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  1. Prepare
  2. Java
  3. Strings
  4. Java Anagrams
  5. Discussions

Java Anagrams

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2516 Discussions

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  • jayeshborase701
     + 0 comments

    Very Simple and Easy logic

    static boolean isAnagram(String a, String b) { if(a.length()!=b.length()) { return false; }

        int [] check = new int[26];
    for(int i =0;i<a.length();i++)
    {
        check[a.toLowerCase().charAt(i)-'a']++;
        check[b.toLowerCase().charAt(i)-'a']--;
    }
    for(int i = 0;i<26;i++)
    {
        if(check[i]!=0){
            return false;
        }
    }
    return true;
  }

  • vineetprash
     + 0 comments
    static boolean isAnagram(String a, String b) {
        // Complete the function
        a = a.toLowerCase();
        b = b.toLowerCase();
        int [] counts_a = new int[27];
        int [] counts_b = new int[27];
        for (int i=0; i<a.length(); i++) counts_a[a.charAt(i)-'a'] += 1;
        for (int i=0; i<b.length(); i++) counts_b[b.charAt(i)-'a'] += 1;
        
        if (a.length() != b.length()) return false;
        for (int i=0; i<a.length(); i++) {
            if (counts_a[a.charAt(i)-'a'] != counts_b[a.charAt(i)-'a'] ) {
                return false;
            }
        }
        return true;
    }

  • bluffer
     + 0 comments

    easy to understand:

    static boolean isAnagram(String a, String b) {
        // Complete the function
        if(a.length()!=b.length()){
            return false;
        }
        
        int[] check=new int[26];
        for(int i=0; i<a.length(); i++){
            check[a.toLowerCase().charAt(i)-'a']++;
            check[b.toLowerCase().charAt(i)-'a']--;
        }
        
        for(int i=0; i<26; i++){
            if(check[i]!=0){
                return false;
            }
        }
        
        return true;
        
    }

  • favio_flores_ol1
     + 0 comments

    For u guys

    static boolean isAnagram(String a, String b) {
        // Complete the function
        if(a.length()!=b.length()){
            return false;
        }
        String[] arrayA = a.toUpperCase().split("");
        String[] arrayB = b.toUpperCase().split("");
        
        boolean isAnagram = true;
        
        for (int i = 0; i <= arrayA.length -1; i ++){
            
            boolean existLetter = false;
            
            for (int x = 0; x < arrayB.length; x ++){
                if(arrayA[i].equals(arrayB[x])){
                    existLetter = true;
                    String[] arrayBTemp = new String[arrayB.length - 1];
                    for(int y = 0, f = 0; y < arrayB.length; y++){
                        if(y!=x){
                            arrayBTemp[f++] = arrayB[y] ;
                        }
                    }

                    arrayB = arrayBTemp;
                    break;
                }
            }
            
            if(!existLetter){
                isAnagram = false;
                break;
            }
            
        }
        
        return isAnagram;
    }

  • hoanmai12345
     + 0 comments

    public class HR080803Frequency { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String a = sc.next(); String b = sc.next(); sc.close();

        boolean ret = isAnagram(a, b);
    System.out.println(ret ? "Anagrams" : "Not Anagrams");
}

public static boolean isAnagram(String a, String b) {
    a = a.toLowerCase();
    b = b.toLowerCase();

    if (a.length() != b.length()) return false;

    int[][] compare = new int[26][2]; // 26 chữ cái a-z

    for (int i = 0; i < a.length(); i++) {
        compare[a.charAt(i) - 'a'][0]++;
        compare[b.charAt(i) - 'a'][1]++;
    }

    for (int i = 0; i < 26; i++) {
        if (compare[i][0] != compare[i][1]) return false;
    }
    return true;
}