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

Happy Ladybugs

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525 Discussions

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  • spriyamalar
     + 0 comments
    public static String happyLadybugs(String str) {

    if(Arrays.stream(str.split("")).distinct().allMatch(s -> s.equals("_"))) return "YES";

    Map<String, Long> map = Arrays.
            stream(str.split("")).
            collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

    if(map.entrySet().stream().filter(en -> !en.getKey().equals("_")).anyMatch(en -> en.getValue()==1L)) 
        return "NO";
    if(str.indexOf("_") == -1) {
        for(int i=0;i<str.length();i++) {
            if(i==0) {
                if(str.charAt(i) != str.charAt(i+1)) return "NO";
            } else if(i==str.length()-1) {
                if(str.charAt(i) != str.charAt(i-1)) return "NO";
            } else {
                if(str.charAt(i) != str.charAt(i+1) && 
                        str.charAt(i) != str.charAt(i-1)) return "NO";
            }
        }
    }

    return "YES";
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, java, c++, c and javascript -https://programmingoneonone.com/hackerrank-happy-ladybugs-problem-solution.html

  • 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";
}

  • patrickgao2020
     + 1 comment

    Here is my Python solution! If we have at least one empty square, then we can move any of the ladybugs anywhere. This means that the only cases that work are if there are more than one empty square and an even number of each type, or if there are no empty squares but all the ladybugs are already happy.

    def ishappy(b):
    for bug in range(1, len(b) - 1):
        if b[bug] == b[bug - 1] or b[bug] == b[bug + 1]:
            continue
        else:
            return False
    return True

def happyLadybugs(b):
    ladybugs = set([string for string in b if string != "_"])
    if list(b).count("_") < 1 and not ishappy(b):
        return "NO"
    for ladybug in ladybugs:
        if list(b).count(ladybug) < 2:
            return "NO"
    return "YES"

  • medyk_pawel
     + 0 comments

    Python3; After realizing 1 empty field gives you possibility to place any existing color anywhere

    from collections import Counter

def are_happy(b):
    for i in range(len(b)):
        happy = False
        if i > 0:
            if b[i-1] == b[i]:
                happy = True
        if i < len(b) - 1:
            if b[i+1] == b[i]:
                happy = True
        if not happy:
            return False
    return True
        

def happyLadybugs(b):
    c = Counter(b)
    if c['_'] < 1:
        if are_happy(b):
            return 'YES'
        else:
            return 'NO'
    
    for key, count in c.most_common()[::-1]:
        if count == 1 and key != '_':
            return 'NO'
        if count > 1:
            return 'YES'
    return 'YES'