# 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, then arr[i] = perm[i / 2].
• If 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.

## Solutions

Solution 1: Find Pattern + Simulation

We observe the change pattern of the numbers and find that:

1. The even-indexed numbers of the new array are the numbers in the first half of the original array in order;
2. The odd-indexed numbers of the new array are the numbers in the second half of the original array in order.

That is, if the index $i$ of a number in the original array is in the range [0, n >> 1), then the new index of this number is i << 1; otherwise, the new index is (i - (n >> 1)) << 1 | 1.

In addition, the path of number movement is the same in each round of operation. As long as a number (except for numbers $0$ and $n-1$) returns to its original position, the entire sequence will be consistent with the previous one.

Therefore, we choose the number $1$, whose initial index is also $1$. Each time we move the number $1$ to a new position, until the number $1$ returns to its original position, we can get the minimum number of operations.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

• class Solution {
public int reinitializePermutation(int n) {
int ans = 0;
for (int i = 1;;) {
++ans;
if (i < (n >> 1)) {
i <<= 1;
} else {
i = (i - (n >> 1)) << 1 | 1;
}
if (i == 1) {
return ans;
}
}
}
}

• class Solution {
public:
int reinitializePermutation(int n) {
int ans = 0;
for (int i = 1;;) {
++ans;
if (i < (n >> 1)) {
i <<= 1;
} else {
i = (i - (n >> 1)) << 1 | 1;
}
if (i == 1) {
return ans;
}
}
}
};

• class Solution:
def reinitializePermutation(self, n: int) -> int:
ans, i = 0, 1
while 1:
ans += 1
if i < n >> 1:
i <<= 1
else:
i = (i - (n >> 1)) << 1 | 1
if i == 1:
return ans


• func reinitializePermutation(n int) (ans int) {
for i := 1; ; {
ans++
if i < (n >> 1) {
i <<= 1
} else {
i = (i-(n>>1))<<1 | 1
}
if i == 1 {
return ans
}
}
}