Intuitive, Readable Python code: O(N) complexity.

def makeAnagram(a, b):
    # Write your code here
    a_list = list(a)
    b_list = list(b)
    remove_b = 0
    for l in b_list: 
        if l in a_list: 
            a_list.remove(l)
        else:
            remove_b +=1 
    remove_a = len(a_list) # unmatched letters remained in this list
    return remove_a + remove_b

Python solution using one dictionary to count all the letters in the first string, remove all the letters in the second string, and them summing up the absolute counts of each letter for the answer.

def makeAnagram(a, b):
    letters = {}
    for _ in a: letters[_] = letters.get(_, 0) + 1
    for _ in b: letters[_] = letters.get(_, 0) - 1
    return sum(map(abs, letters.values()))

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)