Ruby Math Hackerrank
Contents 
Install
Install vim plugin seeingisbelieving and write Gemfile
# Gemfile
source 'https://rubygems.org'
ruby '2.2.3'
# for instant output of commands
gem 'seeing_is_believing'
# irbtools includes interactive_editor (vim inside irb)
# just create ~/.irbrc with
# require 'rubygems'
# require 'irbtools'
gem 'irbtools'
# debugging
gem 'byebug'
# simple test assertions
gem 'minitest'
Links
Math and algorith tasks
 https://www.hackerrank.com
 nice editor, solution have to be memory effiction (it raises Timeout or Run Time Error)
 https://projecteuler.net/
 http://www.beatmycode.com/
 http://codility.com/programmers/
Codility Solutions
 https://github.com/GNakayama/codility/tree/master/src/python and blog http://blog.nkbits.com/
 http://codesays.com/ http://codesays.com/solutionstotrainingbycodility/
 http://www.martinkysel.com/codilitysolutions/
Just enough ruby

read from STDIN or from a file if you run with
ruby script.rb my_data
n = gets.to_i array = gets.strip.split(' ').map(&:to_i)

write to a file
File.open "my_file_name", 'w' do file file.puts "first line (no need for new line char \ n)" file << "second line \n" file.write "third line \n" end
 division is integer division, so that
1/2 == 0
in ruby. You need to convert to floatx = x.to_f
 iterate range
(1..10).each
or1.upto(10).each
 exponent is
2**2 == 4
Example code
We can use pp
standard library to print arrays just require 'pp'
. But if you need to print
matrix, than you can use ppp require './ppp'
# ppp.rb
def ppp(x)
if x.class == Array
puts_array(x)
else
puts x
end
end
def puts_array(x)
need_new_line = false
x.each do e
if e.class == Array
puts e.join(', ')
else
print e.to_s + ', '
need_new_line = true
end
end
puts if need_new_line
end
Here is example for one solution that is submitted on site (it can not
require byebug
) so it use constant PRODUCTION. Test are in the same file.
If the task is to read from file and to submit generated file, than use second
tempalte.
# 0_short_palindrome.rb
# https://www.hackerrank.com/challenges/shortpalindrome
# string s of n lowecase letters s_i 0<=i<n
# (a, b, c, d)
# s_a = s_b chars at a and d are the same
# s_b = s_c
# constrains:
# 0<=a<b<c<d<n
# input: string s
# 1<= s < 10^6
# output: number of (a, b, c, d) tuples, use moduo 10^9 + 7
PRODUCTION = false
if PRODUCTION
def ppp(_arg = nil); end
def pp(_arg = nil); end
else
require 'pp'
require './ppp'
require 'byebug'
end
# rubocop:disable Metrics/MethodLength
# rubocop:disable Metrics/AbcSize
def solution(a)
r = a.length
ppp "r=#{r}"
r
end
if PRODUCTION
n = gets.to_i
a = []
n.times do
# strip is needed since on hackerrank, new line is at the end of gets
r = gets.strip.split('').map { c c == '*' ? 0 : 1 }
a << r
end
puts solution(a)
end # if PRODUCTION
require 'minitest/autorun'
class Test < Minitest::Test
def test_sample
s = 'kkkkkkz'
assert_equal 12, solution(prepare(s))
end
def test_long
s = %(***.*
**..*
***.*
*****
*****)
assert_equal 12, solution(prepare(s))
end
def test_float
asert_in_delta 0.5, solution(1,2) # default delta=0.001
end
def test_that_is_skipped
skip "later"
end
private
def prepare(s)
s
end
end
This template has separate test and actual code
# my_script.rb
# description of a problem
require 'pp'
require './ppp'
require 'byebug'
def solution(aa:, l:, h:)
end
# when we run: ruby my_script.rb data/example.in
# output is saved in data/example.out
if ARGV.length == 1
file_name = ARGV.last.split('.').first
r, _c, l, h = gets.split(' ').map(&:to_i)
aa = []
r.times do
# strip is needed since on hackerrank, new line is at the end of gets
aa << gets.strip.split('').map { c c == 'M' ? 1 : 0 }
end
results = solution(aa: aa, l: l, h: h)
File.open "#{file_name}.out", 'w' do file
# puts results.length
file.puts results.length
results.each do row
# puts row.join(' ')
file.puts row.join(' ')
end
end
end
# test_my_script.rb
require 'minitest/autorun'
require './pizza'
class Test < Minitest::Test
def test_solution
a =
[
1,
]
res = 1
assert_equal res, solution(a)
end
end
If you want to run only one test, than run ruby 0_short_palindrome.rb name
test_sample
If you want to print all local variables, you can use
local_variables.map { vn { vn => eval(vn.to_s) } }
Run with colors ruby my_class_test.rb p
, or ruby r
minitest/pride my_class_test.rb
or put require 'pride'
in test file.
For minitests you can use describe
and it
blocks.
TIPS:
 if you need to perform big number of tests, and you can not calculate solution
but you need to iterate something, than use table of solutions. First call
create_table
of results and query that@table[n]
(if you can, fill just what it needs)  two dimensional 2d array
@table = Array.new(n) { Array.new(n) }
 if stack level too deep then try to use
@table
class variables instead of passing them as arguments (not much help since arguments are passed by reference, but there are less number of arguments link)
Complexity
Ruby Array#find
method has O(n) complexity. If array is big, you should use
bsearch O(log(n)).
Array needs to be sorted. Binary search works in two modes:
 find minimum: block must return true or false and there must be an index
i
so that: block returns false elements whose index is less thani
, block returns true for any element whose index is greater than or equal toi
. This method returnsi
th element.  find any: block returns number and there are must be two indices
i
andj
(0<=i<=j<=ary.size
) so that: block returns positive number for0<=k<i
, returns zero fori<=k<j
and negative forj<=k<ary.size
. This method returns any element whose index is withini..j
.
a = [1, 4, 7, 10]
# find minimum that satisfy block
a.bsearch { x x >= 6 } #=> 7
# find any
ary = [0, 4, 7, 10, 12]
# try to find v such that 4 <= v < 8
ary.bsearch {x 1  x / 4 } #=> 4 or 7
# try to find v such that 8 <= v < 10
ary.bsearch {x 4  x / 2 } #=> nil
Array a.include?
is O(n) but hash a.key?
(or a.include?
) is O(1). So
instead of big time, you can use big space by using Hashtables.
Counting of elements
tutorial if we need to check if some elements exists, we can create counting array in O(N) time and query in constant time
a = [0, 3, 1, 3]
counting = [0]*(a.max+1)
a.each { e counting[e]+=1 }
counting # => [1, 1, 0, 2]
counting[2] == 0 # true in constant time
Prefix sums
tutorial In O(N) time we can calculate p_1, …p_n, where p_k= a_0+a_1+…+a_(k1). Note that p_k does not use a_k, and p_0 == 0 Than we can calculate sum of any slice (x,y) in constant time instead of O(yx) (that is usually needed for query several results). Slice could be defined on 1..n (index starts from 1 and include bounds), slice(x,y) = prefix_sum[y]  prefix_sum[x1] for example
slice(a_0,a_0) = slice(1,1) = prefix_sum[1]prefix_sum[0]
slice(a_1,a_3) = slice(2,4) = prefix_sum[4]prefix_sum[1]
a = [1, 4, 0, 4, 2]
prefix_sum = [0] * (a.length+1)
a.each_with_index do e,i
prefix_sum[i+1] = prefix_sum[i] + e
end
# slice a_1 + a_2 + a_3
prefix_sum[3+1]prefix_sum[1] # => 8
Counting sort O(n+m)
a=[4,2,0,2,3]
c=[0] * (a.max+1)
a.each_with_index do e,i
c[e] += 1
end
s=[]
i=0
c.each_with_index do n,v # => [1, 0, 2, 1, 1]
n.times do
s[i] = v
i+=1
end
end
s # => [0, 2, 2, 3, 4]
Merge sort O(n*log(n))
a=[4,2,0,2,3]
def sort(a)
if a.length == 1
return a # => [4], [2], [0], [2], [3]
else
middle = a.length/2
a[0..middle1] # => [4, 2], [4], [0], [2]
a[middle..1] # => [0, 2, 3], [2], [2, 3], [3]
left = sort(a[0..middle1]) # => [4], [2, 4], [0], [2]
right = sort(a[middle..1]) # => [2], [3], [2, 3], [0, 2, 3]
i = 0
j = 0
r = []
while ! left[i].nil?  ! right[j].nil?
if ! left[i].nil? && (right[j].nil?  left[i] <= right[j])
r << left[i] # => [2, 4], [2], [0], [0, 2], [0, 2, 2, 3, 4]
i += 1
else
r << right[j] # => [2], [2, 3], [0, 2], [0, 2, 3], [0], [0, 2, 2], [0, 2, 2, 3]
j += 1
end
end
return r # => [2, 4], [2, 3], [0, 2, 3], [0, 2, 2, 3, 4]
end
end
sort(a) # => [0, 2, 2, 3, 4]
 leader (value on more than half elements) is the same when we remove pair of different items

maximum slice is a slice (a_p..a_q) with max sum, can be computed in O(n^2)
# max_slice O(n^2) a=[5,7,3,5,2,4,1] res = 0 a.each_with_index do p,i sum = 0 a[i..1].each do q sum += q res = [res,sum].max end end res # => 10

positive ending sum slice is wrapped with negative elements (or bounds first/last) because we can always choose empty slice (sum is 0) we can deduce that positive_ending_sum = max(0, positive_ending_sum_prev + a)
# max_slice O(n) a=[5,7,3,5,2,4,1] res = 0 positive_ending_sum = 0 a.each do p # => [5, 7, 3, 5, 2, 4, 1] positive_ending_sum = [0, positive_ending_sum+ p].max # => 5, 0, 3, 8, 6, 10, 9 res = [res,positive_ending_sum].max # => 5, 5, 5, 8, 8, 10, 10 end res # => 10
Prime Numbers
 count number of divisors of N is O(sqrt(N)) since every divisor a has symetric
divisor b,
n = a * b
n = 30
i = 1
result = 0
while i * i < n
result += 2 if n % i == 0
i += 1
end
result += 1 if i * i == n

finding all primes less than N is in O(Nlog(log(N))) Sieve of Eratosthenes. Find first prime (until sqrt(N) since we do not need to mark greater than N) and mark all multiplies of prime, starting from primeprime (since iprime will be marked with prime divisor of i) untill N, that is n/2 + n/3+ n/5 … = nlog(log(n) Rembember that largest prime factor of N can not be greated than Math.sqrt(N). 0, 1 are not prime numbers.
m = 10 definitely_non_prime = Array.new m + 1, false definitely_non_prime[0] = definitely_non_prime[1] = true 2.upto(Math.sqrt(m)) do i next if definitely_non_prime[i] prime = i non_prime = prime * prime while non_prime <= m definitely_non_prime[non_prime] = true non_prime += prime end end definitely_non_prime.each_with_index { e,i puts i if e == false }

factorization is to find prime numbers and its order. We can create fixed F where F[i] is smallest prime that divides i, ie F[12] = 2, F[6] = 2, F[2] = 0. If we already have that F, than we can factorize in O(log(n)) time
n = 12 f = [0]*(n+1) i=2 while i*i <= n if f[i] == 0 prime = i temp = prime * prime while temp <= n f[temp] = prime if f[temp] == 0 temp += prime end end i += 1 end f # => [0, 0, 0, 0, 2, 0, 2, 0, 2, 3, 2, 0, 2] def factorization(n, f) res = [] while f[n] > 0 res << f[n] # => [2], [2, 2] n /= f[n] # => 6, 3 end res << n # => [2, 2, 3] end factorization(n,f) # => [2, 2, 3]

we can find all divisors, for example 12 has [1,2,3,4,6,12] divisor list, with simply iterate to all numbers and add number (for example 2) to set for product (divisors[4].add(2), divisors[6].add(2) …). Better perfomance is when we stop at
number * number <= n
and we adddivisors[product].add(product/number)
(note that we still iterate to n in inner loop).n = 12 divisors = [] * (n+1) [*1..n].each do i # => [1, 2, 3, 4, 5, 6] divisors[i] = Set.new([1,i]) # => #<Set: {1}>, #<Set: {1, 2}>, #<Set: {1, 3}>, #<Set: {1, 4}>, #<Set: {1, 5}>, #<Set: {1, 6}> end number = 2 while number <= n # number * number <= n product = number # => 2, 3, 4, 5, 6 while product <= n if ! divisors[product].include? number divisors[product].add number # => #<Set: {1, 4, 2}>, #<Set: {1, 6, 2}>, #<Set: {1, 6, 2, 3}> # divisors[product].add(product/number) end product += number end number += 1 end
Euclidean algorithm
Theory:
Def: integer m divides n (mn) if exists r such that m*r = n Def: gcm(m, n) is the largest positive integer d which divides both m and n. That means dm, dn and for any c which cm and cn we have d>=c
The: for a, b nonzero integer, their gcd(a,b) is linear combination of a and b
ie: exists s and t such that sa+tb=gcd(a,b)
Proof: let d be least positive linear combination d = sa+tb . We can show that
da.
We can write a = d*q+r (0 <= r < d )
so `r = a  dq = a  (sa + tb)q = (1
 qs)a + (qt)b
. Since d is least positive and
r<dthat means
r = 0`. Similarly db. We can show it is the greatest common divisor because if d1=gcd(a,b) < d it will be that d1a d1b and from d=sa+t*b that d1d which is contradiction so d1==d
gcd(k, 1) = 1 if ab*c and gcd(a,b)==1 than ac
Lem: if a = bq + r than gcd(a,b) == gcd(b,r) Proof: gcd(a,b)=gcd(bq+r,b)=gcd(r,b) GCD will not change if you can add multiple of b, ie gcd(a,b)=gcd(a+k*b,b)
Def: The integers a and b are relatively prime if and only if there exist integers α and β such that aα + bβ = 1
Def: Lcm(a,b) least common multiple is =  a*b  /gcd(a,b) 
The: ax+by=d has integer solution iff gcd(a,b)  d 
If (x0,y0) is solution, we can add + ab/d ie ax+by + ab/d  ba/d = a(x+b/d) +b(ya/d) so (x+ b/d, ya/d) is solution. General solutions are { (x0+kb/d, y0ka/d) for k in Z}

Found gcd(a,b) with Euclidean algorithm by subtraction in O(a+b) steps, or by division in O(log(a+b))
def gcd(a,b) if a % b == 0 return b else gcd(b, a % b) end end

lcm(a,b) = a*b/gcd(a,b) and lcm(a,b,c) = lcm(a, lcm(b,c))
 Fibonacci numbers F[n] = F[n1] + F[n2] and F[0] = 0 and F[1] = 1 and can be generated in O(N)

faster could be with matrix calculation

binary search for sorted array could be in O(log(N)) time. We found middle element and test condition. Array should be sorted (ascending) so we can test (mid<=target) and pick next item. Result is last middle when condition was true (max element <= target). If it is true go with (bigger) elements, othervise go with lower elements. We do not stop when we found solution since usual we have optimization problem (min or max). When we pass solution and set first = middle +1 (could be equal last) next iteration will be false (or last = new_middle 1 until last < first). If we search for min, then if condition is true we took lower elements (last = mid  1). That is easy to replace in check function but result should be taken from that as well. So general check function store result if test is true with max, or test is false with min. Another nice non general solution is to return last for max, or first for min.
a=[1,2,3,4,5] #[12,15,15,19,24,31,53,59,60,61] target = 3 # 13 def check(a, middle, target, result) if a[middle] <= target result.value = middle true else false end end #a = [0,1,1,0,1,1,0,1,1] #target = 3 # k boards 3 boards #def check(a, middle, target, result) # last_board = 1 # boards = 0 # middle # => 4, 1, 2 # for i in 1..a.length # if a[i] == 1 && last_board < i # boards += 1 # => 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 3 # last_board = i + middle  1 # => 4, 8, 1, 2, 4, 5, 7, 8, 2, 5, 8 # end # end # boards # => 2, 6, 3 # if boards <= target # => true, false, true # # since we search to min, we invert return value # result.value = middle # false # else # true # end #end class Result attr_accessor :value end def binary_search(a, target) first = 0 last = a.length  1 result = Result.new result.value = 1 while first <= last first # => 0, 3 last # => 4, 4 middle = (first+last)/2 # => 2, 3 if check(a, middle, target, result) # => true, false first = middle + 1 # => 3 else last = middle  1 # => 2 end end return result.value end binary_search(a, target) # => 2

caterpillar method. found subseqence which total sum equals s
a = [6,2,7,4,1,3,6] target_sum = 12 back = 0 front = 0 sum = 0 a.each_with_index do e, back while front < a.length && sum + a[front] <= target_suM sum += a[front] front += 1 end if sum == target_sum puts "je" end sum = e end
 greedy algorithm is simplest possible algorithm

dinamic programming
def dynamic_coin_changing(c, k) n = c.length dp = [0] + [1/0.0]* k # => [0, Infinity, Infinity, Infinity, Infinity, Infinity, Infinity] for i in 1..n # => 1..3 for j in c[i1] .. k # => 1..6, 3..6, 4..6 # => 1..6, 3..6, 4..6 dp[jc[i1]]+1 # => 1, 2, 3, 4, 5, 6, 1, 2, 3, 2, 1, 2, 3 dp[j] # => Infinity, Infinity, Infinity, Infinity, Infinity, Infinity, 3, 4, 5, 6, 2, 3, 2 dp[j] = [dp[jc[i1]]+1, dp[j]].min # => 1, 2, 3, 4, 5, 6, 1, 2, 3, 2, 1, 2, 2 end dp # => [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 1, 2, 3, 2], [0, 1, 2, 1, 1, 2, 2] end dp end dynamic_coin_changing([1,3,4],6) # => [0, 1, 2, 1, 1, 2, 2]
Geometry
 Find line that is perpendicular to line
y = 2*x + 11
and contains(6, 7)
video Perpendicular means slope is negative inverse ie1/2
. Note that in ruby you need to usex = x.to_f
since1/2==0
for integer division  maximum area for fixed length sides (edges) polygon has edges on circle (descripted circuit) and orders does not matter link
 triangle area
s = (a+b+c)/2; P = Math.sqrt(s*(sa)*(sb)*(sc));
Math
radius = Math.sqrt((l1/2)**2 + (l2/2)**2)
Tutorial
Constrains
 memory: usually if you need to loop more than 10**6 (1MB) that could be a problem