow is passed to the second person.

Example 2:

Input: n = 3, time = 2
 Output: 3
 Explanation: The pillow delivery situation in the team is: 1 -> 2 -> 3. 
After 2 seconds, the pillow is passed to the third person.

 

hint:

• 2 <= n <= 1000
• 1 <= time <= 1000

Method 1: Calculate

n people pass pillows, and $n-1$ time. In the same way, n − 1 n-1needs to be passed from right to left.$n-1$ time. That is,$2\times (n-1)$ Cycle once.

Therefore, $time$ directly modulo the last$2\times (n-1)$ is equivalent to the result of a single round of transmission.

• If $time\le n-1$ , it means it is passing to the right. Pass$0$ times at$1$ , pass$1$ time in$2$，…，传$time$ times are at$time+1$。

• Otherwise, it means passing to the left. Passed to the left $time-(n-1)$ times. Pass to the left$0$ times in$n$ , pass to the left$1$ time at$n-1$ ,..., pass to the left$time-(n-1)$次处于$n-(time-(n-1))=2\ast n-time-1$。

• Time complexity $O(1)$

• Space complexity $O\left(1\right)$

AC code

C++

class Solution {
    
    
public:
    int passThePillow(int n, int time) {
    
    
        time %= (n - 1) * 2;
        return time <= n - 1 ? time + 1 : 2 * n - time - 1;
    }
};

Python

class Solution:
    def passThePillow(self, n: int, time: int) -> int:
        time %= (n - 1) * 2
        return time + 1 if time <= n - 1 else 2 * n - time - 1

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