HackerRank Anagram problem solution

HackerRank Anagram problem solution – In this HackerRank Anagram problem, two words are anagrams of one another if their letters can be rearranged to form the other word.

Given a string, split it into two contiguous substrings of equal length. Determine the minimum number of characters to change to make the two substrings into anagrams of one another.

Function Description

Complete the anagram function in the editor below.

anagram has the following parameter(s):

• string s: a string

Returns

• int: the minimum number of characters to change or -1.

HackerRank Anagram problem solution in Python.

import sys

def score(word):
        if len(word) & 1:
                return -1
        mid = int(len(word)/2)
        a = word[0:mid]
        b = list(word[mid:])
        num = 0
        for i in range(len(a)):
                try:   
                        b.remove(a[i])
                except ValueError:
                        num += 1
        return num

num = int(sys.stdin.readline())

for i in range(num):
        s = sys.stdin.readline().strip()
        print(score(s))

Anagram problem solution in Java.

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

class Solution {

    public static void main(String[] args) throws IOException{
    	BufferedReader bis = new BufferedReader(new InputStreamReader(System.in));
    	int test = Integer.parseInt(bis.readLine());
    	for(int t=0;t<test;t++){
    		String s = bis.readLine();
    		char temp[]=s.toCharArray();
    	int length = s.length();
    	int fre[][]=new int[26][2];
    	if(length%2!=0){
    		System.out.println("-1");
    		continue;
    	}
    	
    	int mid = length/2;
    	int inc=0;
    	for(int i=0;i<length;i++){
    		if(i==mid)
             	inc++;
    		
                fre[temp[i]-'a'][inc]++;
             }
             int ans=0;
             for(int i=0;i<26;i++){
    		ans=ans+Math.abs(fre[i][0]-fre[i][1]);
    	}
         System.out.println(ans/2);    
             
    		
    	}
    	
        
    }
}

Problem solution in C++.

#include <stdio.h>
#include <string.h>
#include <map>

int main(int argc, char *argv[]) {
  std::map<char,int> a, b;
  static char str[10000+1];
  int n, total;
  size_t len;

  scanf("%d", &n);

  for(int i = 0; i < n; ++i) {
    a.clear();
    b.clear();
    total = 0;

    scanf("%s", str);
    if((len = strlen(str)) % 2 != 0)
      printf("-1n");

    else {
      for(size_t j = 0; j < len/2; ++j)
        a[str[j]] += 1;
      for(size_t j = len/2; j < len; ++j)
        b[str[j]] += 1;

      for(auto it = a.begin(); it != a.end(); ++it) {
        if(b.count(it->first) > 0) {
          if(b[it->first] < it->second) {
            it->second -= b[it->first];
            b[it->first] = 0;
          }
          else {
            b[it->first] -= it->second;
            it->second = 0;
          }
        }
      }

      for(auto it = a.begin(); it != a.end(); ++it)
        total += it->second;

      printf("%dn", total);
    }
  }

  return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int tests,i,temp,len = 0,totalchange = 0;
    char arr[10000];
    int count[26] = {0};
    scanf("%d",&tests);
    while(tests>0){
		getchar();
        scanf("%[^tn]s",arr);
        len = strlen(arr);
        if(len % 2 != 0){
            printf("-1n");
        }else{
            for(i = len/2;i<len;++i){
				temp = (int)arr[i]-97;
                count[temp]++;
            }
            for(i = 0;i<len/2;++i){
				temp = (int)arr[i] - 97;
				if(count[temp] == 0)continue;
                	count[temp]--;
            }
            for(i = 0;i<26;++i){
                totalchange = totalchange + count[i];
            }
            printf("%dn",totalchange);
        }
        --tests;
        for(i = 0;i<26;++i){
			count[i] = 0;
        }
        totalchange = 0;
    }
    return 0;
}