Rotated Version of Two Arrays

Most Frequent Element in an Array

Problem Description

You are given two arrays. Assume that there are no duplicates in each of arrays.
Determine if two arrays are a rotated version of each other.

Example:

Input: A = [1, 2, 3, 4, 5, 6, 7], B = [4, 5, 6, 7, 1 ,2 ,3]
Output: True

Input: A = [1, 2, 3, 4, 5, 6, 7], B = [1, 3, 5, 7, 9, 2, 4]
Output: False

Solutions

One solution that comes up in my mind is to get an element from the first array, then see if that element can be found in the second array. If yes, we use that element as a start point in both arrays and then use two pointers, one for each array, to go through all elements to see if they are all match.

bool isRotation(std::vector<int> a, std::vector<int> b){
    int idx = -1;

    if(a.size() != b.size())
        return false;

    for(int i = 0; i < b.size(); i++){
        if(b[i] == a[0]){
            idx = i;
            break;
        }
    }

    if(idx == -1)
        return false;

    for(int i = 0; i < a.size(); i++){
        if(a[i] != b[idx])
            return false;

        if(++idx == b.size())
            idx = 0;
    }

    return true;
}

Pseudo code in Python

def is_rotation(A, B):
    if A.length != B.length:
        return false

    key = A[0];
    key_i = -1;
    for i from 0 to B.length - 1:
        if B[i] == key;
            key_i = i
            break
    if key_i == -1:
        return false

    for i from 0 to A.length - 1:
        j = (key_i + i) % A.length
        if A[i] != B[j]:
            return false

    return true