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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Happy Ladybugs
  5. Discussions

Happy Ladybugs

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512 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/khDcPEl6to0

    string happyLadybugs(string b) {
    vector<int> occ(26, 0);
    bool happy = true, underscore = false;
    b = "0"+b+"0";
    for(int i = 1; i < b.size()-1; i++){
        if(b[i] != '_' && b[i] != b[i-1] && b[i] != b[i+1]) happy = false;
        if(b[i] == '_') underscore = true;
        else occ[b[i] - 'A']++;
    }
    if(happy) return "YES";
    if(underscore && find(occ.begin(), occ.end(), 1) == occ.end()) return "YES";
    return "NO";
}

  • jmcparland
     + 0 comments

    Haskell

    Slightly sloppy, but good enough.

    module Main where

import Control.Monad (replicateM_)
import Data.List (group)
import Data.Map (Map, delete, fromListWith, (!?))

main :: IO ()
main = do
    cases <- readLn :: IO Int
    replicateM_ cases $ do
        _ <- getLine
        s <- getLine
        putStrLn $ if solve s then "YES" else "NO"

t1 = "RBY_YBR"
t2 = "X_Y__X"
t3 = "B_RRBR"
tests = [t1, t2, t3]

happyEnough :: String -> Bool
happyEnough = all ((>= 2) . length) . group . filter (/= '_')

freq :: String -> Map Char Int
freq = fromListWith (+) . map (\x -> (x, 1))

solve :: String -> Bool
solve s = happyEnough s || (bSpace && bAlpha)
  where
    fs = freq s
    bSpace = maybe False (>= 1) (fs !? '_')
    bAlpha = all (>= 2) $ delete '_' fs

  • yossefibrahim7oo
     + 0 comments

    Python 3 Just filter for happy conditions

    from collections import Counter
def happyLadybugs(b):
    # Write your code here
    if len(b) == 1 and b[0] == '_':
        return 'YES'
    if len(b) == 1 and b[0] != '_':
        return 'NO'
    c = Counter(b)
    flag = 0
    if c['_'] == 0:
        for i in range(len(b)):
            if i == 0:
                if b[i] == b[i+1]:
                    continue
                else:
                    flag+=1
            if i == len(b)-1:
                if b[i] == b[i-1]:
                    continue
                else:
                    flag+=1
            if i > 0 and i < len(b)-1:
                if b[i] == b[i-1] or b[i] == b[i+1]:
                    continue
                else:
                    flag+=1
                    
    for m, n in c.items():
        if n <= 1 and m != '_':
            return 'NO'
            
    return 'YES' if not flag else 'NO'

  • jordanyhchan
     + 0 comments

    Python Solution

    def happyLadybugs(b):
        # Write your code here

        for colour in set(b):
                if colour != "_" and b.count(colour) == 1:
                        return "NO"

        if b.count("_") > 0:
                return "YES"

        for i in range(len(b)):
                if i == 0 and b[0] != b[1]:
                        return "NO"
                if i == len(b) - 1 and b[-1] != b[-2]:
                        return "NO"
                if b[i] != b[i-1] and b[i] != b[i+1]:
                        return "NO"

        return "YES"

  • SVVBathini
     + 0 comments

    for Python3 Platform

    def happyLadyBugs(b):
    for i in b:
        if i != "_" and b.count(i) == 1:
            return "NO"
            break
    
    if (b.count("_") > 0):
        return "YES"
    
    z = list(zip(b, b[1:], b[2:]))
    
    return "YES" if all(map(lambda t: t[1]==t[0] or t[1]==t[2], z)) else "NO"

g = int(input())
for i in range(g):
    n = int(input())
    b = input()
    
    result = happyLadyBugs(b)
    
    print(result)