Strings: Making Anagrams Discussions |

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2 weeks ago+ 0 comments

def makeAnagram(a, b) -> int:
    # Write your code here
    ca, cb = Counter(a), Counter(b)
    diff = (ca|cb) - (ca&cb)
    return sum(diff.values())

2 months ago+ 0 comments

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)

2 months ago+ 0 comments

            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;

2 months ago+ 0 comments

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;
}

3 months ago+ 0 comments

Java15 solution that passes all test cases. Three HashMaps and a dream...

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;
}

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