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  1. Strings: Making Anagrams
  2. Discussions

Strings: Making Anagrams

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  • concealedgloomy
     + 0 comments

    My solution Java 8

    public static int makeAnagram(String a, String b) {

	Map<Character, Integer> charFreq = new HashMap<>();
        int count = 0;

        for (char c : a.toCharArray()) {
            charFreq.put(c, charFreq.get(c) != null ? charFreq.get(c) + 1 : 1);
        }

        for (char c : b.toCharArray()) {
            charFreq.put(c, charFreq.get(c) != null ? charFreq.get(c) - 1 : -1);
        }

        for (char key : charFreq.keySet()) {

            if (charFreq.get(key) != 0) {
                count += charFreq.get(key) < 0 ? charFreq.get(key) * -1 : charFreq.get(key);
            }
        }

        return count;

}

  • abdulquadrioyek1
     + 0 comments
    def makeAnagram(a, b):
    count_a, count_b = Counter(a), Counter(b)
    diff_a = count_a - count_b
    diff_b = count_b - count_a

    return sum(diff_a.values()) + sum(diff_b.values())

  • rctutong
     + 0 comments

    int result = 0; HashMap map1 = new HashMap<>(); HashMap map2 = new HashMap<>(); 

    for(char c:a.toCharArray()){
    map1.put(c, map1.getOrDefault(c, 0)+1);
}
for(char c:b.toCharArray()){
    map2.put(c, map2.getOrDefault(c, 0)+1);
} 
for(Map.Entry<Character,Integer> set : map1.entrySet()){
   if(map2.containsKey(set.getKey())){    
       if(set.getValue()>=map2.get(set.getKey())){
           result += map2.get(set.getKey())*2;
       }else if(set.getValue()<=map2.get(set.getKey())){
           result += set.getValue()*2;
       }
   }
} 
return (a.length()+b.length())- result;

  • nayantara11sure1
     + 0 comments

    C++ solution Using HashMap

    int makeAnagram(string a, string b) {

    map<char,int> m;
int sum=0;
for(int i=0;i<a.size();i++){
    m[a[i]]++;
}
for(int i=0;i<b.size();i++){
    m[b[i]]--;
}
for(auto i:m){
    sum+=abs(i.second);
}
return sum;

    }

  • dylancellison
     + 0 comments

    Javascript solution

    function makeAnagram(a, b) {
  let lettersToRemove = 0;

  for (let i = 0; i < a.length; i++) {
    if (b.includes(a[i])) {
     b = b.replace(a[i], '')
    } else {
      lettersToRemove += 1;
    }
  }

  return lettersToRemove + b.length;
}
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