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;

def makingAnagrams(s1, s2):
    for i in s1:
        if i in s2:
            s1=s1.replace(i,"",1)
            s2=s2.replace(i,"",1)
    return(len(s1)+len(s2))

Here is my simple python 3 solution

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))