In this post, we will solve HackerRank Making Anagrams Problem Solution.
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, $1 and $2, that may not be of the same length, determine the minimum number of character deletions required to make $1 and $2 anagrams. Any characters can be deleted from either of the strings.
Example.
s1 = abc
s2 = amnop
The only characters that match are the a’s so we have to remove bc from $1 and mnop from $2 for a total of 6 deletions.
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.
Sample Input
cde
abc
Sample Output
4
Explanation
Delete the following characters from our two strings to turn them into anagrams:
- Remove
dandefromcdeto getc. - Remove
aandbfromabcto getc.
4 characters have to be deleted to make both strings anagrams.
Making Anagrams C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int a[26];
int b[26];
int i;
for (i = 0; i < 26; i++) {
a[i] = b[i] = 0;
}
char c[10000];
char d[10000];
scanf ("%s", c);
scanf ("%s", d);
for (i = 0; c[i]; i++) {
a[c[i] - 'a']++;
}
for (i = 0; d[i]; i++) {
b[d[i] - 'a']++;
}
int count = 0;
for (i = 0; i < 26; i++) {
count += abs(a[i] - b[i]);
}
printf ("%d\n", count);
return 0;
}Making Anagrams C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
std::string a; std::cin >> a;
std::string b; std::cin >> b;
int freq[26] = {0};
for (char c : a) {
freq[c - 'a'] += 1;
}
for (char c : b) {
freq[c - 'a'] -= 1;
}
int mismatches = 0;
for (int f : freq) {
if (f < 0)
mismatches -= f;
else
mismatches += f;
}
std::cout << mismatches << std::endl;
return 0;
}Making Anagrams C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System;
class Solution {
static void Main(String[] args) {
var firstWordLetters = Console.ReadLine().ToCharArray().GroupBy(e=>e).ToDictionary(e=>e.Key, e=>e.Count());
var secondWordLetters = Console.ReadLine().ToCharArray().GroupBy(e=>e).ToDictionary(e=>e.Key, e=>e.Count());
var allLetters = firstWordLetters.Keys.Concat(secondWordLetters.Keys).Distinct();
var count=0;
foreach(var letter in allLetters)
{
if (firstWordLetters.ContainsKey(letter))
{
if (secondWordLetters.ContainsKey(letter))
{
count += Math.Abs(firstWordLetters[letter] -secondWordLetters[letter]);
}
else
count += firstWordLetters[letter];
}
else
count+= secondWordLetters[letter];
}
Console.WriteLine(count);
}
}Making Anagrams Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'makingAnagrams' function below.
*
* The function is expected to return an INTEGER.
* The function accepts following parameters:
* 1. STRING s1
* 2. STRING s2
*/
public static int makingAnagrams(String s1, String s2) {
// Write your code here
int[] freq = new int[26];
s1.chars().forEach((c) -> { freq[c-97]++; });
s2.chars().forEach((c) -> { freq[c-97]--; });
return Arrays.stream(freq).map(Math::abs).sum();
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
String s1 = bufferedReader.readLine();
String s2 = bufferedReader.readLine();
int result = Result.makingAnagrams(s1, s2);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
bufferedReader.close();
bufferedWriter.close();
}
}Making Anagrams JavaScript Solution
function processData(input) {
input = input.split('\n');
var A = JSON.stringify(input[0]);
var B = JSON.stringify(input[1]);
A = A.split('');
B = B.split('');
//console.log(A);
//console.log(B);
for(var i = 0; i<A.length; i++){
for(var j = 0; j<B.length; j++){
if(A[i] === B[j]){
A.splice(i,1);
B.splice(j,1);
i--;
break;
}
}
}
//console.log(A);
//console.log(B);
console.log(A.length+B.length);
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});Making Anagrams Python Solution
d1 = dict()
d2 = dict()
s1 = input()
s2 = input()
for c in s1:
if c in d1:
d1[c] += 1
else:
d1[c] = 1
for c in s2:
if c in d2:
d2[c] += 1
else:
d2[c] = 1
count = 0
for c, v in d1.items():
if c in d2:
count += abs(d1[c]-d2[c])
d2[c] = 0
else:
count += d1[c]
for c,v in d2.items():
if v:
count += v
print(count)Other Solutions
