Medium
You are given an even integer n
. You initially have a permutation perm
of size n
where perm[i] == i
(0-indexed).
In one operation, you will create a new array arr
, and for each i
:
i % 2 == 0
, then arr[i] = perm[i / 2]
.i % 2 == 1
, then arr[i] = perm[n / 2 + (i - 1) / 2]
.You will then assign arr
to perm
.
Return the minimum non-zero number of operations you need to perform on perm
to return the permutation to its initial value.
Example 1:
Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1st operation, perm = [0,1]
So it takes only 1 operation.
Example 2:
Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1st operation, perm = [0,2,1,3]
After the 2nd operation, perm = [0,1,2,3]
So it takes only 2 operations.
Example 3:
Input: n = 6
Output: 4
Constraints:
2 <= n <= 1000
n
is even.public class Solution {
public int reinitializePermutation(int n) {
final int factor = n - 1;
if (factor < 2) {
return 1;
}
int powerOfTwo = 2;
int ops = 1;
while (powerOfTwo != 1) {
powerOfTwo = ((powerOfTwo << 1) % factor);
ops++;
}
return ops;
}
}