# Strings: Making Anagrams

### Problem Statement :

```A student is taking a cryptography class and has found anagrams to be very useful. Two strings are anagrams of each other if the first string's letters can be rearranged to form the second string. In other words, both strings must contain the same exact letters in the same exact frequency. For example, bacdc and dcbac are anagrams, but bacdc and dcbad are not.

The student decides on an encryption scheme that involves two large strings. The encryption is dependent on the minimum number of character deletions required to make the two strings anagrams. Determine this number.

Given two strings, a and b, that may or may not be of the same length, determine the minimum number of character deletions required to make a and b anagrams. Any characters can be deleted from either of the strings.

Function Description

Complete the makeAnagram function in the editor below.

makeAnagram has the following parameter(s):

string a: a string
string b: another string
Returns

int: the minimum total characters that must be deleted
Input Format

The first line contains a single string, a.
The second line contains a single string, b.

Constraints

1  <=  | a |, | b |  <= 10^4
The strings a and b consist of lowercase English alphabetic letters, ascii[a-z].```

### Solution :

```                            ```Solution in C :

In C :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
char* a = (char *)malloc(512000 * sizeof(char));
scanf("%s",a);
char* b = (char *)malloc(512000 * sizeof(char));
scanf("%s",b);

int ind1,ind2;
int arr1[26]={0};
int arr2[26]={0};
int count=0;
for(int i=0;i<strlen(a);i++)
{
ind1=a[i]-'a';
arr1[ind1]++;
}
for(int j=0;j<strlen(b);j++)
{
ind2=b[j]-'a';
arr2[ind2]++;
}
for(int i=0;i<26;i++)
{

count+=abs(arr1[i]-arr2[i]);

}
printf("%d",count);
return 0;
}```
```

```                        ```Solution in C++ :

In  C ++ :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int number_needed(string a, string b) {
int i,c[26]={0},c1[26]={};
for(i=0;i<a.length();i++)
{
if(97<=a[i]&&a[i]<=123)
c[a[i]-97]++;
}
for(i=0;i<b.length();i++)
{
if(97<=b[i]&&b[i]<=123)
c1[b[i]-97]++;
}
int s=0;
for(i=0;i<26;i++)
{
s=s+abs(c[i]-c1[i]);

}

return (s);

}

int main(){
string a;
cin >> a;
string b;
cin >> b;
cout << number_needed(a, b) << endl;
return 0;
}```
```

```                        ```Solution in Java :

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
public static int numberNeeded(String first, String second) {
int[] charSet = new int[256];

for(int i=0; i<first.length(); i++)
charSet[first.charAt(i)]++;

for(int i=0; i<second.length(); i++)
charSet[second.charAt(i)]--;

int numberNeeded = 0;
for(int i=0; i<256; i++)
numberNeeded += Math.abs(charSet[i]);

return numberNeeded;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String first = in.next();
String second = in.next();
System.out.println(numberNeeded(first, second));
}
}```
```

```                        ```Solution in Python :

In Python3 :

a = input().strip()
b = input().strip()
main_count = 0
total = 0
string1_set = set(a + b)
for i in range(len(string1_set)):
count = a.count(list(string1_set)[i])
count2 = b.count(list(string1_set)[i])
if count > count2:
main_count = count - count2
elif count2 > count :
main_count = count2 - count
elif count2 == count :
main_count = 0
total = total + main_count
print(total)```
```

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