doing the first Project Euler question: summing the multiples of 3 and 5 between 1 and 1000, I came up with this (pretty simple)
sum = 0
1.upto(999) { |i| sum += i if 0 == i%3 || 0 == i%5 }
sum
but I thought this would work but it doesn't, can someone show me what I'm doing wrong, or why it doesn't work?
1.upto(999).inject(0) { |sum, i| sum + i if 0 == i%3 || 0 == i%5 }
thanks!
inject passes the result of the block through to the next iteration as the first argument. Your block will return nil when your if statement is false, which then gets passed back in as sum.
To get the correct answer, the block should return the current sum when it's false:
1.upto(999).inject(0) { |sum, i| (0 == i%3 || 0 == i%5) ? sum + i : sum }
Complementary answer: if you're about to tackle Euler problems you should start to build your own extensions of reusable code. In this case, the first extension would be Enumerable#sum:
module Enumerable
def sum
inject(0, :+)
end
end
And now you can write a solution that separates the condition of the summatory (you can read it out loud and it makes sense, that's typical of functional/declarative style):
1.upto(999).select { |x| x % 3 == 0 || x % 5 == 0 }.sum
You could even push it one step further and create Fixnum#divisible_by? so you can write:
1.upto(999).select { |x| x.divisible_by?(3) || x.divisible_by?(5) }.sum
More: here it's not a problem, but later on strict implementations (those using arrays) will require too much memory. Try then with laziness:
require 'lazy'
1.upto(999).lazy.select { |x| x % 3 == 0 || x % 5 == 0 }.sum
1.upto(999).inject(0) { |sum, i| sum += i if 0 == i%3 || 0 == i%5; sum }
Would also work (note the +=).
Or, use & proc which addresses self.
(1..999).select{|x| x%3==0||x%5==0}.inject &:+
(1..999).to_a.keep_if{|d| d%3 == 0 || d%5 == 0}.reduce(:+) for completeness.
Related
Find the element in an array which has consecutive plus or minus 1. I can achieve in O(n) time by doing a linear search. Is there any efficient way to achieve in less than O(n) time.
What I mean by plus or minus 1 is 2 consecutive number in an array will have either -1 or 1 as the difference
For example sample array inputs
Consecutive element have +1 of difference
arr = [1,2,3,4,5]
Consecutive elements have -1 of difference
arr = [ -1,-2, -3, -4 , -5, ]
Elements have either +1 or -1 of difference
arr = [-5,-4,-3,-4, -3,-2,-1,0,1,2,3,4,3,2,1]
Note: Array is unsorted
Example
arr = [-5,-4,-3,-4,-3,-2,-1,0,1,2,3,2,1,2,3,4,5,6,5,4,3,2,1,0,-1,-2,-3,-4,-3,-2,-1]
while i <= arr.length do
if num == a[i]
puts "found"
break
end
i += 1
end
2nd approach which in my mind is instead of starting from 0 index I have done some custom logic
num = gets.to_i
i = 0
if (num < 0 && arr[0] < 0) || (num > 0 && arr[0] < 0)
i = (num + arr[0].abs).abs
elsif num < 0 && arr[0] > 0
i = num.abs + arr[0]
elsif num > 0 && arr[0] > 0
i = (num > arr[0] ? (num - arr[0]) : (arr[0] - num))
end
while i <= arr.length do
if num == a[i]
puts "found"
break
end
i += 1
end
As Nico Schertler said, O(n) is the best you can do. This is a find operation, and find operations on unsorted data are always O(n), because in the worst case they must check every element. This would be different if your data were sorted. See binary search.
Note also that your example code does not do what you're asking - it simply finds some number num in the list. Your example code is equivalent to:
arr.find { |i| i == num }
What you appear to be asking for is
arr.find { |i| i == num - 1 or i == num + 1 }
I'm trying to create a recursive method sum_of_digits(i) that takes the sum of integers, i.e. '456' = 4+5+6 = 15
However, I receive a NoMethodError for chr.to_i in the following code:
def sum_of_digits(i)
input = i.to_s
if i == 0
return 0
elsif input.length == 1
return i
else
for n in 1..input.length
sum += input[i].chr.to_i % 10^(n-1)
end
end
return sum
end
Thank you!
String indexes are zero-based in ruby. The problem is here:
for n in 1..input.length
it should be written as
for n in 0..input.length-1
BTW, call to chr is superfluous as well, since you already have a string representation of a digit there. As well, sum must be declared in advance and set to zero.
Also, the whole code is not ruby idiomatic: one should avoid using unnecessary returns and for-loop. The modified version (just in case) would be:
def sum_of_digits(i)
input = i.to_s
case
when i == 0 then 0 # return zero
when input.length == 1 then i # return i
else
sum = 0
input.length.times do |index|
sum += input[index].to_i % 10^index
end
sum
end
end
or, even better, instead of
sum = 0
input.length.times do |index|
sum += input[index].to_i % 10^index
end
sum
one might use inject:
input.length.times.inject(0) do |sum, index|
sum += input[index].to_i % 10^index
end
FizzBuzz, a classic problem, returns all the numbers up to N with a slight twist. If a number is divisible by 3, it is replaced with "fizz". If it's divisible by 5, it's replaced with "buzz". If it's divisible by both, it's replaced with "fizzbuzz"
I keep getting this Error message:
comparison of Fixnum with nil failed
Can someone explain this error message to me please? Also why is the code not working?
def fizz_buzz(n)
arr = (1..n).to_a
i = 0
while arr[i] < arr[n]
if i % 3 == 0 && i % 5 == 0
arr[i] = 'fizzbuzz'
elsif i % 3 == 0
arr[i] = 'fizz'
elsif i % 5 == 0
arr[i] = 'buzz'
else
arr[i] = i
i += 1
end
end
return arr
end
fizz_buzz(12)
Your conditions are just a bit off, give this a try:
def fizz_buzz(n)
arr = (1..n).to_a
i = 0
while i < n
if arr[i] % 3 == 0 && arr[i] % 5 == 0
arr[i] = 'fizzbuzz'
elsif arr[i] % 3 == 0
arr[i] = 'fizz'
elsif arr[i] % 5 == 0
arr[i] = 'buzz'
end
i+=1
end
return arr
end
Trying to access arr[n] puts you outside the bounds of the array which returns nil in Ruby.
You can update the code the ruby way, by using blocks and guards, I don't remember last time I used a while loop in ruby :)
Also, Array.new accepts a block as an argument which you can exploit to build your Array in a single step:
def fizz_buzz(n)
Array.new(n) do |index|
x = index + 1
case
when x % 3 == 0 && x % 5 == 0 then "fizzbuzz"
when x % 3 == 0 then "fizz"
when x % 5 == 0 then "buzz"
else x
end
end
end
Notice I used 1 as a base index and not 0, you can just remove x = index + 1 and replace x with index to have it working in a zero index base
A solution with a block instead of the while loop, and guards
def fizz_buzz(n)
arr = (1..n).to_a
0.upto(n - 1) do |i|
arr[i] = "fizzbuzz" and next if i % 3 == 0 && i % 5 == 0
arr[i] = "fizz" and next if i % 3 == 0
arr[i] = "buzz" if i % 5 == 0
end
arr
end
#brad-melanson beat me to the straight-forward answer to your question, so I'll share an answer which uses some common Ruby idioms (passing a range to the Array constructor and map), which simplify things, prevent you from having to do any iteration bookkeeping and prevent the possibility of off-by-one errors, out-of-bounds errors, etc.
def fizz_buzz(n)
Array(1..12).map do |n|
if n % 3 == 0 && n % 5 == 0
'fizzbuzz'
elsif n % 3 == 0
'fizz'
elsif n % 5 == 0
'buzz'
else
n
end
end
end
result = fizz_buzz 12
# result => [1, 2, "fizz", 4, "buzz", "fizz", 7, 8, "fizz", "buzz", 11, "fizz"]
factorial_sum(5) should return 3. The error I'm getting is that "inject is an undefined method". I was also wondering if it's possible to combine the two functions. I wasn't sure as I am just starting out on recursion. Thanks!
def factorial_sum(x)
factorial = factorial(x)
factorial.to_s.split('').collect { |i| i.to_i }
sum = factorial.inject { |sum, n| sum + n }
end
def factorial(x)
if x < 0
return "Negative numbers don't have a factorial"
elsif x == 0
1
else
factorial = x * factorial(x - 1)
end
end
puts factorial_sum(5)
factorial.to_s.split('').collect { |i| i.to_i }
This line is a no-op. You build a list and then throw it away. You probably meant factorial = ...
I have to say though that this would be pretty easy to find with a little effort and some print statements...
By the way, here's a slightly more concise way:
(1..x).reduce(:*).to_s.chars.map(&:to_i).reduce(:+)
A direct way without temporarily converting it into strings, and without recursion.
s, q = 0, 120
while q > 0
q, r = q.divmod(10)
s += r
end
s # => 3
I would like to pass an array of numbers to my is_prime? method and return if the numbers are valid or not. I do not want to use:
require 'prime'
a = [1,2,3,4,5]
Hash[a.zip(a.map(&Prime.method(:prime?)))]
This is learning experience. My current code is only outputing the first number in the array. Can someone help me understand what I am doing wrong? Thanks!
def is_prime?(*nums)
i = 2
nums.each do |num|
while i < num
is_divisible = ((num % i) == 0)
if is_divisible == false
x = "#{num}: is NOT a prime number." #false
else
x = "#{num}: is a prime number." #true
end
i +=1
end
return x
end
end
puts is_prime?(27,13,42)
You are returning in the loop.
A few bugs in your method:
def is_prime?(*nums)
nums.each do |num|
return false if num == 1
next if num == 2 # 2 is the only even prime
i = 2 # needs to be reset for each num
while i < num
return false if num % i == 0 # num is not prime
i += 1
end
end
true # We'll reach here only if all the numbers are prime
end
This will return your results in the same format as your usage of the prime library with the same logic as your custom function:
def is_prime?(*nums)
nums.each_with_object({}) do |num, hsh|
hsh[num] = num > 1 && 2.upto(num - 1).none? { |i| num % i == 0 }
end
end
puts is_prime?(27,13,42)
# => {27=>false, 13=>true, 42=>false}
Since you mention this is just for learning, I'm assuming you know that a sieve is a better way to go for this than brute force iteration.
If you want an explanation of how the above code works or further help understanding why your current code doesn't, let me know in the comments.