Two Characters HackerRank Solution in C, C++, Java, Python

Given a string, remove characters until the string is made up of any two alternating characters. When you choose a character to remove, all instances of that character must be removed. Determine the longest string possible that contains just two alternating letters.

**Example88.

$s = \text{abaacdabd}$

Delete a, to leave bcdbd. Now, remove the character c to leave the valid string bdbd with a length of 4. Removing either b or d at any point would not result in a valid string.

Given a string $s$, convert it to the longest possible string $t$ made up only of alternating characters. Print the length of string $t$ on a new line. If no string $t$ can be formed, print $0$ instead.

Function Description

Complete the alternate function in the editor below.

alternate has the following parameter(s):

  • string s: a string

Returns.

  • int: the length of the longest valid string, or $0$ if there are none

Input Format

The first line contains a single integer that denotes the length of s.

The second line contains string s.

Constraints

  • 1<=lengthofs<=1000
  • S[i] belongs to ascii[a-z]

Sample Input

10

beabeefeab

 

Sample Output

5

 

Explanation

The characters present in s are a, b, e, and f. This means that t must consist of two of those characters and we must delete two others. Our choices for characters to leave are [a,b], [a,e], [a, f], [b, e], [b, f] and [e, f].

If we delete e and f, the resulting string is babab. This is a valid t as there are only two distinct characters (a and b), and they are alternating within the string.

If we delete a and f, the resulting string is bebeeeb. This is not a valid string t because there are consecutive e’s present. Removing them would leave consecutive b’s, so this fails to produce a valid string t.

Other cases are solved similarly.

babab is the longest string we can create.

Two Characters HackerRank Solution in C

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int calculate(char one,char two,char *in){
    int length=strlen(in);
    int a;
    int flag=0;
    int count=0;
    for(a=0;a<length;++a){
        if(in[a]==one || in[a]==two){
            count+=1;
            if(in[a]==one){
                if(flag==-1){
                    return 0;
                }
                else{
                    flag=-1;
                }
            }
            if(in[a]==two){
                if(flag==1){
                    return 0;
                }
                else{
                    flag=1;
                }
            }
        }
    }
    
    return count;
}
int main() {
    int n,a;
    scanf("%d",&n);
    char in[1000];
    scanf("%s",in);
    char list[26];
    for(a=0;a<26;++a){
        list[a]=0;
    }
    int length=strlen(in);
    for(a=0;a<length;++a){
        list[in[a]-'a']=1;
    }
    char one,two;
    int b,value,max;
    for(a=0;a<26;++a){
        if(list[a]==1){
            for(b=a+1;b<26;++b){
               if(list[b]==1){
                   one = a + 'a';
                   two = b + 'a';
                   value=calculate(one,two,in);
                   if(value>max){
                       max=value;
                   }
               } 
            }
            
        }
        
    }
    printf("%d",max);
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

 

Two Characters HackerRank Solution in C++

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> ii;

int valid(string x) {
    const int n = x.size();
    for (int i = 1; i < n; ++i)
        if (x[i] == x[i-1])
            return false;
    return true;
}

int main() {
    int asd;
    cin>>asd;
    string s;
    cin>>s;
    int ans = 0;
    for (char a = 'a'; a <= 'z'; ++a)
    for (char b = 'a'; b <= 'z'; ++b)
    if (a != b)
    {
        if (s.find(a) == string::npos) continue;
        if (s.find(b) == string::npos) continue;
        string x;
        for (const char ch : s)
            if (ch == a || ch == b)
                x.push_back(ch);
        if (valid(x))
            ans = max(ans, (int)x.size());
    }
    printf("%d\n", ans);
}

 

Two Characters HackerRank 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) {
       Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        String s=in.next();
        String res="";
        for(int i=0;i<26;i++){
            for(int j=i+1;j<26;j++){
                char a=(char)('a'+i);
                char b=(char)('a'+j);
                String cur="";
                for(int k=0;k<n;k++){
                    if (s.charAt(k)==a || s.charAt(k)==b) {
                        cur+=s.charAt(k);
                    }
                }
                if (cur.length()<res.length()) continue;
                if (isGood(cur)) res=cur;
            }
        }
        System.out.println(res.length());
    }
    public static boolean isGood(String s){
        if (s.length()==1) return false;
        for(int i=1;i<s.length();i++){
            if (s.charAt(i)==s.charAt(i-1)) return false;
        }
        return true;
    }
}

 

Two Characters HackerRank Solution in Python

from collections import Counter
from itertools import combinations
def is_valid(S):
    c = Counter(S)
    #print c
    if len(c) != 2:
        return False
    for i in xrange(1, len(S)):
        if S[i] == S[i-1]:
            return False
    return True

def keep_letters(lista, keep):
    return filter(lambda x: x in keep, lista)
    
N = int(raw_input())
S = list(raw_input().strip())

letters = {x: 1 for x in S}
letters = list(combinations(letters.keys(), 2))
#print letters
L = list(S)
first = True
m = 0
for keep in letters:
    #print list(S)
    lista = keep_letters(list(S), keep)
    #print lista
    if is_valid(lista):
        m = max(m, len(lista))
    
    
print m
    

 

Two Characters HackerRank Solution in C#

using System;
using System.Collections.Generic;
using System.Linq;
class Solution {
    static void Main(String[] args) {
        int n = int.Parse(Console.ReadLine());
        var line = Console.ReadLine();
        
        if (n == 1) { Console.WriteLine(0); return; }
        int ans = 0;
        for (int i = 0; i < 26; i++) {
            for (int j = 0; j < 26; j++) {
                if (i == j) continue;
                ans = Math.Max(ans, alternating(line, (char)('a' + i), (char)('a' + j)));
            }
        }
        Console.WriteLine(ans);
    }

    static int alternating(string line, char p1, char p2) {
        bool first = true;
        int r = 0;
        foreach (var c in line.Where(x=>x==p1 || x == p2)) {
            if (first) {
                if (c == p2) return -1;
                r++;
            } else {
                if (c == p1) return -1;
                r++;
            }
            first = !first;
        }
        return r;
    }
}

 

Attempt Two Characters HackerRank Challenge

Link – https://www.hackerrank.com/challenges/two-characters/

Next HackerRank Challenge Solution 

Link – https://exploringbits.com/insertion-sort-part-2-hackerrank-solution/

 

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