How do you check if two strings are anagrams in Java?

Q: How do you check if two strings are anagrams in Java?

Check if Two Strings are Anagrams Anagram: Two strings that contain the same characters in the same frequency. "listen" and "silent" → Anagrams "hello" and "world" → Not Anagrams Method 1: Sort and Compare (Simplest) java public static boolean isAnagram(String str1, String str2) { if (str1 == null || str2 == null) return false; if (str1.length() != str2.length()) return false; char arr1 = str1.toLowerCase().toCharArray(); char arr2 = str2.toLowerCase().toCharArray(); Arrays.sort(arr1); Arrays.sort(arr2); return Arrays.equals(arr1, arr2); } System.out.println(isAnagram("listen", "silent")); //

Answer

Check if Two Strings are Anagrams

Anagram: Two strings that contain the same characters in the same frequency. "listen" and "silent" → Anagrams "hello" and "world" → Not Anagrams

Method 1: Sort and Compare (Simplest)

Java

public static boolean isAnagram(String str1, String str2) {
    if (str1 == null || str2 == null) return false;
    if (str1.length() != str2.length()) return false;

    char[] arr1 = str1.toLowerCase().toCharArray();
    char[] arr2 = str2.toLowerCase().toCharArray();

    Arrays.sort(arr1);
    Arrays.sort(arr2);

    return Arrays.equals(arr1, arr2);
}

System.out.println(isAnagram("listen", "silent")); // true
System.out.println(isAnagram("triangle", "integral")); // true
System.out.println(isAnagram("hello", "world"));   // false

Method 2: Character Frequency Array (Optimal O(n))

Java

public static boolean isAnagramFrequency(String str1, String str2) {
    if (str1.length() != str2.length()) return false;

    int[] count = new int[26];  // for lowercase a-z

    for (char c : str1.toLowerCase().toCharArray()) {
        count[c - 'a']++;  // increment for str1
    }
    for (char c : str2.toLowerCase().toCharArray()) {
        count[c - 'a']--;  // decrement for str2
    }

    // If anagram — all counts should be 0
    for (int c : count) {
        if (c != 0) return false;
    }
    return true;
}

Method 3: HashMap (Handles Unicode/spaces)

Java

public static boolean isAnagramMap(String str1, String str2) {
    str1 = str1.toLowerCase().replaceAll("\\s", "");
    str2 = str2.toLowerCase().replaceAll("\\s", "");

    if (str1.length() != str2.length()) return false;

    Map<Character, Integer> freq = new HashMap<>();

    for (char c : str1.toCharArray()) {
        freq.put(c, freq.getOrDefault(c, 0) + 1);
    }
    for (char c : str2.toCharArray()) {
        if (!freq.containsKey(c)) return false;
        freq.put(c, freq.get(c) - 1);
        if (freq.get(c) == 0) freq.remove(c);
    }

    return freq.isEmpty();
}

Method 4: Java 8 Streams

Java

public static boolean isAnagramStream(String str1, String str2) {
    if (str1.length() != str2.length()) return false;

    Map<Integer, Long> freq1 = str1.toLowerCase().chars()
        .boxed()
        .collect(Collectors.groupingBy(c -> c, Collectors.counting()));

    Map<Integer, Long> freq2 = str2.toLowerCase().chars()
        .boxed()
        .collect(Collectors.groupingBy(c -> c, Collectors.counting()));

    return freq1.equals(freq2);
}

Test Cases

Java

@Test
public void testAnagram() {
    // Anagrams
    assertTrue(isAnagram("listen",   "silent"));
    assertTrue(isAnagram("triangle", "integral"));
    assertTrue(isAnagram("Astronomer", "Moon starer"));  // with spaces
    assertTrue(isAnagram("abc",      "bca"));
    assertTrue(isAnagram("DEBIT CARD", "BAD CREDIT"));

    // Not anagrams
    assertFalse(isAnagram("hello",   "world"));
    assertFalse(isAnagram("rat",     "car"));

    // Edge cases
    assertFalse(isAnagram("abc",     "ab"));   // different lengths
    assertFalse(isAnagram(null,      "abc"));
    assertTrue(isAnagram("",        ""));
}