Here is simple C# code, that does the trick. Here I am deleting characters thats are found from both the strings (1 character each). We will be left with characters that are now found in both.

public static int makeAnagram(string a, string b) { string sS1 = ""; string sS2 = "";

    if(a.Length > b.Length)
    {
        sS1 = a;
        sS2 = b;
    }
    else
    {
        sS1 = b;
        sS2 = a;
    }

    for(int i = 0; i < sS1.Length; i++)
    {
        for(int j = 0; j < sS2.Length; j++)
        {
            if(sS1[i] == sS2[j])
            {
                sS1 = sS1.Remove(i,1);
                sS2 = sS2.Remove(j,1);
                i--;
                j--;
                break;
            }
        }
        continue;
    }

    return sS1.Length + sS2.Length;


}

Python solution

def makeAnagram(a, b):
    counter_a = Counter(a)
    counter_b = Counter(b)
    unique_char = set(a).union(set(b))
    
    return sum(abs(counter_a[char] - counter_b[char]) for char in unique_char)

            char[] aArray = a.ToCharArray();
            char[] bArray = b.ToCharArray();  
            Array.Sort(aArray);
            Array.Sort(bArray);  
            int i = 0, j = 0;
            int count = 0;
            while (i < aArray.Length && j < bArray.Length)
            {
                 if (aArray[i] == bArray[j])
                 {
                        i++;
                        j++;
                 }
                 else if (aArray[i] < bArray[j])
                 {
                        i++;
                        count++;
                 }
                else
             {
                     j++;
                     count++;
             }
        }
        if(i < aArray.Length)
        {
             count += aArray.Length - i;
        }
        if(j < bArray.Length)
        {
             count += bArray.Length - j;
        }

        return count;

Well for this one Very simple solution is to add the Char count in hasmap and remove which is there on the first one.

public static int makeAnagram(String a, String b) {
// Write your code here
        int deletecount = 0;
        HashMap<Character,Integer> checkFreq = new HashMap<>();

         for (char c : a.toCharArray()) {
                checkFreq.put(c, checkFreq.getOrDefault(c, 0) + 1);
        }

        // Subtract frequency using second string
        for (char c : b.toCharArray()) {
                checkFreq.put(c, checkFreq.getOrDefault(c, 0) - 1);
        }

        for(int count : checkFreq.values()){
                deletecount += Math.abs(count);
        }

        return deletecount;
}