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

Happy Ladybugs

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517 Discussions

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  • adharshgraju
     + 0 comments

    My Java Solution

    public static String happyLadybugs(String b) {
        String[] bugArr = b.split("");
        if(b.contains("_")){
            Arrays.sort(bugArr);
        }
        bugArr = Arrays.stream(bugArr)
                .filter(s -> !s.equals("_"))
                .toArray(String[]::new);
        boolean isPresent = false;

        if(bugArr.length == 0){
            return "YES";
        }

        if(bugArr.length > 1){
            if(!bugArr[0].equalsIgnoreCase(bugArr[1])
                    || !bugArr[bugArr.length-1].equalsIgnoreCase(bugArr[bugArr.length-2])){
                return "NO";
            }
            for(int i=1; i<bugArr.length-1; i++){
                String currentElement = bugArr[i];
                if(currentElement.equals(bugArr[i-1]) || currentElement.equals(bugArr[i+1])){
                    isPresent = true;
                }else{
                    isPresent = false;
                    break;
                }
            }
        }

        return isPresent ? "YES" : "NO";
    }

  • 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";
}

  • zaidnsour12
     + 0 comments

    Java soluation using HashMap:

     public static String happyLadybugs(String b) {
        HashMap<Character, Integer> colors = new HashMap<>();
        int n = b.length();
        boolean contain_underscore = false;
        boolean same_color_adjacent = true;
        char prev = ' ';
        
        for(int i = 0; i < n; i++){
            char c = b.charAt(i);
            if(c == '_') contain_underscore = true;
         
            else{
              if (i > 0 && !contain_underscore && c != prev && colors.get(prev) <= 1)
                   same_color_adjacent = false; 
              prev = c;
                
                
              if(!colors.containsKey(c))
                    colors.put(c, 1);
              else
                    colors.replace(c, colors.get(c) + 1);  
              }
             
            }
        
        if(!contain_underscore && !same_color_adjacent)
            return "NO";
        
        for(Integer val:colors.values())
            if(val <= 1)
                return "NO";
        
        return "YES"; 
        
        }

  • alex_k5
     + 0 comments

    from collections import Counter

    def happyLadybugs(b):

    count_b = Counter(b)
mc = count_b.most_common()
if len(mc)!=1 and ((mc[-1][1]== 1 and mc[-1][0]!= '_') or mc[-2][1]== 1):
    return 'NO'
else:
    if '_' in b:
        return 'YES'
    for i in mc:
        if count_b[i[0]]==1 or b.rfind(i[0])-b.find(i[0])+1 != count_b[i[0]]:
            return 'NO'
        if i == mc[-1]:
            return 'YES'

  • space59
     + 0 comments

    The solution is a bit tedious due to the number of cases to account for. Not sure I would call this one "easy".