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.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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")

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Output:

YES

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