Anagram checking implementation

Question

Given two strings, check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains same characters, only the order of characters can be different. For example, “act” and “tac” are anagram of each other.

Input:

The first line of input contains an integer T denoting the number of test cases. Each test case consist of two strings in 'lowercase' only, in a separate line.

Output:

Print "YES" without quotes if the two strings are anagram else print "NO".

Constraints:

1 ≤ T ≤ 30

1 ≤ |s| ≤ 100

Example:

Input:

2

geeksforgeeks

forgeeksgeeks

allergy

allergic

Output:

YES

NO

My approach:

/*package whatever //do not write package name here */

import java.util.Scanner;
import java.io.IOException;
import java.util.HashMap;

class GFG {

    private static String isAnagram (String str1, String str2)
        {
            HashMap <Character, Integer>occurs = new HashMap<>();

            if (str1.length() != str2.length())
                {
                    return "NO";
                }

            for (char ch: str1.toCharArray())
                {
                    if (!(occurs.containsKey(ch)))
                        {
                            occurs.put(ch,1);
                        }
                    else
                        {
                            int count = occurs.get(ch);
                            occurs.put(ch,count + 1);
                        }
                }

            for (char ch: str2.toCharArray())
                {
                    if (!(occurs.containsKey(ch)))
                        {
                            return "NO";
                        }
                    else
                        {
                            int count = occurs.get(ch);
                            count  = count - 1;
                            if (count < 0)
                                {
                                    return "NO";
                                }
                            else
                                {
                                    occurs.put(ch,count);
                                }
                        }
                }

        return "YES";    
        }

    public static void main (String[] args) throws IOException {
        Scanner sc = new Scanner (System.in);
        int numTests = sc.nextInt();

        for (int i = 0; i < numTests; i++)
            {
                String str1 = sc.next();
                String str2 = sc.next();
                System.out.println(isAnagram(str1, str2));
            }
    }
}

I have the following questions with regards to the above code:

1) How can I further improve my approach?

2) Is there a better way to solve this question?

3) Are there any grave code violations that I have committed?

4) Can space and time complexity be further improved?

Reference

Answer 1

Your use of the counting system using a HashMap is a functional approach to solving the problem, and algorithmically it's relatively good (has time complexity of \$O(N)\$). But, for a prolem like this, it's easier to normalize each input value in to a sorted array or string, than in to a counting HashMap.

Consider the following function:

public static final String normalize(value) {
    if (value == null) {
        return null;
    }
    char[] chars = value.toCharArray();
    Arrays.sort(chars);
    return new String(chars);
}

Now, all values that are anagrams will have the same normalize result (they have the same characters in the same sorted order).

Your code would simply become:

private static String isAnagram (String str1, String str2) {
    return Objects.equals(normalize(str1), normalize(str2));
}

Note that in this case, Objects is java.util.Objects.

Further, by putting the normalization in to a separate function it reduces the amount of logic duplication in your code.

Now, you ask if there are any grave violations... and I have to answer "yes". In Java it is common practice to put { braces on the same line as the code block definition. For example your code:

        for (char ch: str1.toCharArray())
            {
                if (!(occurs.containsKey(ch)))
                    {
                        occurs.put(ch,1);
                    }
                else
                    {
                        int count = occurs.get(ch);
                        occurs.put(ch,count + 1);
                    }
            }

should be:

        for (char ch: str1.toCharArray()) {
            if (!(occurs.containsKey(ch))) {
                occurs.put(ch,1);
            } else {
                int count = occurs.get(ch);
                occurs.put(ch,count + 1);
            }
        }