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  1. Strings: Making Anagrams
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Strings: Making Anagrams

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2143 Discussions

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  • tan_191
     + 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;
}

  • biren_yadav114
     + 0 comments

    my python solution -

    def makeAnagram(a, b):

    # Write your code here
start = time.time()
count=0
dict_a = {}
dict_b = {}
for i in a:
    if i in dict_a:
        dict_a[i]+=1
    else:
        dict_a[i]=1

for i in b:
    if i in dict_b:
        dict_b[i]+=1
    else:
        dict_b[i]=1

temp_list = list(dict_a.keys()).copy()

for i in temp_list:

    if i in dict_b:
        count += abs(dict_a[i]-dict_b[i])
        dict_a.pop(i)
        dict_b.pop(i)
    else:
        count+=dict_a[i]
        dict_a.pop(i)

count=count+sum(dict_b.values())
print(f'total time: {time.time()-start:.8f}')
return count

  • brkoppad03
     + 0 comments
    function makeAnagram(a: string, b: string): number {
    // Write your code here
    let dictA = new Map<string, number>();
    let dictB = new Map<string, number>();
    for(let ch of a){
        if (dictA.has(ch)){
            dictA.set(ch, dictA.get(ch)+1)
        }else{
            dictA.set(ch, 1)
        }
    }
    for(let ch of b){
        if (dictB.has(ch)){
            dictB.set(ch, dictB.get(ch)+1)
        }else{
            dictB.set(ch, 1)
        }
    }
    let deleteDiff = 0;
    dictA.forEach((v, k) =>{
        if (dictB.has(k)){
            deleteDiff += Math.abs(v - dictB.get(k));
        }else{
            deleteDiff +=v
        }
    })
    dictB.forEach((v, k) =>{
        if (!dictA.has(k)){
            deleteDiff += v;
        }
    })
    
    return deleteDiff
}

  • wilson155079
     + 0 comments
        public static int makeAnagram(String a, String b) {
        int[] arrs = new int['z' + 1];
        
        // Traverse a, b
        for (char c : a.toCharArray()) {
            arrs[c]++;
        }
        for (char c : b.toCharArray()) {
            arrs[c]--;
        }
        
        // Extract the additional values
        return Arrays.stream(arrs) //
                        .reduce(0, (v1, v2) -> v1 + Math.abs(v2));
    }

  • ranechabria
     + 0 comments

    Scala

    def makeAnagram(a: String, b: String): Int = {
        val anagram = a intersect b
        (a.length - anagram.length) + (b.length - anagram.length)
    }