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

02 Jun, 2021
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++
Java
Python3
C#
Javascript
// 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 timesvoid 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 Codeint main(){    string s1 = "you";    string s2 = "ara";     int x = 6;     // Function call    isConversionPossible(s1, s2, x);     return 0;}
// Java implementation to check// that a given string can be// converted to another string// by circular clockwise shift// of each character by atmost// X timesimport 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 timesstatic 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 Codepublic 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 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 timesdef 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 Codeif __name__ == "__main__":    s1 = "you"    s2 = "ara"    x = 6         # Function Call    result = isConversionPossible(s1, s2, x)         if result:        print("YES")    else:        print("NO")
// C# implementation to check// that a given string can be// converted to another string// by circular clockwise shift// of each character by atmost// X timesusing System; class GFG{     // Function to check that all// characters of s1 can be// converted to s2 by circular// clockwise shift atmost X timesstatic 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 Codepublic static void Main (){    String s1 = "you";    String s2 = "ara";     int x = 6;     // Function call    isConversionPossible(s1, s2, x);}} // This code is contributed by chitranayal
<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 timesfunction 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 Codelet s1 = "you";let s2 = "ara";let x = 6; // Function callisConversionPossible(s1, s2, x); // This code is contributed by avanitrachhadiya2155</script>
YES
 
Time Complexity:O(N),N=Length(S1)
Auxiliary Space:O(1)
Ganeshchowdharysadanala
ukasp
aditya7409
avanitrachhadiya2155
ASCII
rotation
Strings
Strings
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