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 :



title-img 


                            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)








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