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Java Exercises: Check if two specified strings are isomorphic or not

Java Basic: Exercise-185 with Solution

Write a Java program to check if two given strings are isomorphic or not.

Two strings are called isomorphic if the letters in one string can be remapped to get the second string. Remapping a letter means replacing all occurrences of it with another letter but the ordering of the letters remains unchanged. No two letters may map to the same letter, but a letter may map to itself.
Example-1: The words "abca" and "zbxz" are isomorphic because we can map 'a' to 'z', 'b' to 'b' and 'c' to 'x'.
Example-1: The words "bar" and "foo" are not isomorphic because we can map 'f' to 'b', 'o' to 'a' and 'o' to 'r'.

Sample Solution:

Java Code:

import java.util.*;
public class Solution {  
   public static void main(String[] args) {
        String str1 = "abca";
		String str2 = "zbxz";
		System.out.println("Is "+str1 +" and "+str2 +" are Isomorphic? "+is_Isomorphic(str1, str2));
    }

  public static boolean is_Isomorphic(String str1, String str2) {
        if (str1 == null || str2 == null || str1.length() != str2.length())
			return false;
        Map<Character, Character> map = new HashMap<>();
        
        for (int i = 0; i < str1.length(); i++) {
            char char_str1 = str1.charAt(i), char_str2 = str2.charAt(i);
            if (map.containsKey(char_str1)) 
			{
                if (map.get(char_str1) != char_str2)
					return false;
            }
			else 
			{
                if (map.containsValue(char_str2)) 
					return false;
             map.put(char_str1, char_str2);
            }
        }
        
        return true;
    }
}

Sample Output:

Is abca and zbxz are Isomorphic? true

Flowchart:

Flowchart: Java exercises: Check if two specified strings  are isomorphic or not.

Java Code Editor:

Company:  LinkedIn

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Java: Tips of the Day

How to remove leading zeros from alphanumeric text?

Regex is the best tool for the job; what it should be depends on the problem specification. The following removes leading zeroes, but leaves one if necessary (i.e. it wouldn't just turn "0" to a blank string).

s.replaceFirst("^0+(?!$)", "")

The ^ anchor will make sure that the 0+ being matched is at the beginning of the input. The (?!$) negative lookahead ensures that not the entire string will be matched.

Test harness:

String[] in = {
    "01234",         // "[1234]"
    "0001234a",      // "[1234a]"
    "101234",        // "[101234]"
    "000002829839",  // "[2829839]"
    "0",             // "[0]"
    "0000000",       // "[0]"
    "0000009",       // "[9]"
    "000000z",       // "[z]"
    "000000.z",      // "[.z]"
};
for (String s : in) {
    System.out.println("[" + s.replaceFirst("^0+(?!$)", "") + "]");
}

Ref: https://bit.ly/2Qdcl8a