Check if two strings can be made equal by reversing a substring of one of the strings

Given two strings X and Y of length N, the task is to check if both the strings can be made equal by reversing any substring of X exactly once. If it is possible, then print “Yes”. Otherwise, print “No”.

Examples:

Input: X = “adcbef”, Y = “abcdef”
Output: Yes
Explanation: Strings can be made equal by reversing the substring “dcb” of string X.

Input: X = “126543”, Y = “123456”
Output: Yes
Explanation: Strings can be made equal by reversing the substring “6543” of string X.

Brute Force Approach:

The brute force approach to solve this problem would be to try every possible substring of X and reverse it, and then check if the reversed substring when replaced in X makes it equal to Y. 

The steps for this approach are:

1. Iterate through every substring of X.
2. Reverse the substring and replace it in X.
3. Check if the modified X is equal to Y. If it is, print “Yes” and return from the function.
4. If the modified X is not equal to Y after iterating through all the substrings, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the strings
// can be made equal or not by
// reversing a substring of X
bool checkString(string X, string Y)
{
    int n = X.length();
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            reverse(X.begin() + i, X.begin() + j + 1);
            if (X == Y) {
                cout << "Yes" << endl;
                return true;
            }
            reverse(X.begin() + i, X.begin() + j + 1);
        }
    }
    cout << "No" << endl;
    return false;
}


// Driver Code
int main()
{
    string X = "adcbef", Y = "abcdef";

    // Function Call
    checkString(X, Y);

    return 0;
}

Java

import java.util.*;

public class Main
{

  // Function to check if the strings
  // can be made equal or not by
  // reversing a substring of X
  static boolean checkString(String X, String Y) {
    int n = X.length();
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        X = reverse(X, i, j);
        if (X.equals(Y)) {
          System.out.println("Yes");
          return true;
        }
        X = reverse(X, i, j);
      }
    }
    System.out.println("No");
    return false;
  }

  // Utility function to reverse the characters
  // in the substring of the given string
  static String reverse(String s, int start, int end) {
    char[] arr = s.toCharArray();
    while (start < end) {
      char temp = arr[start];
      arr[start++] = arr[end];
      arr[end--] = temp;
    }
    return String.valueOf(arr);
  }

  // Driver Code
  public static void main(String[] args) {
    String X = "adcbef", Y = "abcdef";

    // Function Call
    checkString(X, Y);
  }
}

Python3

# Function to check if the strings
# can be made equal or not by
# reversing a substring of X


def checkString(X: str, Y: str) -> bool:
    n = len(X)
    for i in range(n):
        for j in range(i, n):
            X = X[:i] + X[i:j+1][::-1] + X[j+1:]
            if X == Y:
                print("Yes")
                return True
            X = X[:i] + X[i:j+1][::-1] + X[j+1:]
    print("No")
    return False


# Driver Code
if __name__ == '__main__':
    X = "adcbef"
    Y = "abcdef"
    # Function Call
    checkString(X, Y)

C#

// C# program for the above approach
using System;

class Program
{
    // Function to check if the strings
    // can be made equal or not by
    // reversing a substring of X
    static bool CheckString(string X, string Y)
    {
        int n = X.Length;
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            {
                char[] arr = X.ToCharArray();
                Array.Reverse(arr, i, j - i + 1);
                string modifiedX = new string(arr);
                if (modifiedX == Y)
                {
                    Console.WriteLine("Yes");
                    return true;
                }
                Array.Reverse(arr, i, j - i + 1);
            }
        }
        Console.WriteLine("No");
        return false;
    }

    // Driver Code
    static void Main(string[] args)
    {
        string X = "adcbef", Y = "abcdef";

        // Function Call
        CheckString(X, Y);
    }
}

Javascript

  // Function to check if the strings
  // can be made equal or not by
  // reversing a substring of X
  function checkString(X, Y) {
    let n = X.length;
    for (let i = 0; i < n; i++) {
        for (let j = i; j < n; j++) {
            let reversed = X.slice(i, j + 1).split('').reverse().join('');
            let newX = X.slice(0, i) + reversed + X.slice(j + 1);
            if (newX === Y) {
                console.log("Yes");
                return true;
            }
        }
    }
    console.log("No");
    return false;
}

// Driver Code
let X = "adcbef";
let Y = "abcdef";
checkString(X, Y);

Output

Yes

Time Complexity: O(N^2)

Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, follow the steps below to solve the problem:

• Initialize a variable, say L as -1, to store the first index from the left having unequal characters in the two strings.
• Traverse the string X over the range [0, N – 1] using a variable i and if for any index, if the characters in the two strings are found to be unequal, set L = i and break out of the loop.
• Initialize a variable, say R as -1, to store the first index from the right having unequal characters in the two strings.
• Traverse the string X over the range [N – 1, 0] using the variable i and if for any index, if the characters in the two strings are found to be unequal, set R = i and break out of the loop.
• Reverse the characters of the string X over the indices [L, R].
• After completing the above steps, check if both the strings are equal or not. If found to be equal, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the strings
// can be made equal or not by
// reversing a substring of X
bool checkString(string X, string Y)
{
    // Store the first index from
    // the left which contains unequal
    // characters in both the strings
    int L = -1;

    // Store the first element from
    // the right which contains unequal
    // characters in both the strings
    int R = -1;

    // Checks for the first index from
    // left in which characters in both
    // the strings are unequal
    for (int i = 0; i < X.length(); ++i) {

        if (X[i] != Y[i]) {

            // Store the current index
            L = i;

            // Break out of the loop
            break;
        }
    }

    // Checks for the first index from
    // right in which characters in both
    // the strings are unequal
    for (int i = X.length() - 1; i > 0; --i) {
        if (X[i] != Y[i]) {

            // Store the current index
            R = i;

            // Break out of the loop
            break;
        }
    }

    // Reverse the substring X[L, R]
    reverse(X.begin() + L,
            X.begin() + R + 1);

    // If X and Y are equal
    if (X == Y) {
        cout << "Yes";
    }

    // Otherwise
    else {
        cout << "No";
    }
}

// Driver Code
int main()
{
    string X = "adcbef", Y = "abcdef";

    // Function Call
    checkString(X, Y);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if the Strings
// can be made equal or not by
// reversing a subString of X
static void checkString(String X, String Y)
{
    // Store the first index from
    // the left which contains unequal
    // characters in both the Strings
    int L = -1;

    // Store the first element from
    // the right which contains unequal
    // characters in both the Strings
    int R = -1;

    // Checks for the first index from
    // left in which characters in both
    // the Strings are unequal
    for (int i = 0; i < X.length(); ++i) {

        if (X.charAt(i) != Y.charAt(i)) {

            // Store the current index
            L = i;

            // Break out of the loop
            break;
        }
    }

    // Checks for the first index from
    // right in which characters in both
    // the Strings are unequal
    for (int i = X.length() - 1; i > 0; --i) {
        if (X.charAt(i) != Y.charAt(i)) {

            // Store the current index
            R = i;

            // Break out of the loop
            break;
        }
    }

    // Reverse the subString X[L, R]
    X = X.substring(0, L) +
      reverse(X.substring(L, R + 1)) +
      X.substring(R + 1);

    // If X and Y are equal
    if (X.equals(Y)) {
        System.out.print("Yes");
    }

    // Otherwise
    else {
        System.out.print("No");
    }
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}

// Driver Code
public static void main(String[] args)
{
    String X = "adcbef", Y = "abcdef";

    // Function Call
    checkString(X, Y);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to check if the strings
# can be made equal or not by
# reversing a substring of X
def checkString(X, Y):

    # Store the first index from
    # the left which contains unequal
    # characters in both the strings
    L = -1

    # Store the first element from
    # the right which contains unequal
    # characters in both the strings
    R = -1

    # Checks for the first index from
    # left in which characters in both
    # the strings are unequal
    for i in range(len(X)):
        if (X[i] != Y[i]):

            # Store the current index
            L = i

            # Break out of the loop
            break

    # Checks for the first index from
    # right in which characters in both
    # the strings are unequal
    for i in range(len(X) - 1, 0, -1):
        if (X[i] != Y[i]):

            # Store the current index
            R = i

            # Break out of the loop
            break

    X = list(X)

    X = X[:L] + X[R  : L - 1 : -1 ] + X[R + 1:]

    # If X and Y are equal
    if (X == list(Y)):
        print("Yes")

    # Otherwise
    else:
        print("No")

# Driver Code
if __name__ == "__main__" :

    X = "adcbef"
    Y = "abcdef"

    # Function Call
    checkString(X, Y)

# This code is contributed by AnkThon

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if the Strings
// can be made equal or not by
// reversing a subString of X
static void checkString(String X, String Y)
{

    // Store the first index from
    // the left which contains unequal
    // characters in both the Strings
    int L = -1;

    // Store the first element from
    // the right which contains unequal
    // characters in both the Strings
    int R = -1;

    // Checks for the first index from
    // left in which characters in both
    // the Strings are unequal
    for(int i = 0; i < X.Length; ++i)
    {
        if (X[i] != Y[i])
        {

            // Store the current index
            L = i;

            // Break out of the loop
            break;
        }
    }

    // Checks for the first index from
    // right in which characters in both
    // the Strings are unequal
    for(int i = X.Length - 1; i > 0; --i)
    {
        if (X[i] != Y[i])
        {

            // Store the current index
            R = i;

            // Break out of the loop
            break;
        }
    }

    // Reverse the subString X[L, R]
    X = X.Substring(0, L) +
        reverse(X.Substring(L, R + 1 - L)) +
        X.Substring(R + 1);

    // If X and Y are equal
    if (X.Equals(Y))
    {
        Console.Write("Yes");
    }

    // Otherwise
    else
    {
        Console.Write("No");
    }
}

static String reverse(String input)
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;

    for(l = 0; l < r; l++, r--)
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}

// Driver Code
public static void Main(String[] args)
{
    String X = "adcbef", Y = "abcdef";

    // Function Call
    checkString(X, Y);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

      // JavaScript program for the above approach

      // Function to check if the Strings
      // can be made equal or not by
      // reversing a subString of X
      function checkString(X, Y) {
        // Store the first index from
        // the left which contains unequal
        // characters in both the Strings
        var L = -1;

        // Store the first element from
        // the right which contains unequal
        // characters in both the Strings
        var R = -1;

        // Checks for the first index from
        // left in which characters in both
        // the Strings are unequal
        for (var i = 0; i < X.length; ++i) {
          if (X[i] !== Y[i]) {
            // Store the current index
            L = i;

            // Break out of the loop
            break;
          }
        }

        // Checks for the first index from
        // right in which characters in both
        // the Strings are unequal
        for (var i = X.length - 1; i > 0; --i) {
          if (X[i] !== Y[i]) {
            // Store the current index
            R = i;

            // Break out of the loop
            break;
          }
        }

        // Reverse the subString X[L, R]
        X =
          X.substring(0, L) +
          reverse(X.substring(L, R + 1)) +
          X.substring(R + 1);

        // If X and Y are equal
        if (X === Y) {
          document.write("Yes");
        }

        // Otherwise
        else {
          document.write("No");
        }
      }

      function reverse(input) {
        var a = input.split("");
        var l,
          r = a.length - 1;

        for (l = 0; l < r; l++, r--) {
          var temp = a[l];
          a[l] = a[r];
          a[r] = temp;
        }
        return a.join("");
      }

      // Driver Code
      var X = "adcbef",
        Y = "abcdef";

      // Function Call
      checkString(X, Y);

 </script>

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)