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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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1100 Discussions

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  • animeshtyagi2
     + 0 comments
    String target = "hackerrank";
    int j = 0;

    for (int i = 0; i < s.length() && j < target.length(); i++) {
        if (s.charAt(i) == target.charAt(j)) {
            j++;
        }
    }

    return j == target.length() ? "YES" : "NO";

  • spriyamalar
     + 0 comments
    public static String hackerrankInString(String s) {
    String str = "hackerrank";
    for(int i = 0; i < str.length(); i++) {
        int index = s.indexOf(str.charAt(i));
        if(index == -1) return "NO";
        s = s.substring(index+1);
    }
    return "YES";
}

  • ugurkurt90
     + 0 comments
    string hackerrankInString(string s) {
    string searchStr = "hackerrank";
    auto it = searchStr.begin();
    for (int i=0; i<s.size(); ++i) {
        if(s[i] == *it)
            ++it;
    }
    if(it == searchStr.end())
        return "YES";
    return "NO";
}

  • afzal_saiyed931
     + 0 comments
    def hackerrankInString(s):
    # Write your code here
    a = ['h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k']
    
    end = len(s) - 1
    
    while end >= 0:
        if len(a) < 1:
            break
        current = s[end]
        current_match = a.pop()
        if current != current_match:
            a.append(current_match)
        end -= 1
            
    return "NO" if len(a) > 0 else "YES"

  • geovanni_luna
     + 0 comments

    Here is a O(n) time and O(1) time solution in python:

    def hackerrankInString(s):
    hacker = 'hackerrank'
    i = 0

    for c in s:
        if c == hacker[i]:
            i += 1
                
        if i == len(hacker):
            return "YES"
        
    return "NO"