Amazing Logic

nice logic

Commendable

Nice idea to store the next required character in j and increment each time the letter is found. I popped characters from a stack instead, but this is also elegant.

it's nice to know people thought the same way i did lol

good job!

for(int i=0;i<s.length()&&((s.length()-(i-1))>(a.length()-j));i++)

adding this statement as replacement of above can save few ms. I can say ns probably :p

wow...u r good

Very nice logic !!

Good Logic

Amazing solution ..how did you come up with such idea...

loved it

so cool :))

looks like it is failing for input "hackerrankk"

Would it be better to return "YES" if j reaches str.length()? That way, the loop won't have to keep running once a match has been made.

for (int i = 0; i < s.length(); i++)
{
    if (s.charAt(i) == hr.charAt(j))
    {
        j++;
    }
            
    if (j == hr.length())
    {
        return "YES";
    }
}

Superb

zabardast

I thought this was a straight forward method.

Awesome!!

Your code can be optimized further by reducing the size of the string s every time you find a character from hackerrank. This will reduce the time complexity of the algo to nlogn.

I had a similar approach but there is no need to check if the length of s is smaller than "hackerrank" because one of the constraints is that s.length() is atleast 10.

WOW. <\smile :)>

Simple Python function to solve this:

def hackerrankInString(s):
    # Complete this function
    p = 0
    for e in 'hackerrank':
        if e in s[p:]:
            p = s.index(e,p) + 1
        else:
            return 'NO'
    return 'YES'

Java soln using Stacks :)

static String hackerrankInString(String s) {
        // Complete this function
        Stack<Character> stack= new Stack<>();
        stack.push('k');
        stack.push('n');
        stack.push('a');
        stack.push('r');
        stack.push('r');
        stack.push('e');
        stack.push('k');
        stack.push('c');
        stack.push('a');
        stack.push('h');
        
        String result="NO";
        for(Character c:s.toCharArray()){
            if(c==stack.peek())
                stack.pop();
            if(stack.isEmpty()){
                result="YES";
                break;
            } 
        }
        return result;
    }

since j starts from 0 and with "hackerrank" check, j value will be 9, so return (j == str.length() ? "YES" : "NO");

might not work for same cases. It might be like this

return (j+1 == str.length() ? "YES" : "NO");

it does not work in some cases

Wow.Superb logic!!!

One way to improve it would to possibly put the return yes inside the for loop so you dont do any unneeded iterations because right now even if its already confirmed that it contains "hackerrank" in order it will continue through the rest of String s. possibly do this inside the loop instead.

for (int i = 0; i < s.length(); i++) {
            if (j <= str.length()){ 
                if(s.charAt(i) == str.charAt(j)) {
                    j++;
            } else {
                    return "YES";
                }
        }
}
        return "NO";

since it is the same amount of actions as combining the if statement and less if the j integer is already equal to the str length as well as adding a little more power to it.

but i love the

if (s.length() < str.length()) {
            return "NO";
        }

I wish i would have thought of including that in my algorithm

Yep, ended up with the same in C++ also:

string hackerrankInString(string s) {
	// Complete this function
	int ss[] = {104,97,99,107,101,114,114,97,110,107};
	int ss_l = 10;
	int ss_ind = 0;
	for ( int i=0 ; i<s.size() && ss_ind<ss_l ; i++) {
		if ( ss_ind<ss_l && (int)s[i]==ss[ss_ind] ) {
			ss_ind++;
		}
	}
	return (ss_ind==ss_l)?"YES":"NO";
}

string hackerrankInString(string s) {
// Complete this function
string h = "hackerrank";
int len = s.length();
int j=0;
for (int i=0; i<len; i++) {
    if (s[i] == h[j]) {
        j++;
    }
}
if (j==h.length())
    return "YES";
else
    return "NO";

}

Amazing! You nailed it man.

This is not as efficientas it scans to the end of the array. One should insted of j++ say

if (++j == sttr.length()) return "YES";

Java 8

    static String hackerrankInString(String s) {
        String hr = "hackerrank";
        int index = 0;
        for (int i = 0; i < hr.length(); i++) {
            int current = s.indexOf(hr.charAt(i), index);
            if (current < index)
                return "NO";
            index = current + 1;
        }
        return "YES";
    }

very nice solution

Love you buddy for this solution! :)

Saving some computational time.

The loop will run till the end of the s.length() even if we have already found our given subsequence.

For example for above variables, consider s = "hraaa" and str = "hr", the loop will go through 5 iterations even when we found our subsequence after 2, so better to put an if condition inside loop checking if ( j == str.length) and return "YES" and return "NO" outside loop in case the subsequence is not found.

But it should break out of the loop once j is equal to str.length(), he had done some wasted empty iterations

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

amazing

you can break the loop when found enough word "hackerrank". It'll avoid go through all String s

As soon as str.length() == j -> you can return "YES" so you don't have to iterate the rest of the loop.

Nice guard clause.. Also for extemely large strings, you can return "YES" as soon as j==str.length(), that way you don't have to iterate through the entire string.

I got the same logic. You can improve this a little bit.

Rather than declaring 'j' outside the for loop, declare it inside.

Return "YES" when the j == str.length() within the for loop itself. So that that it won't have to loop through other characters once you know that "hackerrank" is already present.

Return "NO" outside the for loop.

Thank you so much for the logic.

Here is my JS solution:

    let hR = `hackerrank`;
    let j = 0;

    for(let i = 0; i < hR.length; i++){
        j = s.indexOf(hR.charAt(i), j) + 1;
        if(j === 0){ return `NO`; }
    }
    return `YES`;

bhai format kaise kiya ? bta pleaseeeeeeeeeeeee

why j

super

Sorry pal!! wrong in many cases...

it does not check the order of letters

osm logic

vector<char> hr{ 'h','a','c','k','e','r','r','a','n','k' };
reverse(hr.begin(),hr.end());
stack<char, vector<char>> stk(hr);
for (int i = 0; i < s.length(); i++)
{
    if(stk.top()==s[i]){
        stk.pop();
    }
}

return (stk.empty()?"YES":"NO");

can you plzz explain the need of j

noice

Same logic...

For me that string is too much work, so I used

'.*?'.join(list("hackerrank"))

Okay! I am going to use to learn regex more. Anyone give me good sites for it?

Anyway, this is what i did:

I didn't find anyone using this type of code.

q = int(raw_input().strip())
for a0 in xrange(q):
    s = raw_input().strip()
    # your code goes here
    valid = 'YES'
    x=s.find('h',0)
    for i in 'ackerrank':
        #print i, x
        if x!= -1 :
            x = s.find(i,x+1)
        else:
            valid = 'NO'
            break
    print valid

my very pretty one-liner:

print( (re.search('.*'.join("hackerrank"), s) and "YES") or "NO")

import re 
def hackerrankInString(s):
        return ('YES' if re.search(r'.*h.*a.*c.*k.*e.*r.*r.*a.*n.*k.*', s) else 'NO')

I think you do not need to use for loop because it automatically iterate.

This is cheating lol...take an upvote

Why do you iterate over the string when re already does the iteration in the background? just

print('YES' if re.search(thepattern, thestring) else 'NO')

would be fine

Really, most effective way

this is just brilliant!!!

sungmkim80, genius man! Very elegant solution.

Tyűha!

Wow this is a nice thought!

This indeed is elegant and O(n). I have been trying to get non regex and something better than O(n^2) going.

I also did very similar thing.