Isomorphic String in Java

In this tutorial, we will understand what is meant by isomorphic String in java. We will also see a Java program to find out if the string is isomorphic or not. It is one of the popular interview questions for top-rated IT companies like Amazon, Accenture, Google, etc. This problem enhances the thinking power and problem-solving ability of students. The intention is to check the thinking – critical, and logical of the student.

Isomorphic String

The string is referred to as isomorphic if the letters of one string can be mapped to obtain the second string. By Mapping, we mean replacing all existences of a letter with another letter without changing the order of the letters.It should benoted that no two letters may map to the identical letter but a letter may map to itself.

Now, let us try to grasp it through an example.

Let us consider, string1 as DEDFE and string2 as WVZ. Now try to map string1 to string2.

Character	Mapping
D	W
E	V
F	Z

Replace the letters of the string

D ->W, E ->V, D ->W, F -> Z, E ->V.

Recalling the definition of isomorphic, we observe that string1 and string2 are isomorphic to one another.

Example 2:

Let us consider, string1 is EFEFGE and string2 is WVZU. Now try to map string1 to string2.

Character	Mapping
E	W
F	V
G	Z
E	U

Replacing the letters,

E ->W, F ->V, E ->W, F -> V, G ->Z, E ->U.

Recalling the definition, we can say that string1 and string2 are not isomorphic to one another.

Example 3:

Let us consider, string1 as SERVICE and string2 asTUVWXYZ. Now try to map string1 to string2.

Character	Mapping
S	T
E	U
R	V
V	W
I	X
C	Y
E	Z

Replacing the letters,

S ->T, E ->U, R ->V, I ->X, C ->Y, E ->Z.

Recalling the definition, we can say that string1 and string2 are not isomorphic to one another. Two Es in string 1 is mapped with two different letters in string 2.

Set of Rules

1. A mapping table is to be taken into accountthat would map every character from the first string to the only character in the second string.
2. Another table is considered; call it mapped before the table to record every character from the second string which is already connected to a character from the second string.
3. The first stringis read character by character.
4. Get the mapping if the character gets presented in the mapping table.
5. The equivalent character from the second string is read.
6. Return false if the mapping and the studied character from the 2^nd string are different.
7. Otherwise, check if the studied character from the first string does not exist in the mapping table and take the following action:
   1. Read the equivalent character from the subsequent string.
   1. Return false if the character resides in the table named mappedBefore.
   1. Otherwise, include the read character from the first table and the read character from the second table in the mapping table.
8. Go to step no. 3 and continue.
9. Return true.

Implementation:

Now, let us see a java program to understand the explained concept.

import java.util.*;  
// creation of the class named IsomorphicString
public class IsomorphicString
{    


public static void main(String args[])   
{  
//strings to check whether they are isomorphic in nature or not    


String strg1 = "SERVICE", strg2 = "TUVWXYZ";  


System.out.println("Are "+strg1 +" and "+strg2 +" Isomorphic? "+isIsomorphicString(strg1, strg2));  


}  
//function to check if the given string is isomorphic or not  


public static booleanisIsomorphicString(String strg1, String strg2)   
{  


if (strg1 == null || strg2 == null || strg1.length() != strg2.length())  


//it returns false if string1 and string2 are null and of varying lengths
return false;  


//creation of an instance of Map  
Map<Character, Character> map = new HashMap<>();  
for (int j = 0; j < strg1.length(); j++)   
{  
//finding the character of the string at i-th position  
char char_strg1 = strg1.charAt(j), char_strg2 = strg2.charAt(j);  
//checks whether a particular key (i-th character) is being mapped into the Map or not  


if (map.containsKey(char_strg1))   
{  
if (map.get(char_strg1) != char_strg2)  
//returns false if the character of the first string and the character of the second string are not equal
return false;  
}  
else   
{  
//checking if the specific keyi.e (i-th character) is being mapped into the Map or not      
if (map.containsValue(char_strg2))            
return false;  
//adding key and value to the map  
.put(char_strg1, char_strg2);  
}  
}  
return true;  
}  
}  

Output:

Explanation: The given two strings are not isomorphic as it violates the definition of an isomorphic string. There are two E in string 1 mapped to different letters in string 2.