Hackerrank - Java Solution - Java Anagrams Solution - Online Judge Solution

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Tuesday, December 12, 2017

Hackerrank - Java Solution - Java Anagrams Solution

Hackerrank - Java Solution - Java Anagrams Solution


Problem Name - Java Anagrams
Problem Link -  Java Anagrams 
Level - String

Java Anagram Full Code
mport java.io.*;
import java.util.*;

public class Solution {
    static boolean isAnagram(String a, String b) {
        if ((a == null || b == null) || (a.length() != b.length())) {
            return false;
        }
        final char[] ARRAY_A = a.toUpperCase().toCharArray();
        final Map map = new HashMap<>();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_A[i])) {
                map.put(ARRAY_A[i], (map.get(ARRAY_A[i]) + 1));
            } else {
                map.put(ARRAY_A[i], 1);
            }
        }
        final char[] ARRAY_B = b.toUpperCase().toCharArray();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_B[i])) {
                Integer count = map.get(ARRAY_B[i]);
                if (count == 0) {
                    return false;
                } else {
                    count --;
                    map.put(ARRAY_B[i], count);
                }
            } else {
                return false;
            }
        }
        return true;
    }
    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" );
    }
}


Java Anagram Only Submitted Code
static boolean isAnagram(String a, String b) {
        if ((a == null || b == null) || (a.length() != b.length())) {
            return false;
        }
        final char[] ARRAY_A = a.toUpperCase().toCharArray();
        final Map map = new HashMap<>();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_A[i])) {
                map.put(ARRAY_A[i], (map.get(ARRAY_A[i]) + 1));
            } else {
                map.put(ARRAY_A[i], 1);
            }
        }
        final char[] ARRAY_B = b.toUpperCase().toCharArray();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_B[i])) {
                Integer count = map.get(ARRAY_B[i]);
                if (count == 0) {
                    return false;
                } else {
                    count --;
                    map.put(ARRAY_B[i], count);
                }
            } else {
                return false;
            }
        }
        return true;
    }


Output: 
Hackerrank - Java Solution - Java Anagrams Solution


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