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 ith the 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 ith 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 games of Happy Ladybugs, determine if it’s possible to make all the ladybugs happy. For each game, print YES on a new line if all the ladybugs can be made happy through some number of moves. Otherwise, print NO.
As an example,b=[YYR_B_BR] . You can move the rightmost B and R to make b=[YYRRBB_] and all the ladybugs are happy.
Function Description
Complete the happyLadybugs function in the editor below. It should return an array of strings, either ‘YES’ or ‘NO’, one for each test string.
happyLadybugs has the following parameters:
b: an array of strings that represents the initial positions and colors of the ladybugs
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 describing the n cells of the board.
Constraints
Output Format
For each game, print YES on a new line if it is possible to make all the ladybugs happy. Otherwise, print NO.
Sample Input 0
4 7 RBY_YBR 6 X_Y__X 2 __ 6 B_RRBR
Sample Output 0
YES NO YES YES
Happy Ladybugs HackerRank Solution in C
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int Q,i;
scanf("%d",&Q);
for(int a0 = 0; a0 < Q; a0++){
int n,a[27]={0},empty=0,f=0,l;
scanf("%d",&n);
char* b = (char *)malloc(512000 * sizeof(char));
scanf("%s",b);
for(i=0;i<n;i++)
{
if(b[i]!='_')
a[b[i]-64]++;
if(b[i]=='_')
empty++;
}
//for(i=1;i<=26;i++)
//printf("%d ",a[i]);
for(i=1;i<=26;i++)
{
if(a[i]==1)
{
f=1;
break;
}
}
if(f==1)
printf("NO\n");
else if(empty==0)
{
f=0;
for(i=1;i<n-1;i++)
{
if(b[i]!=b[i-1] && b[i]!=b[i+1])
{
f=1;
break;
}
}
if(f==1)
printf("NO\n");
else
printf("YES\n");
}
else
printf("YES\n");
}
return 0;
}
Happy Ladybugs HackerRank Solution in C++
#include <stdio.h>
int cnt[26];
int main(){
int g;
scanf("%d", &g);
for(int test=0; test<g; test++){
int n;
char s[105];
scanf("%d", &n);
scanf("%s", &s[1]);
s[0] = '.';
int fr=0;
int valid=1;
for(int i=0; i<26; i++) cnt[i] = 0;
for(int i=1; s[i]; i++) cnt[s[i]-'A']++;
for(int i=0; i<26; i++) if(cnt[i] == 1) valid = 0;
for(int i=1; s[i]; i++) if(s[i] == '_') fr=1;
if(!valid){ printf("NO\n"); continue; }
if(fr){ printf("YES\n"); continue; }
for(int i=1; s[i]; i++){
if(s[i] != s[i-1] && s[i] != s[i+1]) valid = 0;
}
if(valid) printf("YES\n");
else printf("NO\n");
}
}
Happy Ladybugs 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 Q = in.nextInt();
for(int a0 = 0; a0 < Q; a0++){
char[] counts = new char[256];
int n = in.nextInt();
String b = in.next();
for (char c : b.toCharArray()) {
counts[c]++;
}
if (counts['_']==0) {
String pr = "YES";
for (int i = 0; i < n; i++) {
if ((i==0||b.charAt(i)!=b.charAt(i-1))&&(i==n-1||b.charAt(i)!=b.charAt(i+1)))
pr = "NO";
}
System.out.println(pr);
} else {
String pr = "YES";
for (int i = 0; i < 256; i++) {
if (i != (int)'_' && counts[i]==1)
pr = "NO";
}
System.out.println(pr);
}
}
}
}
Happy Ladybugs HackerRank Solution in Python
#!/bin/python
import sys
from collections import Counter
from string import ascii_uppercase
Q = int(raw_input().strip())
for a0 in xrange(Q):
n = int(raw_input().strip())
b = raw_input().strip()
c = Counter(b)
balanced = True
for key, value in c.iteritems():
if key in ascii_uppercase and value > 1:
continue
elif key == '_':
continue
else:
balanced = False
if not balanced:
print 'NO'
continue
if '_' not in b:
bounds = []
for char in b:
if bounds == []:
bounds.append([char])
continue
if char == bounds[-1][-1]:
bounds[-1].append(char)
else:
bounds.append([char])
balanced = True
for bound in bounds:
if len(bound) < 2:
balanced = False
break
print 'YES' if balanced else 'NO'
else:
print 'YES'
Happy Ladybugs HackerRank Solution in C#
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('_'))
{
int[] counts = new int[26];
foreach(char c in b)
{
if (c != '_') counts[(int)(c - 'A')]++;
}
foreach(int c in counts)
{
if (c == 1)
{
happy = false;
break;
}
}
}else
{
for(int i=0;i<n;i++)
{
if (b[i] != '_')
{
bool friendleft = i > 0 && b[i - 1] == b[i];
bool friendright = i < n - 1 && b[i + 1] == b[i];
if (!friendleft && !friendright)
{
happy = false;
break;
}
}
}
}
Console.WriteLine(happy ? "YES" : "NO");
}
}
}
Attempt Happy Ladybugs HackerRank Challenge
Link – https://www.hackerrank.com/challenges/happy-ladybugs/
Next HackerRank Challenge Solution
Link – https://exploringbits.com/strange-counter-hackerrank-solution/
Aayush Kumar Gupta is the founder and creator of ExploringBits, a website dedicated to providing useful content for people passionate about Engineering and Technology. Aayush has completed his Bachelor of Technology (Computer Science & Engineering) from 2018-2022. From July 2022, Aayush has been working as a full-time Devops Engineer.