HackerRank Making Anagrams Solution

Hello Programmers, In this post, you will learn how to solve HackerRank Making Anagrams Solution. This problem is a part of the HackerRank Algorithms Series.

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HackerRank Making Anagrams Solution

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

HackerRank Making Anagrams Solution

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We consider two strings to be anagrams of each other if the first strings 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.

Example

s1 = abc
s2 = amnop

The only characters that match are the a‘s so we have to remove bc from s1 and mnop from s2 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.

Constraints

  • 1 <= |s1|, |s2| <= 104
  • It is guaranteed that s1 and s2 consist of lowercase English letters, ascii[az].

Sample Input

cde
abc

Sample Output

4

Explanation

Delete the following characters from our two strings to turn them into anagrams:

  1. Remove d and e from cde to get c.
  2. Remove a and b from abc to get c.

 characters have to be deleted to make both strings anagrams.

HackerRank Making Anagrams Solution

Making Anagrams Solution 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;
}

Making Anagrams Solution in Cpp

#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;
}

Making Anagrams Solution 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) {
        /* Enter your code here. Read input from STDIN. 
        Print output to STDOUT. Your class should be named Solution. */
        
        Scanner in = new Scanner(System.in);
        String src = (in.nextLine());
        String tar = in.nextLine();
        int length = 0;       
		Map<Character, Integer> an= new HashMap <Character, Integer>();	
		
		for ( int i=0; i< src.length(); i++){			
			char c = src.charAt(i);			
			if (an.containsKey(c) )			
				an.put(c, an.get(c)+1);
			else
				an.put(c, 1);			
		}
		
		for ( int j=0; j< tar.length(); j++){			
			char c = tar.charAt(j);					
			if (an.containsKey(c) && an.get(c)!= 0 ) {
				an.put(c,an.get(c) -1);
                length+=2;
            }
            
		}	
        
        System.out.println(src.length() + tar.length()- length );
            
      }
    
}
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Making Anagrams Solution in Python

w1=raw_input()
w2=raw_input()
total=0
for letter in "abcdefghijklmnopqrstuvwxyz":
    total+=abs(w1.count(letter)-w2.count(letter))
print total

Making Anagrams Solution using JavaScript

  var processData, solve;
  processData = function(input) {
    var A, B, answer, _ref;
    _ref = input.split('\n'), A = _ref[0], B = _ref[1];
    answer = solve(A, B);
    console.log(answer);
    return answer;
  };
  solve = function(A, B) {
    var hash, i, key, sum, _i, _j, _ref, _ref1;
    hash = {};
    A = A.split('');
    for (i = _i = 0, _ref = A.length; 0 <= _ref ? _i < _ref : _i > _ref; i = 0 <= _ref ? ++_i : --_i) {
      if (A[i] in hash) {
        hash[A[i]] += 1;
      } else {
        hash[A[i]] = 1;
      }
    }
    for (i = _j = 0, _ref1 = B.length; 0 <= _ref1 ? _j < _ref1 : _j > _ref1; i = 0 <= _ref1 ? ++_j : --_j) {
      if (B[i] in hash) {
        hash[B[i]] -= 1;
      } else {
        hash[B[i]] = -1;
      }
    }
    sum = 0;
    for (key in hash) {
      sum += Math.abs(hash[key]);
    }
    return sum;
  };
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 Solution in Scala

import java.util.Scanner
object Solution {
  def main(args: Array[String]) {
    val sc = new Scanner(System.in);
    val n1 = sc.nextLine().toList.groupBy(e => e).map(e => (e._1, e._2.size)).withDefaultValue(0)
    val n2 = sc.nextLine().toList.groupBy(e => e).map(e => (e._1, e._2.size)).withDefaultValue(0)
    var minDel = 0
    for (key <- n1.keySet ++ n2.keySet) {
      minDel += Math.max(n1(key), n2(key)) - Math.min(n1(key), n2(key))
    }
    System.out.println(minDel)
  }
}

Making Anagrams Solution in Pascal

{$H+}
uses math;
const
 tfi='';
 tfo='';
var
 fi,fo:text;
 n,t:longint;
 a,b:string;
 d1,d2:array['a'..'z'] of longint;
procedure nhap;
 var i:longint;
 begin
     readln(fi,a);
     readln(fi,b);
 end;
procedure process;
 var i,j,tg:longint;
     ch:char;
 begin
     for i:=1 to length(a) do inc(d1[a[i]]);
     for i:=1 to length(b) do inc(d2[b[i]]);
     for ch:='a' to 'z' do
      t:=t+max(d1[ch],d2[ch])-min(d1[ch],d2[ch]);
     writeln(fo,t); 
 end;
BEGIN
    assign(fi,tfi);reset(fi);
    assign(fo,tfo);rewrite(fo);
      nhap;
      process;
    close(fi);close(fo);
END.

Disclaimer: This problem (Making Anagrams) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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