Java Program To Check Whether Two Strings Are Anagram

Write a function to check whether two given strings are anagrams of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different.

For example, “abcd” and “dabc” are an anagram of each other.

Program to Check Two Strings Are Anagram Of Each Other in Java

Below is the Program to check if two strings are anagrams Java using Sorting:

Java

// Java program to check whether two strings
// are anagrams of each other
import java.io.*;
import java.util.Arrays;
import java.util.Collections;

class GFG {
    // Function to check whether two strings
    // are anagram of each other
    static boolean areAnagram(char[] str1, char[] str2)
    {
        // Get lengths of both strings
        int n1 = str1.length;
        int n2 = str2.length;

        // If length of both strings is not
        // same, then they cannot be anagram
        if (n1 != n2)
            return false;

        // Sort both strings
        Arrays.sort(str1);
        Arrays.sort(str2);

        // Compare sorted strings
        for (int i = 0; i < n1; i++)
            if (str1[i] != str2[i])
                return false;

        return true;
    }

    // Driver Code
    public static void main(String args[])
    {
        char str1[] = { 't', 'e', 's', 't' };
        char str2[] = { 't', 't', 'e', 'w' };

        // Function Call
        if (areAnagram(str1, str2))
            System.out.println("The two strings are"
                               + " anagram of each other");
        else
            System.out.println("The two strings are not"
                               + " anagram of each other");
    }
}

Output

The two strings are not anagram of each other

Explanation of the Program:

In this method, we first sort both strings and then compare both of them. If both string are equal that means both are Anagram else both strings are not Anagram.

The complexity of the above method

Time Complexity: O(NLogN)
Auxiliary space: O(1). 

We strongly recommend that you click here and practice it, before moving on to the solution.

Other Methods to Write Anagram Program in Java

There are certain methods to write an anagram program in Java as mentioned below:

1. Count Characters
2. Count Characters using one array
3. Using HashMap()

1. Count characters for Checking Anagram

This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bits and there can be 256 possible characters. 

1. Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
2. Iterate through every character of both strings and increment the count of characters in the corresponding count arrays.
3. Compare count arrays. If both count arrays are the same, then return true.

Below is the implementation of the above idea:

Java

// Java program to check if two strings
// are anagrams of each other
import java.io.*;
import java.util.*;

// Driver Class
class GFG {
    static int NO_OF_CHARS = 256;

    // Function to check whether two strings
    // are anagram of each other
    static boolean areAnagram(char str1[], char str2[])
    {
        // Create 2 count arrays and initialize
        // all values as 0
        int count1[] = new int[NO_OF_CHARS];
        Arrays.fill(count1, 0);
        int count2[] = new int[NO_OF_CHARS];
        Arrays.fill(count2, 0);
        int i;

        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i < str1.length && i < str2.length;
             i++) {
            count1[str1[i]]++;
            count2[str2[i]]++;
        }

        // If both strings are of different length.
        // Removing this condition will make the
        // program fail for strings like "aaca"
        // and "aca"
        if (str1.length != str2.length)
            return false;

        // Compare count arrays
        for (i = 0; i < NO_OF_CHARS; i++)
            if (count1[i] != count2[i])
                return false;

        return true;
    }

    // Driver code
    public static void main(String args[])
    {
        char str1[] = ("geeksforgeeks").toCharArray();
        char str2[] = ("forgeeksgeeks").toCharArray();

        // Function call
        if (areAnagram(str1, str2))
            System.out.println("The two strings are"
                               + "anagram of each other");
        else
            System.out.println("The two strings are not"
                               + " anagram of each other");
    }
}

Output

The two strings areanagram of each other

The complexity of the above method:

Time Complexity: O(n)
Auxiliary space: O(n)

2. Count Characters Using one Array

The above implementation can be further to use only one count array instead of two. We can increment the value in the count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are anagrams of each other. Thanks to Ace for suggesting this optimization. 

Below is the implementation of the above method:

Java

// Java program to check if two strings
// are anagrams of each other

// Driver Class
class GFG {
    static int NO_OF_CHARS = 256;

    // Function to check if two strings
    // are anagrams of each other
    static boolean areAnagram(char[] str1, char[] str2)
    {
        // Create a count array and initialize
        // all values as 0
        int[] count = new int[NO_OF_CHARS];
        int i;

        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i < str1.length; i++) {
            count[str1[i] - 'a']++;
            count[str2[i] - 'a']--;
        }

        // If both strings are of different
        // length. Removing this condition
        // will make the program fail for
        // strings like "aaca" and "aca"
        if (str1.length != str2.length)
            return false;

        // See if there is any non-zero
        // value in count array
        for (i = 0; i < NO_OF_CHARS; i++)
            if (count[i] != 0) {
                return false;
            }
        return true;
    }

    // Driver code
    public static void main(String[] args)
    {
        char str1[] = "geeksforgeeks".toCharArray();
        char str2[] = "forgeeksgeeks".toCharArray();

        // Function call
        if (areAnagram(str1, str2))
            System.out.print("The two strings are "
                             + "anagram of each other");
        else
            System.out.print("The two strings are "
                             + "not anagram of each other");
    }
}

Output

The two strings are anagram of each other

The complexity of the above method

Time Complexity: O(n)
Auxiliary space: O(n)

3. Using HashMap()

We can optimize the space complexity of the above method by using HashMap instead of initializing 256 characters array. So in this approach, we will first count the occurrences of each unique character with the help of HashMap for the first string. Then we will reduce the count of each character while we encounter them in the second string. Finally, if the count of each character in the hash map is 0 then it means both strings are anagrams else not.

Below is the code for the above approach:

Java

// Java program to check if two
// strings are anagrams of each other
import java.io.*;
import java.util.*;

// Driver Class
class GFG {
    // main function
    public static boolean areAnagram(String a, String b)
    {
        // Check if both string has same length or not
        if (a.length() != b.length()) {
            return false;
        }

        // Creating a HashMap containing Character as Key
        // and Integer as Value. We will be storing
        // character as Key and count of character as Value.
        HashMap<Character, Integer> map = new HashMap<>();

        // Loop over all character of first string and put
        // in HashMap.
        for (int i = 0; i < a.length(); i++) {
            // Check if HashMap already contain the current
            // character or not
            if (map.containsKey(a.charAt(i))) {
                // If contains then increase count by 1
                map.put(a.charAt(i),
                        map.get(a.charAt(i)) + 1);
            }
            else {
                // else put that character in map and set
                // count to 1 as character is encountered
                // first time
                map.put(a.charAt(i), 1);
            }
        }

        // Now loop over String b
        for (int i = 0; i < b.length(); i++) {

            // Check if HashMap already contain the current
            // character or not
            if (map.containsKey(b.charAt(i))) {

                // If contains reduce count of that
                // character by 1 to indicate that current
                // character has been already counted as
                // idea here is to check if in last count of
                // all characters in last is zero which
                // means all characters in String a are
                // present in String b.
                map.put(b.charAt(i),
                        map.get(b.charAt(i)) - 1);
            }
        }

        // Extract all keys of HashMap/map
        Set<Character> keys = map.keySet();

        // Loop over all keys and check if all keys are 0
        // as it means that all the characters are present
        // in equal count in both strings.
        for (Character key : keys) {
            if (map.get(key) != 0) {
                return false;
            }
        }
        // Returning True as all keys are zero
        return true;
    }
    public static void main(String[] args)
    {
        String str1 = "geeksforgeeks";
        String str2 = "forgeeksgeeks";

        // Function call
        if (areAnagram(str1, str2))
            System.out.print("The two strings are "
                             + "anagram of each other");
        else
            System.out.print("The two strings are "
                             + "not anagram of each other");
    }
}

Output

The two strings are anagram of each other

The complexity of the above method:

Time Complexity: O(n)
Auxiliary space: O(n) because using HashMap

Reference

Please refer complete article on Check whether two strings are anagram of each other for more details!