HackerRank in a String!


Problem Statement :


We say that a string contains the word hackerrank if a subsequence of its characters spell the word hackerrank. Remeber that a subsequence maintains the order of characters selected from a sequence.

More formally, let  p[0] , p[1], . . . , p[9]  be the respective indices of h, a, c, k, e, r, r, a, n, k in string s. If p[0] < p[1] < p[2] , . . . , < p[9]  is true, then s contains hackerrank.

For each query, print YES on a new line if the string contains hackerrank, otherwise, print NO.


Function Description

Complete the hackerrankInString function in the editor below.

hackerrankInString has the following parameter(s):

string s: a string

Returns

string: YES or NO


Input Format

The first line contains an integer q, the number of queries.
Each of the next q lines contains a single query string s.

Constraints


2  <=   q   <=  10^2
10  <=  length of s  <=  10^4


Solution :



title-img 


                            Solution in C :

In   C++  :






#include <bits/stdc++.h>

using namespace std;

int main(){
    int q;
    cin >> q;
    for(int a0 = 0; a0 < q; a0++){
        string s;
        cin >> s;
        string cur = "hackerrank";
        int st = 0;
        for (int i= 0; i < s.size() && st < cur.size(); i++) {
            if (s[i] == cur[st]) {
                st++;
            }
        }
        if (st == cur.size()) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
        // your code goes here
    }
    return 0;
}







In   Java  :







import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            String s = in.next();
            // your code goes here
            System.out.println(hasSubseq(s,"hackerrank") ? "YES" : "NO"); 
        }
    }
    
    private static boolean hasSubseq(String source, String target) {
        if (target.length() == 0) return true; 
        if (source.length() == 0) return false;
        
        int x = 0; 
        for (char c : source.toCharArray()) {
            if (c == target.charAt(x)) 
                x++;
            if (x == target.length()) 
                return true; 
        }
        return false; 
    }
}








In  C :







#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

char *hackerrank(char *string)
{
    int total;
    int i;
    int j;    
    char *comp = "hackerrank\0";
    
    i = 0;
    j = 0;
    while (string[i] != '\0')
    {
        if (string[i] == comp[j])
            j += 1;
        i += 1;
    }
    if (strlen(comp) == j)
        return ("YES");
    else
        return ("NO");
}

int main(){
    int q; 
    scanf("%d",&q);
    for(int a0 = 0; a0 < q; a0++){
        char* s = (char *)malloc(512000 * sizeof(char));
        scanf("%s",s);
        printf("%s\n", hackerrank(s));
        // your code goes here
    }
    return 0;
}








In   Python3  :







#!/bin/python3

import sys
import re

q = int(input().strip())
for a0 in range(q):
    s = input().strip()
    
    pattern = ".*h.*a.*c.*k.*e.*r.*r.*a.*n.*k.*"
    m = re.search(pattern, s)
    if m is not None:
        print("YES")
    else:
        print("NO")








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