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”  

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. 

Follow the below steps to implement the above idea:

• Declare integer variables diff and n.
• Assign the length of string s1 to variable n.
• Iterate over each character of the strings.
  • If the characters at the current index are the same, continue to the next iteration of the loop.
  • Calculate the difference between the ASCII values of the characters at the current index of both strings.
  • To handle the circular shift, add 26 to the difference and then take the result modulo 26.
  • Check if the resulting difference is greater than x. If yes, output “NO” and return from the function.
• If the loop completes without returning, output “YES” to indicate that the conversion is 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 <bits/stdc++.h>
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")

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
using System;

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[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)
        {
            Console.Write("NO");
            return;
        }
    }
    Console.Write("YES");
}

// Driver Code
public static void Main ()
{
    String s1 = "you";
    String s2 = "ara";

    int x = 6;

    // Function call
    isConversionPossible(s1, s2, x);
}
}

// This code is contributed by chitranayal

Javascript

<script>
// Javascript implementation to check
// that a given string can be
// converted to another string
// by circular clockwise shift
// of each character by atmost
// X times


// Function to check that all
// characters of s1 can be
// converted to s2 by circular
// clockwise shift atmost X times
function isConversionPossible(s1,s2,x)
{
    let 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(let 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 = ((s2[i].charCodeAt(0) -
                     s1[i].charCodeAt(0)) + 26) % 26;

       // If difference exceeds X
       if (diff > x)
       {
           document.write("NO<br>");
           return;
       }
    }
    document.write("YES<br>");
}

// Driver Code
let s1 = "you";
let s2 = "ara";
let x = 6;

// Function call
isConversionPossible(s1, s2, x);

// This code is contributed by avanitrachhadiya2155
</script>

Output:

YES

Time Complexity:O(N),N=Length(S1)
Auxiliary Space:O(1)