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  1. Prepare
  2. Algorithms
  3. Strings
  4. Making Anagrams
  5. Discussions

Making Anagrams

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763 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/B4pZbX0VzzU

    #include <bits/stdc++.h>

using namespace std;

int main()
{
    string fstr, sstr;
    cin >> fstr; cin >> sstr;
    vector<int>fsc(26,0), ssc(26,0);
    int i, result = 0;
    for(i = 0; i < fstr.size(); i++) fsc[fstr[i] - 'a']++;
    for(i = 0; i < sstr.size(); i++) ssc[sstr[i] - 'a']++;
    for(i = 0; i < 26; i++) result += abs(fsc[i] - ssc[i]);
    cout << result << endl;
    return 0;
}

  • iamaradhyakesha1
     + 0 comments

    int makingAnagrams(string s1, string s2) { unordered_map f1,f2;

      for(int i=0;i<s1.length();i++){
       f1[s1[i]]++; }

  for(int i=0;i<s2.length();i++){
       f2[s2[i]]++;}

  int ans=0 ;
  for(char i ='a';i<='z';i++){
    ans+= abs(f1[i]-f2[i]);  
  }
return ans;

    }

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def makingAnagrams(s1, s2):
    same = 0
    letters = set(list(s1) + list(s2))
    for letter in letters:
        print(list(s1).count(letter), list(s2).count(letter), letter)
        same += min(list(s1).count(letter), list(s2).count(letter))
    return len(s1) + len(s2) - 2 * same

  • highsoft_soi
     + 0 comments

    Perl:

    sub makingAnagrams {
    my $s1 = shift;
    my $s2 = shift;
    my (%h1, %h2);
    my $cnt = 0;
    my $sum = 0;
    
    foreach (split("", $s1)) {
        $h1{$_} += 1;
    }
    foreach (split("", $s2)) {
        $h2{$_} += 1;
    }
    foreach my $k (keys %h1) {
        if (exists($h2{$k})) {
            $cnt += abs($h1{$k} - $h2{$k});
            delete $h1{$k};
            delete $h2{$k};
        }
        else {
            $cnt += $h1{$k};
            delete $h1{$k};
        }        
    }
    foreach my $k2 (keys %h2) {
        $sum += $h2{$k2};
    }
    return $cnt + $sum;
}

  • vuhuutuan0
     + 0 comments

    My answer with Typescript, simple

    function makingAnagrams(s1: string, s2: string): number {
    // 0. define hash [m]
    let m = new Map<string, number>()

    // 1. counting [c]+ in [s1]
    // 2. counting [c]- in [s2]
    for (let c of s1) m.set(c, (m.get(c) || 0) + 1)
    for (let c of s2) m.set(c, (m.get(c) || 0) - 1)

    // 3. return sum of diff of each character between [s1]&[s2] that counted in [m]
    return Array.from(m.values()).reduce((p, c) => p + Math.abs(c), 0)
}