**Making Anagrams**

### Problem Statement :

We consider two strings to be 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. Alice is taking a cryptography class and finding anagrams to be very useful. She decides on an encryption scheme involving two large strings where encryption is dependent on the minimum number of character deletions required to make the two strings anagrams. Can you help her find this number? Given two strings, s1 and s2, that may not be of the same length, determine the minimum number of character deletions required to make s1 and s2 anagrams. Any characters can be deleted from either of the strings. Function Description Complete the makingAnagrams function in the editor below. makingAnagrams has the following parameter(s): string s1: a string string s2: a string Returns int: the minimum number of deletions needed Input Format The first line contains a single string, s1. The second line contains a single string, s2 Constraints 1 <= | s1 | , | s2 | <= 10^4 It is guaranteed that s1 and s2 consist of lowercase English letters, ascii[a-z]..

### Solution :

` ````
Solution in C :
In C++ :
#include <cmath>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
char s1[10010],s2[10010];
cin>>s1>>s2;
int a[26]={0};
for(int i=0;i<strlen(s1);i++)
a[s1[i]-'a']++;
for(int i=0;i<strlen(s2);i++)
a[s2[i]-'a']--;
long long int ans = 0;
for(int i=0;i<26;i++)
ans += abs(a[i]);
cout<<ans<<endl;
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2 = br.readLine();
int[] counts1 = new int[256];
int[] counts2 = new int[256];
for(int i=0; i<str1.length();i++)
{
int index = (int)(str1.charAt(i) - '\0');
counts1[index] += 1;
}
for(int i=0; i<str2.length();i++)
{
int index = (int)(str2.charAt(i) - '\0');
counts2[index] += 1;
}
int ans = 0;
for(int i=0; i<256;i++)
{
ans += Math.abs(counts1[i] - counts2[i]);
}
br.close();
System.out.println(ans);
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char str1[20000];
char str2[20000];
scanf("%s", str1);
scanf("%s", str2);
int n1 = strlen(str1);
int n2 = strlen(str2);
int i, j;
char s1[26] = {0};
for (i = 0; i < n1; ++i) {
s1[str1[i] - 97] += 1;
}
for (i = 0; i < n2; ++i) {
s1[str2[i] - 97] -= 1;
}
int count = 0;
for (i = 0; i < 26; ++i) {
count += abs(s1[i]);
}
printf("%d\n", count);
return 0;
}
In Python3 :
import sys
from functools import *
a = sys.stdin.readline()
b = sys.stdin.readline()
x = {}
def f(w, cf = 1):
for c in w:
if c not in 'abcdefghijklmnopqrstuvwxyz':
continue
if c not in x:
x[c] = 0
x[c] += cf
f(a)
f(b, -1)
res = reduce(lambda a, b: a + abs(x[b]), x.keys(), 0)
print(res)
```

## View More Similar Problems

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →