Check if two arrays are cyclic permutations

Question

Given two arrays, how do you check if one is a cyclic permutation of the other?

For example, given a = [1, 2, 3, 1, 5], b = [3, 1, 5, 1, 2], and c = [2, 1, 3, 1, 5] we have that a and b are cyclic permutations but c is not a cyclic permutation of either.

Note: the arrays may have duplicate elements.

Answer 1

The standard trick here is to concatenate one of the arrays with itself, and then try to find the 2nd array in the concatenated array.

For example, 'a' concatenated with itself is:

[1, 2, 3, 1, 5, 1, 2, 3, 1, 5]

Since you do see 'b' in this array starting from the 3rd element then a and b are cyclic permutations.

Answer 2

If A and B are cyclic permutations of each other, A will be found in doubled list BB (as will B in AA).

Answer 3

The efficient way to handle large amounts of data, is to transform each of them into a 'canonical' form then compare to see of they're equal. For this problem you can choose as the canonical form of all the rotated permutations the one that 'sorts smallest'.

So the canonical form for 'a' and 'b' is [1, 2, 3, 1, 5] which are equal so they are acyclic permutations.

The canonical form for 'c' is [1, 3, 1, 5, 2] which is different.

Answer 4

Here is simple adhoc approach to findout cyclic permutations with O(n) time complexity.

a = [1, 2, 3, 1, 5], b = [3, 1, 5, 1, 2]

Find index of b[0] in a[], lets say index is 'x'.Then start navigating in both the array's. a[] starts from index 'x' and b[] starts from '0'. Such that both of them must have same values. If not, they are not cyclic. Here is the sample code.

 public class CyclicPermutation {

    static char[] arr = { 'A', 'B', 'C', 'D' };
    static char[] brr = { 'C', 'D', 'K', 'B' };
    boolean dec = false;

    public static void main(String[] args) {
        boolean avail = true;
        int val = getFirstElementIndex(brr[0]);
        if(val ==Integer.MIN_VALUE){
            avail = false; 
            return;
            }

        for (int i = val, j = 0; j <= brr.length-1; ) {
            if (i > arr.length-1) {
                i = 0;
            }
            if (arr[i] == brr[j]) {
                i++;

                j++;

            } else {
                avail = false;
                System.out.println(avail);
                return;
            }


        }

   System.out.println(avail);
    }

    public static int getFirstElementIndex(char c) {

        for (int i = 0; i <= arr.length; i++) {
            if (arr[i] == c) {
                return i;
            }
        }
        return Integer.MIN_VALUE;
    }


}