April 22, 2018 May 18, 2018 7 min reading

Warm-up Challenges Repeated String

I'm posting about my solutions to the HackerRank challenges as a way to improve my learning, reinforce my knowledge, and establish an understanding of the concepts covered. If I help someone by sharing along the way, even better!

Repeated String is one of the initial challenges with 'EASY' difficulty.

What's the challenge?

Lilah has a string, s, of lowercase English letters that are repeated infinitely. Given an integer, n, find and return the number of letters a in the first n letters of Lilah's infinite string.

Example

If Lilah's string is s = 'abcac' and n = 10, the substring we are considering is abcacabcac, the first 10 characters of her infinite string. There are 4 occurrences of the letter a in the substring.

What do we have to do?

A function that returns an integer number of occurrences of the letter a in the prefix of length n in the infinitely repeating string.

Parameters

• s = a string to repeat

• n = the number of characters to consider

Sample Input

aba
10

Sample Output

7

Solution

function repeatedString(s, n) {
	let amount = parseInt(n / s.length),
		count = s.split('').filter(c => c === 'a').length,
		rest = s.slice(0, n % s.length).split('').filter(c => c === 'a').length;

	return amount * count + rest;
}

Logic

let amount = parseInt(n / s.length),
		count = s.split('').filter(c => c === 'a').length,
		rest = s.slice(0, n % s.length).split('').filter(c => c === 'a').length;

All logic is contained in these three variables:

amount is the amount of times the string s appears based on n of Lilah's infinite string;

count is the total amount of a contained in the string s

rest is the total amount of a contained in the string resulting from the remainder of dividing the given length n with the string s

Finally, I finalize the function by returning the final count.

Extra Solutions

Other languages that I speak

Ruby

def repeatedString(s, n)
	amount = n / s.size
	count = s.count('a')
	rest = s.slice(0, n % s.size).count('a')

	amount * count + rest
end

PYTHON 2 & 3

def repeatedString(s, n):
	amount = n // len(s)
	count = s.count('a')
	rest = s.count('a', 0, n % len(s))

	return amount * count + rest

GOLANG

func repeatedString(s string, n int64) int64 {
	var rest, amount, count int64
	rest = n % int64(len(s))
	amount = n / int64(len(s))
	count = int64(strings.Count(s, "a"))
	total := amount * count + int64(strings.Count(s[0:rest], "a"))

	return total
}

SCALA

def repeatedString(s: String, n: Long): Long = {
	val rest = n % s.length
	val amount = n / s.length
	val count = s.count(_ == 'a')
	val total = amount * count + s.substring(0, rest.toInt).count(_ == 'a')

	return total
}

JAVA 7 & 8

static int repeatedString(String s, long n) {
	long rest = n % s.length();
	long amount = n / s.length();
	long count = s.chars().filter(c -> c == 'a').count();
	long total = amount * count + s.substring(0, (int) rest).chars().filter(c -> c == 'a').count();

	return total;
}

PHP

function repeatedString($s, $n) {
	$rest = substr_count(substr($s, 0, $n % strlen($s)), "a");
	$amount = intval($n / strlen($s));
	$count = substr_count($s, "a");

	return $amount * $count + $rest;
}

Thought: People who know little are usually great talkers, while men who know much say little.