public static int makeAnagram(String a, String b) {
    int count = 0;
    
    HashMap<Character, Integer> aCharFreq = new HashMap<Character, Integer>();
    HashMap<Character, Integer> bCharFreq = new HashMap<Character, Integer>();
    HashMap<Character, Integer> toDelCharFreq = new HashMap<Character, Integer>();
    
    for (int i = 0; i < a.length(); i++) { //Get the character frequencies for string a.
        char c = a.charAt(i);
        if (aCharFreq.containsKey(c)) aCharFreq.replace(c, aCharFreq.get(c) + 1);
        else aCharFreq.put(c, 1);
    }
    
    for (int i = 0; i < b.length(); i++) { //Get the character frequencies for string b.
        char d = b.charAt(i);
        if (bCharFreq.containsKey(d)) bCharFreq.replace(d, bCharFreq.get(d) + 1);
        else bCharFreq.put(d, 1);
    }
    
    for (char keyA : aCharFreq.keySet()) { //Now check the character frequencies for string a.
        if (bCharFreq.containsKey(keyA)) toDelCharFreq.put(keyA,Math.abs(bCharFreq.get(keyA) - aCharFreq.get(keyA))); //If b contains the character in a that we're looking at, add the difference between the number of times that character shows up in b and a.
        else toDelCharFreq.put(keyA,aCharFreq.get(keyA)); //Add every character that shows up in a but not b.
    }
            
    for (char keyB : bCharFreq.keySet()) { //Then check the character frequencies for string b.
        if (aCharFreq.containsKey(keyB) && toDelCharFreq.get(keyB) == 0) toDelCharFreq.remove(keyB); //If a and b have an equal number of instances of a character, that character will be part of the anagram set and must be removed from the target character frequencies.
        else if (!aCharFreq.containsKey(keyB)) toDelCharFreq.put(keyB,bCharFreq.get(keyB)); //Add every character that shows up in b but not a.
    }
    
    for (char key : toDelCharFreq.keySet()) count += toDelCharFreq.get(key); //Finally, add up the frequencies of all characters remaining to get the retur value.
    
    return count;
}