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# C++ Program to 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 ``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;``}`

Output:

`Yes`

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

### Approach – String Rotation Check

• Check if the length of both strings is the same. If not, return false.
• If d is greater than the length of the string, set d to d % length of string.
• Create two temporary strings, left_rotate and right_rotate.
• For left_rotate, copy the first d characters of the original string to the end of left_rotate, and then copy the remaining characters to the beginning of left_rotate.
• For right_rotate, copy the last d characters of the original string to the beginning of right_rotate, and then copy the remaining characters to the end of right_rotate.
• Check if either left_rotate or right_rotate is equal to the second string. If either one is equal, return true. Otherwise, return false.

## C++

 `#include ``#include ` `using` `namespace` `std;` `bool` `canRotate(string str1, string str2, ``int` `d) {``   ``int` `n = str1.length();` `   ``// Check if the length of both strings is the same``   ``if` `(n != str2.length()) {``       ``return` `false``;``   ``}` `   ``// If d is greater than the length of the string, set d to d % length of string``   ``d = d % n;` `   ``// Create temporary strings for left and right rotations``   ``string left_rotate = str1.substr(d) + str1.substr(0, d);``   ``string right_rotate = str1.substr(n - d) + str1.substr(0, n - d);` `   ``// Check if either left_rotate or right_rotate is equal to str2``   ``if` `(left_rotate == str2 || right_rotate == str2) {``       ``return` `true``;``   ``}` `   ``return` `false``;``}` `int` `main() {``   ``string str1 = ``"abcdefg"``;``   ``string str2 = ``"cdefgab"``;``   ``int` `d = 2;` `   ``if` `(canRotate(str1, str2, d)) {``       ``cout << ``"Yes"` `<< endl;``   ``} ``else` `{``       ``cout << ``"No"` `<< endl;``   ``}` `   ``return` `0;``}`

Output

```Yes
```

The time complexity is O(n), where n represents the size of the given string.
The auxiliary space is O(n)

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