Question

Background An anagram is a word phrase, or name formed by rearranging the letters of another, such as the word "cinema", formed from 'iceman". Problem Statement Given two strings s and t, write a function to determine ift is an anagram of S. A sample main function is provided so that you may test your code on sample inputs. Inputs can be provided in the box below your code, please format your input as follows: Example input nose soen Requirements • Spaces, tabs, and casing should be ignored. • The expected input will only contain characters a-z, A-Z, tabs, and spaces. • You are limited to 10 submission attempts. • The inputs are NOT guaranteed to be the same length.

Answer 1

Code You provided is working fine .I explained the code in comments part .

Source code:

import java.util.*;
public class Solution
{
    public boolean isAnagram(String s,String t)
    {
        s=s.toLowerCase(); //converting string into toLowerCase
        t=t.toLowerCase(); //converting string into to lowercase
        int count[]=new int[26]; //count occurance of each Character
        //intially count array is 0 ===>[0,0,0,0,0.......0(26 0's)]
        for(int i=0;i<s.length();i++) //loop-1
        {
            //if any Character in between a-z occurs correspondig index is incremented by one
            //example present character is 'a' count[s.charAt[i]-'a']++=>count[0]++
            //similary present character is 'z' count[s.charAt[i]-'a']++=>count[25]++
            if(s.charAt(i)>='a' && s.charAt(i)<='z')
                count[s.charAt(i)-'a']++;
        }
        for(int i=0;i<t.length();i++) //loop-2
        {
            //if any Character in between a-z occurs correspondig index is decremented by one
            //example present character is 'b' count[s.charAt[i]-'a']++=>count[1]--
            //similary present character is 'y' count[s.charAt[i]-'a']++=>count[24]--
            if(t.charAt(i)>='a' && t.charAt(i)<='z')
                count[t.charAt(i)-'a']--;
        }
        //for Anagram words after performing loop1 ,loop2 count arrays becomes zero [0,0,0,0,0,.....(26 0's)]
        //below loop used to check Anagram conditon
        for(int i=0;i<26;i++)
        {
            //we check whether any extra characters present or not
            //count array must be 0 otherwise either String s or String t contains extra characters so it is not Anagram word
            if(count[i]!=0)
            {
                return false;
            }
        }
        return true;
    }
   public static void main(String[] args) {
       Scanner scnr=new Scanner(System.in);
        Solution test=new Solution();
       String a=scnr.nextLine(); //reading string a
       String b=scnr.nextLine(); //reaing string b
       boolean out=(test.isAnagram(a,b))?true:false; //calling isAnagram function
       System.out.println((out));
   }
}

Code screenshot:

1 import java.util.*; 2 public class Solution 3- { public boolean isAnagram(String s,String t) S=s.toLowerCase(); //converting string into tolowercase t=t.toLowerCase(); //converting string into to lowercase int count[]=new int[26]; //count occurance of each Character //intially count array is ===>[0,0,0,0,0.......0(26 O's)] for(int i=0;i<s.length();i++) // Loop-1 //if any character in between a-z occurs correspondig index is incremented by one //example present character is 'a' count[s.charAt[i]-'a']++=>count ++ //similary present character is 'z' count[s.charAt[i]-'a']++=>count [25]++ if(s.charAt(i)>='a'_&& s.charAt(i)<='Z') count[s.charAt(i)-'a']++; for(int i=0;i<t.length();i++) //loop-2 //if any character in between a-z occurs correspondig index is decremented by one //example present character is 'b' count[s.charAt[i]-'a']++=>count[1]-- //similary present character is 'y' count[s.charAt[i]-'a']++=>count [24]-- if(t.charAt(i)>='a'_&& t.charAt(i)<='2') count[t.charAt(i)-'a']--; //for Anagram words after performing Loopi Loop2 count arrays becomes zero [0,0,0,0,0,..... (26 O's) //below Loop used to check Anagram conditon for(int i=0;i<26;i++) //we check wheter any extra characters present or not //count array must be o otherwise either String s or string t contains extra characters so it is not Anagram word, if(count[i]!=0) return false; return true;

public static void main(String[] args) { Scanner scnr=new Scanner(System.in); Solution test=new solutions string_a=scr.nextLine(); //reading string a string b=scnr.nextLine(); //reaing string by boolean out=(test.isAnagram(a,b))?true:false; //calling isAnagram function System.out.println((out));

Output

run-1:

race Care true -

run-2:

assa ssaas false