还原排列的最少操作步数

/*
 * Problem: Given an even number n, there exists a permutation perm of length n
 * where perm[i] == i (0-indexed). In one step, a new array arr is created as follows:
 * - If i % 2 == 0, then arr[i] = perm[i / 2]
 * - If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2]
 * The task is to find the minimum number of steps to return perm back to its original state.
 *
 * Solution using Java Streams:
 * - Initialize perm with values 0 to n-1.
 * - Create arr using stream operations for each transformation.
 * - Keep track of steps until the permutation is restored.
 */

import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public int reinitializePermutation(int n) {
        int[] perm = IntStream.range(0, n).toArray();
        int steps = 0;
        int[] arr = new int[n];
        while (true) {
            steps++;
            arr = IntStream.range(0, n).map(i -> i % 2 == 0 ? perm[i / 2] : perm[n / 2 + (i - 1) / 2]).toArray();
            if (Arrays.equals(arr, IntStream.range(0, n).toArray())) break;
            perm = arr;
        }
        return steps;
    }
}