Check if a string can be obtained by rotating another string d places

Given two strings str1 and str2 and an integer d, the task is to check whether str2 can be obtained by rotating str1 by d places (either to the left or to the right).

Examples:

Input: str1 = “abcdefg”, str2 = “cdefgab”, d = 2 
Output: Yes 
Rotate str1 2 places to the left.

Input: str1 = “abcdefg”, str2 = “cdfdawb”, d = 6 
Output: No 

Approach: An approach to solve the same problem has been discussed here. In this article, reversal algorithm is used to rotate the string to the left and to the right in O(n). If any one of the rotations of str1 is equal to str2 then print Yes else print No.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to reverse an array from left
// index to right index (both inclusive)
void ReverseArray(string& arr, int left, int right)
{
    char temp;
    while (left < right) {
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

// Function that returns true if str1 can be
// made equal to str2 by rotating either
// d places to the left or to the right
bool RotateAndCheck(string& str1, string& str2, int d)
{

    if (str1.length() != str2.length())
        return false;

    // Left Rotation string will contain
    // the string rotated Anti-Clockwise
    // Right Rotation string will contain
    // the string rotated Clockwise
    string left_rot_str1, right_rot_str1;
    bool left_flag = true, right_flag = true;
    int str1_size = str1.size();

    // Copying the str1 string to left rotation string
    // and right rotation string
    for (int i = 0; i < str1_size; i++) {
        left_rot_str1.push_back(str1[i]);
        right_rot_str1.push_back(str1[i]);
    }

    // Rotating the string d positions to the left
    ReverseArray(left_rot_str1, 0, d - 1);
    ReverseArray(left_rot_str1, d, str1_size - 1);
    ReverseArray(left_rot_str1, 0, str1_size - 1);

    // Rotating the string d positions to the right
    ReverseArray(right_rot_str1, 0, str1_size - d - 1);
    ReverseArray(right_rot_str1, str1_size - d, str1_size - 1);
    ReverseArray(right_rot_str1, 0, str1_size - 1);

    // Comparing the rotated strings
    for (int i = 0; i < str1_size; i++) {

        // If cannot be made equal with left rotation
        if (left_rot_str1[i] != str2[i]) {
            left_flag = false;
        }

        // If cannot be made equal with right rotation
        if (right_rot_str1[i] != str2[i]) {
            right_flag = false;
        }
    }

    // If both or any one of the rotations
    // of str1 were equal to str2
    if (left_flag || right_flag)
        return true;
    return false;
}

// Driver code
int main()
{

    string str1 = "abcdefg";
    string str2 = "cdefgab";

    // d is the rotating factor
    int d = 2;

    // In case length of str1 < d
    d = d % str1.size();

    if (RotateAndCheck(str1, str2, d))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Python3

# Python3 implementation of the approach

# Function to reverse an array from left
# index to right index (both inclusive)
def ReverseArray(arr, left, right) :

    while (left < right) :
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left += 1;
        right -= 1;

# Function that returns true if str1 can be
# made equal to str2 by rotating either
# d places to the left or to the right
def RotateAndCheck(str1, str2, d) :

    if (len(str1) != len(str2)) :
        return False;

    # Left Rotation string will contain
    # the string rotated Anti-Clockwise
    # Right Rotation string will contain
    # the string rotated Clockwise
    left_rot_str1 = []; right_rot_str1 = [];
    left_flag = True; right_flag = True;
    str1_size = len(str1);

    # Copying the str1 string to left rotation string
    # and right rotation string
    for i in range(str1_size) :
        left_rot_str1.append(str1[i]);
        right_rot_str1.append(str1[i]);

    # Rotating the string d positions to the left
    ReverseArray(left_rot_str1, 0, d - 1);
    ReverseArray(left_rot_str1, d, str1_size - 1);
    ReverseArray(left_rot_str1, 0, str1_size - 1);

    # Rotating the string d positions to the right
    ReverseArray(right_rot_str1, 0, str1_size - d - 1);
    ReverseArray(right_rot_str1,
                 str1_size - d, str1_size - 1);
    ReverseArray(right_rot_str1, 0, str1_size - 1);

    # Comparing the rotated strings
    for i in range(str1_size) :

        # If cannot be made equal with left rotation
        if (left_rot_str1[i] != str2[i]) :
            left_flag = False;

        # If cannot be made equal with right rotation
        if (right_rot_str1[i] != str2[i]) :
            right_flag = False;

    # If both or any one of the rotations
    # of str1 were equal to str2
    if (left_flag or right_flag) :
        return True;
    return False;

# Driver code
if __name__ == "__main__" :

    str1 = list("abcdefg");
    str2 = list("cdefgab");

    # d is the rotating factor
    d = 2;

    # In case length of str1 < d
    d = d % len(str1);

    if (RotateAndCheck(str1, str2, d)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by AnkitRai01

C#

using System;

// C# program to check if a string is two time
// rotation of another string.
public class Test {

  static string ReverseArray(char[] arr, int left, int right)
  {
    char temp;
    while (left < right) {
      temp = arr[left];
      arr[left] = arr[right];
      arr[right] = temp;
      left++;
      right--;
    }
    return String.Join("",arr);
  }

  // Method to check if string2 is obtained by
  // string 1
  public static bool RotateAndCheck(string str1, string str2, int d)
  {
    if (str1.Length != str2.Length) {
      return false;
    }

    // Left Rotation string will contain
    // the string rotated Anti-Clockwise
    // Right Rotation string will contain
    // the string rotated Clockwise
    string left_rot_str1="", right_rot_str1="";
    bool left_flag = true, right_flag = true;
    int len1 = str1.Length;

    // Copying the str1 string to left rotation string
    // and right rotation string
    for (int i = 0; i < len1; i++) {
      left_rot_str1+=str1[i];
      right_rot_str1+=str1[i];
    }

    // Rotating the string d positions to the left
    left_rot_str1=ReverseArray(left_rot_str1.ToCharArray(), 0, d - 1);
    left_rot_str1=ReverseArray(left_rot_str1.ToCharArray(), d, len1 - 1);
    left_rot_str1=ReverseArray(left_rot_str1.ToCharArray(), 0, len1 - 1);

    // Rotating the string d positions to the right
    right_rot_str1=ReverseArray(right_rot_str1.ToCharArray(), 0, len1 - d - 1);
    right_rot_str1=ReverseArray(right_rot_str1.ToCharArray(), len1 - d, len1 - 1);
    right_rot_str1=ReverseArray(right_rot_str1.ToCharArray(), 0, len1 - 1);

    // Comparing the rotated strings
    for (int i = 0; i < len1; i++) {

      // If cannot be made equal with left rotation
      if (left_rot_str1[i] != str2[i]) {
        left_flag = false;
      }

      // If cannot be made equal with right rotation
      if (right_rot_str1[i] != str2[i]) {
        right_flag = false;
      }
    }

    // If both or any one of the rotations
    // of str1 were equal to str2
    if (left_flag || right_flag)
      return true;
    return false;
  }

  // Driver code
  public static void Main(string[] args)
  {
    string str1 = "abcdefg";
    string str2 = "cdefgab";

    // d is the rotating factor
    int d = 2;

    // In case length of str1 < d
    d = d % str1.Length;

    Console.WriteLine(RotateAndCheck(str1, str2,d) ? "Yes"
                      : "No");
  }
}

// This code is contributed by Aarti_Rathi

Javascript

<script>

// JavaScript implementation of the approach

// Function to reverse an array from left
// index to right index (both inclusive)
function ReverseArray(arr, left, right)
{
    var temp;
    while (left < right)
    {
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

// Function that returns true if str1 can be
// made equal to str2 by rotating either
// d places to the left or to the right
function RotateAndCheck(str1, str2, d)
{
    if (str1.length !== str2.length)
        return false;

    // Left Rotation string will contain
    // the string rotated Anti-Clockwise
    // Right Rotation string will contain
    // the string rotated Clockwise
    var left_rot_str1 = [];
    var right_rot_str1 = [];
    var left_flag = true,
    right_flag = true;
    var str1_size = str1.length;

    // Copying the str1 string to left rotation string
    // and right rotation string
    for(var i = 0; i < str1_size; i++)
    {
        left_rot_str1.push(str1[i]);
        right_rot_str1.push(str1[i]);
    }

    // Rotating the string d positions to the left
    ReverseArray(left_rot_str1, 0, d - 1);
    ReverseArray(left_rot_str1, d, str1_size - 1);
    ReverseArray(left_rot_str1, 0, str1_size - 1);

    // Rotating the string d positions to the right
    ReverseArray(right_rot_str1, 0, str1_size - d - 1);
    ReverseArray(right_rot_str1, str1_size - d,
                 str1_size - 1);
    ReverseArray(right_rot_str1, 0, str1_size - 1);

    // Comparing the rotated strings
    for(var i = 0; i < str1_size; i++)
    {
        // If cannot be made equal with left rotation
        if (left_rot_str1[i] !== str2[i])
        {
            left_flag = false;
        }

        // If cannot be made equal with right rotation
        if (right_rot_str1[i] !== str2[i])
        {
            right_flag = false;
        }
    }

    // If both or any one of the rotations
    // of str1 were equal to str2
    if (left_flag || right_flag)
        return true;

    return false;
}

// Driver code
var str1 = "abcdefg";
var str2 = "cdefgab";

// d is the rotating factor
var d = 2;

// In case length of str1 < d
d = d % str1.length;

if (RotateAndCheck(str1, str2, d))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by rdtank

</script>

Output:

Yes

Time Complexity: O(n), where n represents the size of the string.
Auxiliary Space: O(n), where n represents the size of the string.