Hello Programmers, In this post, you will learn how to solve HackerRank in a String Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank in a String Solution
Task
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.
s = haacckkerrannkk
This contains a subsequence of all of the characters in the proper order. Answer YES
s = haacckkerrannk
This is missing the second ‘r’. Answer NO.
s = hccaakkerrannkk
There is no ‘c’ after the first occurrence of an ‘a’, so answer NO.
Function Description
Complete the hackerrankInString function in the editor below.
hackerrankInString has the following parameter(s):
- string s: a string
Returns
- string:
YESorNO
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 <= 102
- 10 <= length of s <= 104
Sample Input 0
2 hereiamstackerrank hackerworld
Sample Output 0
YES NO
Explanation 0
We perform the following q = 2 queries:
- s = hereiamstackerrank
The characters ofhackerrankare bolded in the string above. Because the string contains all the characters inhackerrankin the same exact order as they appear inhackerrank, we returnYES. - s = hackerworld does not contain the last three characters of
hackerrank, so we returnNO.
Sample Input 1
2 hhaacckkekraraannk rhbaasdndfsdskgbfefdbrsdfhuyatrjtcrtyytktjjt
Sample Output 1
YES NO
HackerRank in a String Solution
HackerRank in a String Solutions 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;
}HackerRank in a String Solutions in Cpp
#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;
}HackerRank in a String Solutions 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;
}
}HackerRank in a String Solutions in Python
import sys
needle = 'hackerrank'
n = len(needle)
q = int(sys.stdin.readline())
for _ in xrange(q):
s = sys.stdin.readline().strip()
i = 0
for c in s:
if c == needle[i]: i += 1
if i == n: break;
if i == n: print "YES"
else: print "NO"HackerRank in a String Solutions using JavaScript
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function containsString(string, substring) {
let charIndex = 0;
for(let i = 0; i < string.length; i += 1) {
const char = substring[charIndex];
if(string[i] === char) {
charIndex += 1;
}
}
return charIndex === substring.length;
}
function main() {
var q = parseInt(readLine());
for(var a0 = 0; a0 < q; a0++){
var string = readLine();
const result = containsString(string, 'hackerrank');
console.log(result ? 'YES' : 'NO');
}
}HackerRank in a String Solutions in Scala
import scala.io.StdIn
object Solution {
val hackerrank = List('h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k')
def main(args: Array[String]): Unit = {
val q = StdIn.readInt()
(0 until q).foreach(i => {
val s = StdIn.readLine().toList
println(if (contains(s, hackerrank)) "YES" else "NO")
})
}
def contains(s: List[Char], h: List[Char]): Boolean = h match {
case Nil => true
case hHead :: hTail => s match {
case Nil => false
case sHead :: sTail => if (sHead == hHead) contains(sTail, hTail) else contains(sTail, h)
}
}
}HackerRank in a String Solutions in Pascal
Disclaimer: This problem (HackerRank in a String) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.
Next: HackerRank Weighted Uniform Strings Solution
