Formatted question description: https://leetcode.ca/all/1806.html
1806. Minimum Number of Operations to Reinitialize a Permutation
Level
Medium
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: nums = [0,1] initially.
After the 1st operation, nums = [0,1]
So it takes only 1 operation.
Example 2:
Input: n = 4
Output: 2
Explanation: nums = [0,1,2,3] initially.
After the 1st operation, nums = [0,2,1,3]
After the 2nd operation, nums = [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
Simply simulate the process. Create two arrays init
of perm
, each of which has length n
, and initialize init[i] = perm[i] = i
for 0 <= i < n
. Then update perm
each time until perm
equals init
, and return the number of operations.
class Solution {
public int reinitializePermutation(int n) {
int[] init = new int[n];
int[] perm = new int[n];
for (int i = 0; i < n; i++) {
init[i] = i;
perm[i] = i;
}
int ops = 0;
while (true) {
ops++;
update(n, perm);
if (Arrays.equals(perm, init))
break;
}
return ops;
}
public void update(int n, int[] perm) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
arr[i] = perm[i / 2];
else
arr[i] = perm[n / 2 + (i - 1) / 2];
}
System.arraycopy(arr, 0, perm, 0, n);
}
}