Check whether two strings are equivalent or not according to given condition

Given two strings A and B of equal size. Two strings are equivalent either of the following conditions hold true:
1) They both are equal. Or,
2) If we divide the string A into two contiguous substrings of same size A1 and A2 and string B into two contiguous substrings of same size B1 and B2, then one of the following should be correct:

  • A1 is recursively equivalent to B1 and A2 is recursively equivalent to B2
  • A1 is recursively equivalent to B2 and A2 is recursively equivalent to B1

Check whether given strings are equivalent or not. Print YES or NO.

Examples:



Input : A = “aaba”, B = “abaa”
Output : YES
Explanation : Since condition 1 doesn’t hold true, we can divide string A into “aaba” = “aa” + “ba” and string B into “abaa” = “ab” + “aa”. Here, 2nd subcondition holds true where A1 is equal to B2 and A2 is recursively equal to B1

Input : A = “aabb”, B = “abab”
Output : NO

Naive Solution : A simple solution is to consider all possible scenarios. Check first if the both strings are equal return “YES”, otherwise divide the strings and check if A1 = B1 and A2 = B2 or A1 = B2 and A2 = B1 by using four recursive calls. Complexity of this solution would be O(n2), where n is the size of the string.

Efficient Solution : Let’s define following operation on string S. We can divide it into two halves and if we want we can swap them. And also, we can recursively apply this operation to both of its halves. By careful observation, we can see that if after applying the operation on some string A, we can obtain B, then after applying the operation on B we can obtain A. And for the given two strings, we can recursively find the least lexicographically string that can be obtained from them. Those obtained strings if are equal, answer is YES, otherwise NO. For example, least lexicographically string for “aaba” is “aaab”. And least lexicographically string for “abaa” is also “aaab”. Hence both of these are equivalent.

Below is the implementation of the above approach.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP Program to find whether two strings
// are equivalent or not according to given
// condition
#include <bits/stdc++.h>
using namespace std;
  
// This function returns the least lexicogr
// aphical string obtained from its two halves
string leastLexiString(string s)
{
    // Base Case - If string size is 1
    if (s.size() & 1)
        return s;
  
    // Divide the string into its two halves
    string x = leastLexiString(s.substr(0,
                                        s.size() / 2));
    string y = leastLexiString(s.substr(s.size() / 2));
  
    // Form least lexicographical string
    return min(x + y, y + x);
}
  
bool areEquivalent(string a, string b)
{
  return (leastLexiString(a) == leastLexiString(b));
  
// Driver Code
int main()
{
    string a = "aaba";
    string b = "abaa";
    if (areEquivalent(a, b))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
  
    a = "aabb";
    b = "abab";
    if (areEquivalent(a, b))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find whether two strings
// are equivalent or not according to given
// condition
class GfG 
{
  
// This function returns the least lexicogr
// aphical String obtained from its two halves
static String leastLexiString(String s)
{
    // Base Case - If String size is 1
    if (s.length() == 1)
        return s;
  
    // Divide the String into its two halves
    String x = leastLexiString(s.substring(0,
                                        s.length() / 2));
    String y = leastLexiString(s.substring(s.length() / 2));
  
    // Form least lexicographical String
    return String.valueOf((x + y).compareTo(y + x));
}
  
static boolean areEquivalent(String a, String b)
{
    return !(leastLexiString(a).equals(leastLexiString(b)));
  
// Driver Code
public static void main(String[] args) 
{
    String a = "aaba";
    String b = "abaa";
    if (areEquivalent(a, b))
        System.out.println("Yes");
    else
        System.out.println("No");
  
    a = "aabb";
    b = "abab";
    if (areEquivalent(a, b))
        System.out.println("Yes");
    else
        System.out.println("No");
    }
}
  
/* This code contributed by PrinciRaj1992 */

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 Program to find whether two strings
# are equivalent or not according to given
# condition
  
# This function returns the least lexicogr
# aphical string obtained from its two halves
def leastLexiString(s):
      
    # Base Case - If string size is 1
    if (len(s) & 1 != 0):
        return s
  
    # Divide the string into its two halves
    x = leastLexiString(s[0:int(len(s) / 2)])
    y = leastLexiString(s[int(len(s) / 2):len(s)])
  
    # Form least lexicographical string
    return min(x + y, y + x)
  
def areEquivalent(a,b):
    return (leastLexiString(a) == leastLexiString(b))
  
# Driver Code
if __name__ == '__main__':
    a = "aaba"
    b = "abaa"
    if (areEquivalent(a, b)):
        print("YES")
    else:
        print("NO")
  
    a = "aabb"
    b = "abab"
    if (areEquivalent(a, b)):
        print("YES")
    else:
        print("NO")
  
# This code is contributed by
# Surendra_Gangwar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find whether two strings 
// are equivalent or not according to given 
// condition 
using System;
class GFG 
  
// This function returns the least lexicogr- 
// aphical String obtained from its two halves 
static String leastLexiString(String s) 
    // Base Case - If String size is 1 
    if (s.Length == 1) 
        return s; 
  
    // Divide the String into its two halves 
    String x = leastLexiString(s.Substring(0, 
                               s.Length / 2)); 
    String y = leastLexiString(s.Substring(
                               s.Length / 2)); 
  
    // Form least lexicographical String 
    return ((x + y).CompareTo(y + x).ToString()); 
  
static Boolean areEquivalent(String a, String b) 
    return !(leastLexiString(a).Equals(
             leastLexiString(b))); 
  
// Driver Code 
public static void Main(String[] args) 
    String a = "aaba"
    String b = "abaa"
    if (areEquivalent(a, b)) 
        Console.WriteLine("YES"); 
    else
        Console.WriteLine("NO"); 
  
    a = "aabb"
    b = "abab"
    if (areEquivalent(a, b)) 
        Console.WriteLine("YES"); 
    else
        Console.WriteLine("NO"); 
  
// This code is contributed by PrinciRaj1992 

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP Program to find whether two strings 
// are equivalent or not according to given 
// condition 
  
// This function returns the least lexicogr 
// aphical string obtained from its two halves 
function leastLexiString($s
    // Base Case - If string size is 1 
    if (strlen($s) & 1) 
        return $s
  
    // Divide the string into its two halves 
    $x = leastLexiString(substr($s, 0,floor(strlen($s) / 2))); 
    $y = leastLexiString(substr($s,floor(strlen($s) / 2),strlen($s))); 
  
    // Form least lexicographical string 
    return min($x.$y, $y.$x); 
  
function areEquivalent($a, $b
    return (leastLexiString($a) == leastLexiString($b)); 
  
    // Driver Code 
    $a = "aaba"
    $b = "abaa"
    if (areEquivalent($a, $b)) 
        echo "YES", "\n"
    else
        echo "NO", "\n"
  
    $a = "aabb"
    $b = "abab"
    if (areEquivalent($a, $b)) 
        echo "YES", "\n"
    else
        echo "NO","\n"
  
// This code is contributed by Ryuga
?>

chevron_right


Output:

YES
NO

Time Complexity : O(n), where n is the size of the string.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.