HackerRank Happy Ladybugs Problem Solution

In this post, we will solve HackerRank Happy Ladybugs Problem Solution.

Happy Ladybugs is a board game having the following properties:

The board is represented by a string, b, of length n. The ith character of the string, b[i], denotes the ¿th cell of the board.
If b[i] is an underscore (i.e., _), it means the ith cell of the board is empty.
If b[i] is an uppercase English alphabetic letter (ascii[A-Z]), it means the ¿th cell contains a ladybug of color b[i].
String b will not contain any other characters.
A ladybug is happy only when its left or right adjacent cell (i.e., b[i+1]) is occupied by another ladybug having the same color.
In a single move, you can move a ladybug from its current position to any empty cell.
Given the values of n and b for g games of Happy Ladybugs, determine if it’s possible to make all the ladybugs happy. For each game, return YES if all the ladybugs can be made happy through some number of moves. Otherwise, return NO.
Example
b = [YYR_B_BR]
You can move the rightmost B and R to make b = [YYRRBB_] and all the ladybugs are happy. Return YES.

Function Description

Complete the happyLadybugs function in the editor below.

happyLadybugs has the following parameters:

string b: the initial positions and colors of the ladybugs

Returns

string: either YES or NO

Input Format
The first line contains an integer g, the number of games.
The next g pairs of lines are in the following format:

The first line contains an integer n, the number of cells on the board.
The second line contains a string b that describes then cells of the board.

Sample Input 0

4
7
RBY_YBR
6
X_Y__X
2
__
6
B_RRBR

Sample Output 0

YES
NO
YES
YES

Explanation 0

The four games of Happy Ladybugs are explained below:

Initial board:

Now all the ladybugs are happy, so we print YES on a new line.

There is no way to make the ladybug having color y happy, so we print NO on a new line.
There are no unhappy ladybugs, so we print YES on a new line.
Move the rightmost B and R to form b = [BBRRR___].

Sample Input 1

5
5
AABBC
7
AABBC_C
1
_
10
DD__FQ_QQF
6
AABCBC

Sample Output 1

NO
YES
YES
YES
NO

HackerRank Happy Ladybugs Problem Solution

Happy Ladybugs C Solution

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

char arr[101];

int compare (const void * a, const void * b)
{
  return ( *(char*)a - *(char*)b );
}

int main(){
    int G,N,whitespaces,i,unhappy;
    scanf("%d",&G);
    while(G--)
    {
        unhappy=0;
        whitespaces=0;
        scanf("%d\n",&N);
        i=N;
        while(i--)
        {
            scanf("%c",&arr[N-i-1]);
            if(arr[N-i-1] == '_')
                whitespaces++;
        }
        
        if(whitespaces > 0)
            qsort(arr,N,sizeof(char), compare);
        
        //for(i=0;i<N;i++)
        //    printf("%c",arr[i]);
        //printf("\n");
        
        for(i = 0; i<N && !unhappy; i++)
        {
            if (arr[i] != '_' &&
                (  (i==0 && i+1<N && arr[i+1] != arr[i])
                 || (i==N-1 && i-1>=0 && arr[i-1] != arr[i])
                 || (arr[i]!=arr[i-1] && arr[i]!=arr[i+1])))
                 unhappy++;
        }
        
        if(unhappy)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}

Happy Ladybugs C++ Solution

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <ctype.h>
#include <deque>
#include <queue>
#include <cstring>
#include <set>
#include <list>
#include <map>
#include <random>
#include <unordered_map>
#include <stdio.h>

using namespace std;

typedef long long ll;
typedef std::vector<int> vi;
typedef std::vector<bool> vb;
typedef std::vector<string> vs;
typedef std::vector<double> vd;
typedef std::vector<long long> vll;
typedef std::vector<std::vector<int> > vvi;
typedef vector<vvi> vvvi;
typedef vector<vll> vvll;
typedef std::vector<std::pair<int, int> > vpi;
typedef vector<vpi> vvpi;
typedef std::pair<int, int> pi;
typedef std::pair<ll, ll> pll;
typedef std::vector<pll> vpll;

const long long mod = 1000000007;

#define all(c) (c).begin(),(c).end()
#define sz(c) (int)(c).size()
#define forn(i, a, b) for(int i = a; i < b; i++)

#define pb push_back
#define mp make_pair

int main()
{

    int t;
    scanf("%d\n", &t);
    forn(agag,0,t) {
        int n;
        scanf("%d\n", &n);
        string s;
        getline(cin, s);
        int cem =0;
        vi cnt(26,0);
        forn(i,0,n) {
            if(s[i] == '_') cem++;
            else cnt[s[i]-'A']++;
        }
        if(cem==0) {
            string ans = "YES";
            forn(i,0,n) {
                bool f= false;
                if(i<n-1 && s[i+1]==s[i]) f = true;
                if(i>0 && s[i-1]==s[i]) f = true;
                if(!f) ans = "NO";
            }
            cout<<ans<<endl;
            
        }
        else {
            string ans = "YES";
            forn(i,0,26) if(cnt[i] == 1) ans = "NO";
            cout<<ans<<endl;
        }
    }
}

Happy Ladybugs C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {

    static void Main(String[] args) {
        int Q = Convert.ToInt32(Console.ReadLine());
            for (int a0 = 0; a0 < Q; a0++)
            {
                int n = Convert.ToInt32(Console.ReadLine());
                string b = Console.ReadLine();

                bool happy = true;

                if (!b.Contains("_"))
                {
                    for(int i = 0; i < b.Length; i++)
                    {
                        char left;
                        if (i > 0)
                            left = b[i - 1];
                        else
                            left = '_';

                        char right;
                        if (i < b.Length-1)
                            right = b[i + 1];
                        else
                            right = '_';

                        if (b[i] != left && b[i] != right)
                        {
                            happy = false;
                            break;
                        }
                    }
                }
                else
                {
                    Dictionary<char, int> dict = new Dictionary<char, int>();

                    foreach (char c in b)
                    {
                        if (c != '_')
                        {
                            if (dict.ContainsKey(c))
                            {
                                dict[c] = dict[c] + 1;
                            }
                            else
                            {
                                dict.Add(c, 1);
                            }
                        }
                    }

                    foreach (KeyValuePair<char, int> pair in dict)
                    {
                        if (pair.Value < 2)
                        {
                            happy = false;
                            break;
                        }
                    }
                }

                Console.WriteLine(happy ? "YES" : "NO");
            }
            
    }
}

Happy Ladybugs 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 'happyLadybugs' function below.
     *
     * The function is expected to return a STRING.
     * The function accepts STRING b as parameter.
     */

    public static String happyLadybugs(String b) {
        // Write your code here
        int [] cnts = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
        
        for (int i = 0; i < b.length(); i++) {
            if (b.charAt(i) != '_')
                cnts[b.charAt(i) - 'A']++;
        }
        
        boolean byes = true;
        for (int i = 0; i < cnts.length; i++) {
            if (cnts[i] == 1) {
                byes = false;
                break;
            }
        }
        
        if (byes) {
            if (b.indexOf("_") >= 0) {
                return "YES";
            } else {
                for (int i = 0; i < b.length(); i++) {
                    if (i == 0) {
                        if (b.charAt(i) != b.charAt(i+1)) {
                            return "NO";
                        }
                    } else if (i == b.length() - 1) {
                        if (b.charAt(i) != b.charAt(i-1)) {
                            return "NO";
                        }
                    } else {
                        if (b.charAt(i) != b.charAt(i-1) && b.charAt(i) != b.charAt(i+1)) {
                            return "NO";
                        }
                    }
                }
                return "YES";
            }
        } else {
            return "NO";
        }
    }

}

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")));

        int g = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, g).forEach(gItr -> {
            try {
                int n = Integer.parseInt(bufferedReader.readLine().trim());

                String b = bufferedReader.readLine();

                String result = Result.happyLadybugs(b);

                bufferedWriter.write(result);
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Happy Ladybugs JavaScript Solution

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
    var Q = parseInt(readLine());
    for(var a0 = 0; a0 < Q; a0++){
        var n = parseInt(readLine());
        var b = readLine();
        var result = null;
        if(b.includes('_')){
            var ladybugs = {};
            for(var i = 0; i < n; i++){
                var ladybug = b[i];
                if(ladybug != '_'){
                    ladybugs[ladybug] = ladybugs[ladybug] + 1 || 1;
                }
            }
            result = Object.keys(ladybugs).reduce(function(x,y){
                return x && ladybugs[y] > 1;
            }, true);
        } else{
            result = b.split('').reduce(function(total, value, index){
                return total && (value == b[index-1] || value == b[index + 1]);
            }, true);
        }
        console.log(result ? 'YES' : 'NO');
    }

}

Happy Ladybugs Python Solution

#!/bin/python3

import sys
from collections import Counter

Q = int(input().strip())
for a0 in range(Q):
    n = int(input().strip())
    b = input().strip()
    myList = list(b)
    d = Counter(myList)

    if len(d) < 2:
        if d["_"] == 0:
            for k in d:
                if d[k] > 1:
                    print("YES")
                else:
                    print("NO")
        else:
            print("YES")
    if len(d)>=2:
        if d["_"] == 0:
            count = 1
            for k in range(1, n):
                if myList[k] == myList[k-1]:
                    count+=1
                else:
                    if count==1:
                        print("NO")
                        break
                    else:
                        count = 1
                if k == n-1 and count==1:
                    print("NO")
                elif k == n-1 and count!=1:
                    print("YES")
        else:
            c = False
            for x in d:
                if d[x]==1 and x!="_":
                    c = True
                    print ("NO")
                    break
            if c == False:
                print("YES")