# Check if a string can be formed from another string by at most X circular clockwise shifts

Given an integer X and two strings S1 and S2, the task is to check that string S1 can be converted to the string S2 by shifting characters circular clockwise atmost X times.

Input: S1 = “abcd”, S2 = “dddd”, X = 3
Output: Yes
Explanation:
Given string S1 can be converted to string S2 as-
Character “a” – Shift 3 times – “d”
Character “b” – Shift 2 times – “d”
Character “c” – Shift 1 times – “d”
Character “d” – Shift 0 times – “d”

Input: S1 = “you”, S2 = “ara”, X = 6
Output: Yes
Explanation:
Given string S1 can be converted to string S2 as –
Character “y” – Circular Shift 2 times – “a”
Character “o” – Shift 3 times – “r”
Character “u” – Circular Shift 6 times – “a”

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to traverse the string and for each index and find the difference between the ASCII values of the character at the respective indices of the two strings. If the difference is less than 0, then for a circular shift, add 26 to get the actual difference. If for any index, the difference exceeds X, then S2 can’t be formed from S1, otherwise possible.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to check ` `// that a given string can be ` `// converted to another string ` `// by circular clockwise shift ` `// of each character by atmost ` `// X times ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check that all ` `// characters of s1 can be ` `// converted to s2 by circular ` `// clockwise shift atmost X times ` `void` `isConversionPossible(string s1, ` `                          ``string s2, ``int` `x) ` `{ ` `    ``int` `diff, n; ` `    ``n = s1.length(); ` ` `  `    ``// Check for all characters of ` `    ``// the strings whether the ` `    ``// difference between their ` `    ``// ascii values is less than ` `    ``// X or not ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If both the characters ` `        ``// are same ` `        ``if` `(s1[i] == s2[i]) ` `            ``continue``; ` ` `  `        ``// Calculate the difference ` `        ``// between the ASCII values ` `        ``// of the characters ` `        ``diff = (``int``(s2[i] - s1[i]) ` `                ``+ 26) ` `               ``% 26; ` ` `  `        ``// If difference exceeds X ` `        ``if` `(diff > x) { ` `            ``cout << ``"NO"` `<< endl; ` `            ``return``; ` `        ``} ` `    ``} ` ` `  `    ``cout << ``"YES"` `<< endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s1 = ``"you"``; ` `    ``string s2 = ``"ara"``; ` ` `  `    ``int` `x = 6; ` ` `  `    ``// Function call ` `    ``isConversionPossible(s1, s2, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to check  ` `// that a given string can be  ` `// converted to another string  ` `// by circular clockwise shift  ` `// of each character by atmost  ` `// X times  ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to check that all  ` `// characters of s1 can be  ` `// converted to s2 by circular  ` `// clockwise shift atmost X times  ` `static` `void` `isConversionPossible(String s1,  ` `                                 ``String s2, ` `                                 ``int` `x)  ` `{  ` `    ``int` `diff = ``0``, n;  ` `    ``n = s1.length();  ` `     `  `    ``// Check for all characters of  ` `    ``// the strings whether the  ` `    ``// difference between their  ` `    ``// ascii values is less than  ` `    ``// X or not  ` `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `         `  `       ``// If both the characters  ` `       ``// are same  ` `       ``if` `(s1.charAt(i) == s2.charAt(i)) ` `           ``continue``;  ` `        `  `       ``// Calculate the difference  ` `       ``// between the ASCII values  ` `       ``// of the characters  ` `       ``diff = ((``int``)(s2.charAt(i) -  ` `                     ``s1.charAt(i)) + ``26``) % ``26``;  ` `        `  `       ``// If difference exceeds X  ` `       ``if` `(diff > x) ` `       ``{  ` `           ``System.out.println(``"NO"``);  ` `           ``return``;  ` `       ``}  ` `    ``}  ` `    ``System.out.println(``"YES"``); ` `}  ` `     `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``String s1 = ``"you"``;  ` `    ``String s2 = ``"ara"``;  ` ` `  `    ``int` `x = ``6``;  ` ` `  `    ``// Function call  ` `    ``isConversionPossible(s1, s2, x);  ` `} ` `} ` ` `  `// This code is contributed by Ganeshchowdharysadanala `

## Python3

 `# Python3 implementation to check ` `# that the given string can be ` `# converted to another string  ` `# by circular clockwise shift ` ` `  `# Function to check that the  ` `# string s1 can be converted ` `# to s2 by clockwise circular ` `# shift of all characters of  ` `# str1 atmost X times ` `def` `isConversionPossible(s1, s2, x): ` `    ``n ``=` `len``(s1) ` `    ``s1 ``=` `list``(s1) ` `    ``s2 ``=` `list``(s2) ` `     `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Difference between the  ` `        ``# ASCII numbers of characters ` `        ``diff ``=` `ord``(s2[i]) ``-` `ord``(s1[i]) ` `         `  `        ``# If both characters  ` `        ``# are the same ` `        ``if` `diff ``=``=` `0``: ` `            ``continue` `         `  `        ``# Condition to check if the  ` `        ``# difference less than 0 then ` `        ``# find the circular shift by  ` `        ``# adding 26 to it ` `        ``if` `diff < ``0``: ` `            ``diff ``=` `diff ``+` `26` `        ``# If difference between  ` `        ``# their ASCII values ` `        ``# exceeds X ` `        ``if` `diff > x: ` `            ``return` `False` `     `  `    ``return` `True` `     `  ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``s1 ``=` `"you"` `    ``s2 ``=` `"ara"` `    ``x ``=` `6` `     `  `    ``# Function Call ` `    ``result ``=` `isConversionPossible(s1, s2, x) ` `     `  `    ``if` `result: ` `        ``print``(``"YES"``) ` `    ``else``: ` `        ``print``(``"NO"``) `

Output:

```YES
```

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

3

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.