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

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2506 Discussions

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  • shinatip_t1
     + 0 comments
          Scanner sc = new Scanner(System.in);
        String a = sc.next();
        String b = sc.next();
        HashMap<String, Integer> mapA = new HashMap<>();
        HashMap<String, Integer> mapB = new HashMap<>();
        for (int i = 0; i < a.length(); i++) {
            String letter = String.valueOf(a.charAt(i)).toLowerCase();
            if(mapA.containsKey(letter)){
                Integer sum = mapA.get(letter);
                mapA.put(letter, sum += 1);
            }else{
                mapA.put(letter, 1);
            }
        }
        for (int i = 0; i < b.length(); i++) {
            String letter = String.valueOf(b.charAt(i)).toLowerCase();
            if(mapB.containsKey(letter)){
                Integer sum = mapB.get(letter);
                mapB.put(letter, sum += 1);
            }else{
                mapB.put(letter, 1);
            }
        }
        
        boolean isAnagrams = true;
        for(Map.Entry entry : mapA.entrySet()){
            if(entry.getValue() != mapB.get(entry.getKey()) || mapA.size() != mapB.size()){
                isAnagrams = false;
            }
        }
        System.out.println(isAnagrams ? "Anagrams" : "Not Anagrams");

  • maragouni_saiki1
     + 1 comment

    import java.io.; import java.util.;

    public class Solution {

    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String a=sc.nextLine().toLowerCase();
    String b=sc.nextLine().toLowerCase();
    if(a.length()!=b.length()){
        System.out.println("Not Anagrams");
        return;
    }
    int chararr[] = new int[26];
    int charbrr[] = new int[26];
    for(int i=0;i<26;i++)
        chararr[i]=0;
    for(int i=0;i<a.length();i++){
        chararr[a.charAt(i)-97]++;
        charbrr[b.charAt(i)-97]++;

    }
    for(int i=0;i<a.length();i++){
        if(chararr[a.charAt(i)-97]!=charbrr[a.charAt(i)-97]){
            System.out.println("Not Anagrams");
            break;
        }else{
            System.out.println("Anagrams");
            break;
        }
    }
}

    }

  • dhdhdcbsndnvcb
     + 0 comments

    Hello! I put here my solution! I know maybe it's not the best solution but with I'm open to opinions! :)

    import java.util.Scanner;

    public class Solution {

    static boolean isAnagram(String a, String b) {
    boolean check;
    String a1 = a.toLowerCase();
    String b2 = b.toLowerCase();
    char[] sortA1 = a1.toCharArray();
    char[] sortB2 = b2.toCharArray();
    java.util.Arrays.sort(sortA1); 
    java.util.Arrays.sort(sortB2); 
     a1 = new String(sortA1);
     b2 = new String(sortB2);
    if (a1.equals(b2) ) {
        check = true;
    } else {
        check = false;
    }
    return check;
}

public static void main(String[] args) {

    Scanner scan = new Scanner(System.in);
    String a = scan.next();
    String b = scan.next();
    scan.close();
    boolean ret = isAnagram(a, b);
    System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
}

    }

  • vinayipod12
     + 1 comment

    why it not covering test cases failed 3/17

    static boolean isAnagram(String a, String b) {
        // Complete the function
        String c=a.toLowerCase();
        String d=b.toLowerCase();
        String e= new StringBuilder(b.toLowerCase()).reverse().toString();   
         return (c.equalsIgnoreCase(e)||c.equalsIgnoreCase(d)? true : false);
            
        }

  • TracyHIbbert34
     + 0 comments

    This is a great exercise to reinforce understanding of string manipulation, character frequency, and case insensitivity in coding! 11xplayblack