In this guide, you will get 2582. Pass the Pillow LeetCode Solution with the best time and space complexity. The solution to Pass the Pillow problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
Note: This question is the same as 3178: Find the Child Who Has the Ball After K Seconds.
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(1)
2582. Pass the Pillow LeetCode Solution in C++
class Solution {
public:
int passThePillow(int n, int time) {
// Repeat every (n - 1) * 2 seconds.
time %= (n - 1) * 2;
if (time < n) // Go forward from 1.
return 1 + time;
return n - (time - (n - 1)); // Go backward from n.
}
}; /* code provided by PROGIEZ */
2582. Pass the Pillow LeetCode Solution in Java
class Solution {
public int passThePillow(int n, int time) {
// Repeat every (n - 1) * 2 seconds.
time %= (n - 1) * 2;
if (time < n) // Go forward from 1.
return 1 + time;
return n - (time - (n - 1)); // Go backward from n.
}
} // code provided by PROGIEZ
2582. Pass the Pillow LeetCode Solution in Python
class Solution:
def passThePillow(self, n: int, time: int) -> int:
# Repeat every (n - 1) * 2 seconds.
time %= (n - 1) * 2
if time < n: # Go forward from 1.
return 1 + time
return n - (time - (n - 1)) # Go backward from n. # code by PROGIEZ
