Hackerrank Lazy Evaluation Solution
Lazy evaluation is an evaluation strategy that delays the assessment of an expression until its value is needed.
Ruby introduced a lazy enumeration feature. Lazy evaluation increases performance by avoiding needless calculations, and it has the ability to create potentially infinite data structures.
Example:
power_array = -> (power, array_size) do
1.upto(Float::INFINITY).lazy.map { |x| x**power }.first(array_size)
end
puts power_array.(2 , 4) #[1, 4, 9, 16]
puts power_array.(2 , 10) #[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
puts power_array.(3, 5) #[1, 8, 27, 64, 125]
In this example, lazy
avoids needless calculations to compute power_array.
If we remove lazy from the above code, then our code would try to compute all ranging from to Float::INFINITY.
To avoid timeouts and memory allocation exceptions, we use lazy
. Now, our code will only compute up to first(array_size).
Task
Your task is to print an array of the first palindromic prime numbers.
For example, the first palindromic prime numbers are .
Input Format.
A single line of input containing the integer .
Constraints
You are not given how big is.
Output Format.
Print an array of the first palindromic primes.
Sample Input.
5
Sample Output.
[2, 3, 5, 7, 11]
Solution in ruby
Approach 1.
# Enter your code here. Read input from STDIN. Print output to STDOUT
require 'prime'
n = gets.to_i
p Prime.each.lazy.select {|i| i.to_s == i.to_s.reverse}.first(n)
Approach 2.
# Enter your code here. Read input from STDIN. Print output to STDOUT
require 'prime'
def yield_var(var_num)
sum = (2..Float::INFINITY).lazy.select{
|x| x.to_s.reverse.to_i == x && Prime.prime?(x) }.first(var_num.to_i)
return sum
end
num = gets.chomp
print yield_var(num)
Approach 3.
# Enter your code here. Read input from STDIN. Print output to STDOUT
def prime?(num)
num >= 2 && (2..Math.sqrt(num)).none? { |div| (num % div).zero? } ? true : false
end
def palindrome?(num)
num.to_s == num.to_s.reverse
end
p 2.upto(Float::INFINITY).lazy.select { |num| palindrome?(num) && prime?(num) }
.first(gets.to_i)