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  1. Prepare
  2. Algorithms
  3. Strings
  4. HackerRank in a String!
  5. Discussions

HackerRank in a String!

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1105 Discussions

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  • abhishekrajvir66
     + 0 comments

    Java solution

        public static String hackerrankInString(String s) {
    // Write your code here
        int[] arr = { 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k' };
        
        int target = 0;
        
        for(int i = 0; i<s.length(); i++){
            if(target==8){
                return "YES";
            }
            if(s.charAt(i)==arr[target]){
                target++;
            }
        }
        return "NO";
    }

  • yashdeora98294
     + 0 comments

    Here is problem solution in python, java, c++, c and javascript programming - https://programmingoneonone.com/hackerrank-in-a-string-problem-solution.html

  • mrashidcit
     + 0 comments

    My Solution in kotlin cover all the test cases.

    fun hackerrankInString(s: String): String {
    val target = "hackerrank"
    val input = s.lowercase()
    var output = ""
    
    val lastIndex = target.length - 1
    var index = 0
    for(ch in input) {
       
       if (target[index] == ch) {
            index++
            output += ch
       }
       
       if (index > lastIndex) break
    }
    
    return if (output == target)
            "YES"
        else
            "NO"

}

  • kumbhar_saurabh1
     + 0 comments
    def hackerrankInString(s):
    Match_String = "hackerrank"
    j = 0
    for i in range(len(s)):
        if(j<len(Match_String) and s[i] == Match_String[j] ):
            j = j + 1
    
    if(j == len(Match_String)):
        return "YES"
    else :
        return "NO"

  • rezl_jaws
     + 0 comments

    C++ solution

    string hackerrankInString(string s) { string s1("hackerrank"); string str("");

    for (int i = 0; i < s1.size(); i++)
{
    for (int j = 0; j < s.size(); j++)
    {
        if (s1[i] == s[j])
        {
            str += s[j];
            i++;

            if (str == s1)
                return "YES";
        }
    }
}
return "NO";

    }