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")








View More Similar Problems

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →