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 then - 1th
person, then to then - 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
Approach 01:
-
C++
-
Python
#include <bits/stdc++.h> using namespace std; class Solution { public: int passThePillow(int num_people, int time_elapsed) { // Calculate the effective time within one complete cycle of passing the pillow int cycle_time = (num_people - 1) * 2; int effective_time = time_elapsed % cycle_time; // If the effective time is less than the number of people, the pillow is moving forward if (effective_time < num_people) { return 1 + effective_time; } // If the effective time is greater than or equal to the number of people, the pillow is moving backward return num_people - (effective_time - (num_people - 1)); } };
class Solution: def passThePillow(self, num_people: int, time_elapsed: int) -> int: # Calculate the effective time within one complete cycle of passing the pillow cycle_time = (num_people - 1) * 2 effective_time = time_elapsed % cycle_time # If the effective time is less than the number of people, the pillow is moving forward if effective_time < num_people: return 1 + effective_time # If the effective time is greater than or equal to the number of people, the pillow is moving backward return num_people - (effective_time - (num_people - 1))
Time Complexity
-
Modulo Operation:
Calculating
effective_time
using modulo operation takes \( O(1) \) time. -
Conditional Checks:
Checking the condition
if effective_time < num_people
and the subsequent operations inside if-else block take \( O(1) \) time. -
Overall Time Complexity:
The overall time complexity is \( O(1) \) since all operations are constant time operations.
Space Complexity
-
Space Usage:
The algorithm uses a constant amount of extra space for variables
cycle_time
andeffective_time
, which takes \( O(1) \) space. -
Overall Space Complexity:
The overall space complexity is \( O(1) \) as there are no additional data structures used that grow with input size.