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. Afer five seconds, the pillow is given to the 2nd person.
Example 2:
Input: n = 3, time = 2 Output: 3 Explanation: People pass the pillow in the following way: 1 -> 2 -> 3. Afer two seconds, the pillow is given to the 3rd person.
Constraints:
2 <= n <= 1000
1 <= time <= 1000
Solution: Math
It takes n – 1 seconds from 1 to n and takes another n – 1 seconds back from n to 1.
So one around takes 2 * (n – 1) seconds. We can mod time with 2 * (n – 1).
After that if time < n – 1 answer is time + 1, otherwise answer is n – (time – (n – 1))
Time complexity: O(1)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 |
// Author: Huahua class Solution { public: int passThePillow(int n, int time) { time %= (2 * (n - 1)); return time > n - 1 ? n - (time - (n - 1)) : time + 1; } }; |
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