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

Anagram

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  • 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;
}

  • quangluan0305201
     + 0 comments

    include

    using namespace std;

    int test(string s) { if (s.length() % 2 == 1) return -1; int n = s.length(); map mp; for (int i = 0; i < n / 2; i++) mp[s[i]]++; for (int i = n / 2; i < n; i++) if (mp[s[i]] > 0) mp[s[i]]--; int dem = 0; for (auto it : mp) dem += it.second; return dem; }

    int main() { int t; cin >> t; while (t--) { string s; cin >> s; cout << test(s) << endl; } return 0; }

  • eduardo_ojeda_p1
     + 0 comments

    Java 8 This is one of the easiest way to solve this problem.

     public static int anagram(String s) {
        int n = s.length();
        
        if(n % 2 != 0){
            return -1;
        }
        
        //Split the main string
        String left = s.substring(0, n/2);
        String right = s.substring(n/2, n);
        
        //Iterate the left string to verify if the letters are contained into the 
        // right string
        for(int i = 0; i < n/2; i++){
            if(right.contains(left.substring(i, i + 1))){
                right = right.replaceFirst(left.substring(i, i + 1), "");
            }
        }
        return right.length();
        
        

    }

  • vbthani
     + 0 comments

    for Python3 Platform

    def anagram(s):
    if (len(s) % 2 != 0):
        return -1
    else:
        lst = list(s[len(s)//2:])
        
        for i in s[:len(s)//2]:
            if (i in lst):
                lst.remove(i)
        
        return len(lst)

q = int(input())
for j in range(q):
    s = input()
    
    result = anagram(s)
    
    print(result)

  • ricardo1512
     + 0 comments
    def anagram(s):
    n = len(s)
    if n % 2 != 0:
        return -1
    A = s[:n//2]
    B = list(s[n//2:])
    for a in A:
        if a in B:
            B.remove(a)
    return len(B)