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  1. Prepare
  2. Algorithms
  3. Strings
  4. Anagram
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Anagram

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  • pradeep_tarakar
     + 0 comments

    My Intuition based easy C++ solution:

    int anagram(string s) {
    if(s.length()%2 != 0)return -1;
    unordered_map<char,int> mp1;
    unordered_map<char,int> mp2;
    int changes = 0;
    string s1 = s.substr(0,s.size()/2);
    string s2 = s.substr(s.size()/2);
    for(int i=0; i<s1.size();i++){
        mp1[s1[i]]++;
        mp2[s2[i]]++;
    }
    for(auto key : mp1){
        if(mp2.find(key.first) == mp2.end()){
            changes += key.second;
        }else if(key.second > mp2[key.first]){
            changes += key.second - mp2[key.first];
        }
    }
    return changes;
}

  • alban_tyrex
     + 0 comments

    My c++ solution using map, here is the explanation : https://youtu.be/0-xHzWDVAME

    int anagram(string s) {
    if(s.size() % 2 == 1) return -1;
    map<char, int> mp;
    int ans = 0;
    for(int i = 0; i < s.size() / 2; i++) mp[s[i]]++;
    for(int i = s.size() / 2; i < s.size(); i++){
        if(mp[s[i]] != 0) mp[s[i]]--;
        else ans++;
    }
    return ans;
}

  • emhhacker
     + 1 comment

    My Java solution:

    public static int anagram(String s) {
        if (s.length() % 2 != 0) return -1; // If length is odd, return -1

        int[] frequency = new int[26]; // Array for character frequencies

        int mid = s.length() / 2;
        for (int i = 0; i < mid; i++) {
            frequency[s.charAt(i) - 'a']++;  // Count chars in first half
            frequency[s.charAt(i + mid) - 'a']--; // Subtract chars in second half
        }

        int changes = 0;
        for (int count : frequency) {
            if (count > 0) changes += count; // Count excess chars that need replacement
        }

        return changes;
    }

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def anagram(s):
    if len(s) % 2 == 1:
        return -1
    s1 = list(s[:len(s) // 2])
    s2 = list(s[len(s) // 2:])
    letters = set(s1 + s2)
    same = 0
    for letter in letters:
        same += min(s1.count(letter), s2.count(letter))
    return (len(s1) + len(s2) - 2 * same) // 2

  • vutrantrung14821
     + 0 comments
    simple and fullly
def anagram(s):
    t=0
    n=len(s)
    if n%2!=0:
        return -1
    a=s[:n//2]
    b=s[n//2:]
    dicta=dict()
    dictb=dict()
    for i in range(n//2):
        if a[i] not in dicta:
            dicta[a[i]]=1
        else:
            dicta[a[i]]+=1
        if b[i] not in dictb:
            dictb[b[i]]=1
        else:
            dictb[b[i]]+=1
    for i in dicta:
       if i in dictb:
        if dicta[i]>=dictb[i]:
            t+=dictb[i]
        else:
            t+=dicta[i]
    t=n//2-t
    return t