Java implementation of the problem!!

import java.io.*;

import java.util.*;

import java.text.*;

import java.math.*;

import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
    Scanner sc=new Scanner(System.in);
    String s1=sc.nextLine();
    String s2=sc.nextLine();
    int cArr[]=new int[26];
    int cArr1[]=new int[26];
    for(int i=0;i<s1.length();i++)cArr[s1.charAt(i)-97]++;
    for(int i=0;i<s2.length();i++)cArr1[s2.charAt(i)-97]++;
    int count=0;
    for(int i=0;i<26;i++)count+=Math.abs(cArr[i]-cArr1[i]);
    System.out.println(count);
}

}

4 clean lines - python3

from collections import Counter
counts = Counter(input())
counts.subtract(input())
print(sum(abs(x) for x in counts.values()))

passed all test cases in 0s

string in1,in2;
cin>>in1>>in2;
int array[26] = {0};
for(int i=0;i<in1.length();i++)
    array[in1[i]-'a']++;
for(int i=0;i<in2.length();i++)
    array[in2[i]-'a']--;
int sum = 0;
for(int i=0;i<26;i++)
    sum+=abs(array[i]);
cout<<sum;
return 0;

simple

def makingAnagrams(s1, s2):
    # Complete this function
    k=[c for c in s1]
    l=list()
    for m in s2:
        if m in k:
            k.remove(m)
        else:
            l.append(m)
    
    
    return (len(k)+len(l))

java solution satisfying all test cases

static int makingAnagrams(String s1, String s2){
    int l1,l2,i,j,c=0,e=0;
    l1=s1.length();
    l2=s2.length();
    char[] c1= s1.toCharArray();
    char[] c2 = s2.toCharArray();
    for(i=0;i<l1;i++) {
        for(j=0;j<l2;j++) {
            if(c1[i]==c2[j]){
                c++;
                 e++;
                c2[j]='@';
                break;
            }



        }
        c=0;
    }
     **Strong Text**
   return (l1-e)+(l2-e);
}

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String s1 = in.next();
    String s2 = in.next();
    int result = makingAnagrams(s1, s2);
    System.out.println(result);

}

}

Short javascript version

function makingAnagrams(s1, s2) {
    let cnt = {};
    [...s1].forEach(v => { cnt[v] = (cnt[v] || 0) + 1 });
    [...s2].forEach(v => { cnt[v] = (cnt[v] || 0) - 1 });
    return Object.values(cnt).reduce((a, v) => a + Math.abs(v), 0);
}

def Making_Anagrams(s1, s2):

same = 0
s1 = list(s1)
s2 = list(s2)
for i in s1:
    if i in s2:
        s2.remove(i)
        same += 1
return (len(s1) - same) + len(s2)

working java solution

static int makingAnagrams(String s1, String s2){
    // Complete this function
    int[] temp = new int[26];
    for(int i=0,j=0;i<s1.length() || j<s2.length();i++,j++) 
    {            
        if(i<s1.length())                
            temp[s1.charAt(i)-'a']++;            
        if(j<s2.length())                
            temp[s2.charAt(j)-'a']--;        
    }           
    int count=0;        
    for(int i=0;i<26;i++)            
        count+=Math.abs(temp[i]);        
    return count;

}

I took a slightly less compact approach while leveraging defaultdict, which inits each letter to zero before it's counted. The count of each letter in the first string is totaled, then the count of each letter in the second string is subtracted. Any count that is nonzero represents the net surplus or deficit of letters between the two strings, the total of which is the answer.

(It is trivial to then show exactly what letters were in that counted symmetric difference set.)

def makingAnagrams(s1, s2):
    d = defaultdict(int)
    for c in s1:
        d[c] += 1
    for c in s2:
        d[c] -= 1
    cnt = 0
    for c in d:
        cnt += abs(d[c])
    return cnt