Anagram

                            Solution in C :

In  C++  :






#include <string> 
#include <cmath>  
#include <cstdlib>  
#include <algorithm>  
#include <vector>  
#include <string.h>  
#include <utility>  
#include <queue>
#include <stack>
#include <iostream>  
#include <iomanip>  
#include <ctype.h>  
#include <sstream>  
#include <map> 
#include <set> 
#include <stdio.h>
#include <ctype.h>

using namespace std;

#define INF = 2000000000
#define FOR(i,n) for(int i = 0;i < n;i++)
#define CLEAR(x) memset((x),0,sizeof(x))
#define REP(i,a,b) for(int i = (a);i<(b);++i)
#define MP make_pair
#define ALL(a) (a).begin(),(a).end()
#define PB push_back
#define PII pair<int,int>
#define sz(a) (int)(a).size()

typedef long long LL;

int main(){
    ios_base::sync_with_stdio(0);
    int n;
    cin >> n;
    string s;
    int a[30];
    FOR(i,n){
        cin >> s;
        if (s.length()%2 == 1){
            cout << -1 << endl;
            continue;
        }
        
        CLEAR(a);
        FOR(j,s.length()/2){
            a[s[j]-'a']++;
        }
        int cnt = 0;
        FOR(j,s.length()/2){
            if (a[s[s.length()-j-1]-'a'] > 0 ){
                a[s[s.length()-j-1]-'a']--;
            }else
            cnt++;
        }
        cout << cnt << endl;
    }

    return 0;
}







In   Java  :





import java.io.*;
import java.util.*;

public class Solution {

	static void solve() throws IOException {
		int tests = nextInt();
		while (tests-- > 0) {
			String s = nextToken();
			int answer = solve(s);
			out.println(answer);
		}
	}

	private static int solve(String s) {
		if ((s.length() & 1) != 0) {
			return -1;
		}
		int k = s.length() >> 1;
		char[] c1 = s.substring(0, k).toCharArray();
		char[] c2 = s.substring(k, 2 * k).toCharArray();
		int[] cnt1 = get(c1);
		int[] cnt2 = get(c2);
		int result = 0;
		for (int i = 0; i < 256; i++) {
			result += Math.abs(cnt1[i] - cnt2[i]);
		}

		return result >> 1;
	}

	private static int[] get(char[] c1) {
		int[] ret = new int[256];
		for (char cc : c1) {
			++ret[cc];
		}
		return ret;
	}

	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		InputStream input = System.in;
		PrintStream output = System.out;
		File file = new File("a.in");
		if (file.exists() && file.canRead()) {
			input = new FileInputStream(file);
		}
		br = new BufferedReader(new InputStreamReader(input));
		out = new PrintWriter(output);
		solve();
		out.close();
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}
}








In    C:







#include<string.h>
#include<stdio.h>
int main()
{
    int t;
    scanf("%d",&t);
    
    while(t--)
    {
              char str[10005];
              int fre_a[27]={0},fre_b[27]={0},i,sum=0;
              scanf("%s",str);
              
              int len=strlen(str);
              
              if(len%2)
              printf("-1\n");
              
              else{
                   for(i=0;i<len/2;i++)
                   fre_a[str[i]-'a']++;
                   
                   for(i=len/2;i<len;i++)
                   fre_b[str[i]-'a']++;
                   
                   for(i=0;i<27;i++)
                   if(fre_b[i] && fre_b[i]>fre_a[i])
                   sum+=(fre_b[i]-fre_a[i]);
                   printf("%d\n",sum);
                   }
    }
    
    return 0;
}








In  Python3  :







T = int(input())
for i in range(0,T):
	al= (input())
	l = len(al)
	if (l%2==0):
		half = int(l/2)
		a = al[:half]
		b = al[half:]
		count = 0
		for c in a:
			if c in b:
				pos = b.find(c)
				b=b[:pos]+b[pos+1:]
			else:
			 count += 1
		print (count)
	else:
		print (-1)