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.
Naive Approach: The simplest approach to solve the problem is to reversing every possible substring of the string X and for each reversal, check if both the strings are equal. If found to be true after any reversal, then print “Yes”. Otherwise, print “No”.
Time Complexity: O(N2)
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 |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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