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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
(0indexed).
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 nonzero 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 1^{st} 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 1^{st} operation, perm = [0,2,1,3] After the 2^{nd} 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:
 The evenindexed numbers of the new array are the numbers in the first half of the original array in order;
 The oddindexed 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 $n1$) 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 } } }