make two strings anagrams

Question

i solving a challenge of Hacker rank . it about how anagrams. i give two string input and i have to find ...

Print a single integer denoting the number of characters you must delete to make the two strings anagrams of each other.

i have detected if it's anagrams or not and difference. but now can do the rest of it dont have any ideas.please help.

function main() {
    var a = readLine();
    var b = readLine();
    var sum1 = 0 ;
    var sum2 = 0 ;

    for (var i= 0; i<= a.length-1; i++ )
        {
            sum1 = sum1 + a.charCodeAt(i);
        }

       console.log(sum1);


        for (var i= 0; i<= b.length-1; i++ )
        {
            sum2 = sum2 + b.charCodeAt(i);
        }

       console.log(sum2);

    if(sum1== sum2)
        {
            console.log("anagrams");
        }
    else
        {
            console.log("not anagram");
            var diff = sum1 - sum2;
            console.log(diff);

            /// what to do now ?   
        }


}

Answer 1

Here's another approach using Map() of all the 26 alphabets in each string

  function makeAnagram(a, b) {
  let result = 0;
  const alphabets = 'abcdefghijklmnopqrstuvwxyz'.split('');

  const getCharCountMap = str => {
    const strArray = str.split('');
    let charMap = new Map();
    alphabets.forEach(alphabet => charMap.set(alphabet, 0));
    strArray.forEach(letter => charMap.set(letter, charMap.get(letter) + 1));
    return charMap;
  };

  const aMap = getCharCountMap(a);
  const bMap = getCharCountMap(b);

  alphabets.forEach(alphabet => {
    result = result + Math.abs(aMap.get(alphabet) - bMap.get(alphabet));
  });

  return result;
}

Answer 2

You are really just looking for the sum of the differences in later frequency. You can just count the frequencies into an 26 item array (the rules tell you there will only be lower case numbers). Then subtract each array from the other item by item but item and add the whole thing up. Seems like it only need about four lines of code:

function makeAnagram(a, b) {
  const makeCountArray = (str) => [...str].reduce((a, c) => (a[c.charCodeAt(0) - 97]++, a), Array(26).fill(0))
  let a1 = makeCountArray(a)
  let a2 = makeCountArray(b)
  return a1.reduce((a, c, i) => a + Math.abs(c - a2[i]), 0)
}

// test case expected: 30
let count = makeAnagram('fcrxzwscanmligyxyvym', 'jxwtrhvujlmrpdoqbisbwhmgpmeoke')
console.log(count)

Answer 3

I have solved this on hackerRank by using the object approach to count the frequency of letters if you are still looking for a reasonable solution.

function makeAnagrams(a,b){
  let charA=buildcharMap(a)
  let charB=buildcharMap(b)
  let characters=[]
  let counter=0

 for(let char in charA){
   if(charA[char] && charB[char]){
     if(charA[char]===charB[char]){ //same frequency
        continue;
     }
     else{
       if(charA[char]>charB[char]){
         counter=counter+ charA[char]-charB[char]
       }
       else{
         counter=counter+ charB[char]-charA[char]
       }
     } 
    }
    else{
      counter=counter+charA[char]
    }
 }

  for(let char in charB){
    if(charB[char] && charA[char]===undefined){
        counter=counter+charB[char]
    }
  }
  return counter;
}


function buildcharMap(str){
  var charMap={}
  for(let char of str){
    if (charMap[char]===undefined){
      charMap[char]=1
    }
    else{
      charMap[char]+=1
    }
  }
  return charMap
}

console.log(makeAnagrams('cde','abc'))

Answer 4

I have solve this question on hackerearth, i took slightly different approach.

What i have done in this code is that i have check for all the characters and replace the character with "@" sign if both characters in string are same, then i count all the "@" signs in one of the strings that is modified, and then subtracted value of that count multiplied by two(because... they are similar in both strings) from total length of both strings.

Here is my code, hope you can convert it into javascript. ->

import java.util.Scanner;

public class Anagrams {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int Testcases = scanner.nextInt();

        String[] str1 = new String[1000];
        String[] str2 = new String[1000];

        //  Taking input
        for (int i = 0; i < Testcases; i++) {
            str1[i] = scanner.next().toLowerCase();
            str2[i] = scanner.next().toLowerCase();
        }

        //  For Total Length of both strings
        int TotalLength[] = new int[Testcases];

        //  For counting "@" signs
        int count[] = new int[Testcases];

        //  Loop through TestCases
        for (int i = 0; i < Testcases; i++) {

            TotalLength[i] = str1[i].length() + str2[i].length();

            for (int j = 0; j < str1[i].length(); j++) {
                for (int k = 0; k < str2[i].length(); k++) {
                    //  If both characters are similar, then replace those characters with "@" signs
                    if (str1[i].charAt(j) == str2[i].charAt(k)) {
                        str1[i] = str1[i].replaceFirst(Character.toString(str1[i].charAt(j)),"@");
                        str2[i] = str2[i].replaceFirst(Character.toString(str2[i].charAt(k)),"@");
                    }
                }
            }
        }

        //  Counting "@" signs from one string
        for (int i = 0; i < Testcases; i++) {
            count[i] = 0;
            char[] c1 = str1[i].toCharArray();
            for (char c: c1) {
                if(c == '@'){
                    count[i]++;
                }
            }
        }

        //  Output
        for (int i = 0; i < Testcases; i++) {   
            System.out.println(TotalLength[i] - 2*count[i]);
        }

        scanner.close();
    }
}