1806. Minimum Number of Operations to Reinitialize a Permutation
Description
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
:
- If
i % 2 == 0
, thenarr[i] = perm[i / 2]
. - If
i % 2 == 1
, thenarr[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.
Solution
minimum-number-of-operations-to-reinitialize-a-permutation.py
class Solution:
def reinitializePermutation(self, n: int) -> int:
arr = [i for i in range(n)]
desired = [i for i in range(n)]
count = 0
while count == 0 or arr != desired:
tmp = [0] * n
for i in range(n):
if i % 2 == 0:
tmp[i] = arr[i // 2]
else:
tmp[i] = arr[n // 2 + (i-1) // 2]
arr = tmp
count += 1
return count