I am trying to populate an array according to the Collatz sequence. The constraints for the sequence are as follows:
positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Example Output
3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
Ideally, I wanted to construct a recursive call that would populate the array according to the constraints of the sequence. However, I believe my logic for the recursive call is extremely flawed. The intended behavior is to iterate over the nested array, manipulating only the last element of each sub array until the element reaches 1. I am trying to build my understanding of recursion and would appreciate any suggestions on how to fix this problem.
def collatzSum(maxNumber)
sequenceHash = Hash.new(0)
i = maxNumber
until i == 0 do
if i.even?
sequenceHash[i] = [(i), (i / 2)]
elsif i.odd? && i != 1
sequenceHash[i] = [(i), (3 * i + 1)]
elsif i == 1
sequenceHash[i] = [i]
end
i -= 1
end
p sequenceHash
helper_method recursion. Method should take in hash values and iterate according to if statements.
=begin
desired output
hash = {5=>[5,16, 8, 4, 2,1],
4=>[4,2,1],
3=>[3,10,5,16,8,4,2,1],
2=>[2,1],
1=>[1]}
=end
Code:
collatzChain = lambda do |k|
j = 0
k = j[-1]
until k == 1 do
if k.even?
sequenceHash[k] << (k / 2)
elsif k.odd?
sequenceHash[k] << (3 * k + 1)
end
end
j += 1
end
collatzChain.call(sequenceHash.values)
sequenceHash
end
collatzSum(5)
So you mention that you wanted a recursive algorithm, your current approach looks iterative to me. To be recursive, you need to call the method you're in with values closer and closer to a base condition and then, once you hit the base condition, you return back out, up the call chain building up your return values. So, for the Collatz sequence a recursive approach would look like:
def build_collatz_chain(max_number)
return_value = [max_number]
# our base condition is when the number passed in is equal to 1, so
# when we get 1 as the max_number, we'll return an array looking like
# [1]
return return_value if max_number == 1
if max_number.even?
# here, with an even max_number, we'll recurse and call our method
# again, passing in the new max_number, which is the current
# max_number / 2.
return_value + build_collatz_chain(max_number / 2)
else
# same as above, but we're odd, so we'll recurse with 3 * max_number + 1
return_value + build_collatz_chain(3 * max_number + 1)
end
end
and now when we call this with a value of 5, what will end up happening is something like:
call build_collatz_chain(5)
call build_collatz_chain(16)
call build_collatz_chain(8)
call build_collatz_chain(4)
call build_collatz_chain(2)
call build_collatz_chain(1)
We have hit the base condition! return with [1]
return from 2 with [2, 1]
return from 4 with [4, 2, 1]
return from 8 with [8, 4, 2, 1]
return from 16 with [16, 8, 4, 2, 1]
return from 5 with [5, 16, 8, 4, 2, 1]
So, now if you want a hash of all numbers up to the passed in max_number with their Collatz chains as values you can use a helper to call this for each value, up to max (this helper is iterative, but could be made recursive...exercise for the viewer if you want it recursive):
def collatz_sum(max_number)
{ }.tap do |sequence_hash|
max_number.downto(1) do |i|
sequence_hash[i] = build_collatz_chain(i)
end
end
end
and then when you call collatz_sum(5) you get back:
{5=>[5, 16, 8, 4, 2, 1], 4=>[4, 2, 1], 3=>[3, 10, 5, 16, 8, 4, 2, 1], 2=>[2, 1], 1=>[1]}
The reason your approach is iterative is in the collatzChain lambda, you are setting a value (j) and then incrementing it and just looping through until k is equal to 1. It's also an infinite loop because you initially set k as:
j = 0
k = j[-1]
and so k == 0, and then you iterate until k == 1 and then you never update what the value of k is again.
It's not clear that a recursive operation is necessary here since this seems to be a straightforward mapping between a value x and f(x). By switching to a simple array output you can achieve what you want with:
def collatz_sum(max)
(2..max).map do |i|
[
i,
if (i.even?)
i / 2
else
3 * i + 1
end
]
end.reverse + [ [ 1 ] ]
end
For example, consider array A:
[3, 1, 2, 4, 3]
I can split this array in four places:
P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7
The minimal difference is 1.
Each element of array is an integer within the range [−1,000..1,000].
Below is the code I wrote. My code is correct for this exercise, and is correct for the given input but incorrect in some other cases. I do not understand why.
def minimaldifference(a)
sumArr = []
sumArr = sumfromstarting(a)
sum = sumArr.last
sumRev = sumFromReverse(a, sum)
size = sumArr.size-2
min = 0
v = 0
for i in 0..size
if(i==0)
min = (sumArr[i] - sumRev[i]).abs
else
v = (sumArr[i] - sumRev[i]).abs
if( v < min)
min = v
end
end
end
min
end
def sumfromstarting(a)
sumArr = []
sum = 0
a.each do |i|
sum += i
sumArr.push(sum)
end
sumArr
end
def sumFromReverse(a, sum)
sumArr = []
rsum = 0
a.each do |i|
rsum += i
sumArr.push(sum - i)
end
sumArr
end
a = [3, 1, 2, 4, 3]
puts minimaldifference(a)
This solution requires that all elements of the array be non-negative, which I understand is the case.
Consider
a = [4,1,3,1,2,4,3,1,5]
The possible partitions of this array are as follows:
Partition Totals Diff Abs Diff
[4], [1,3,1,2,4,3,1,5] [4, 20]
[4,1], [3,1,2,4,3,1,5] [4+1, 20-1] =>[5, 19] -14 14
[4,1,3], [1,2,4,3,1,5] [5+3, 19-3] =>[8, 16] -8 8
[4,1,3,1], [2,4,3,1,5] [8+1, 16-1] =>[9, 15] -6 6
[4,1,3,1,2], [4,3,1,5] [9+2, 15-2] =>[11, 13] -2 2
[4,1,3,1,2,4], [3,1,5] [11+4, 13-4]=>[15, 9] 6 6
[4,1,3,1,2,4,3], [1,5] [15+3, 9-3] =>[18, 6] 12 12
[4,1,3,1,2,4,3,1], [5] [18+1, 6-1] =>[19, 5] 14 14
In each row of this table "Diff" equals the difference between the totals for the two sets. Because all elements of a are non-negative, these values are monotonically non-decreasing. The absolute differences are therefore non-increasing then non-decreasing. It follows that to obtain the minimum absolute difference of totals we simply step through this array, stopping when the absolute difference would increase if the next element were moved from the "right" array to the "left" array.
We can implement this in code as follows.
def minimize_difference(arr)
raise ArgumentError, "array must contain at least two elements" if arr.size < 2
left, *right = arr
last = left - right.reduce(:+)
left = [left]
while right.any?
test = last + 2*right.first
break if test.abs > last.abs
last = test
left << right.shift
end
[left, right, last.abs]
end
minimize_difference [3, 1, 2, 4, 3]
#=> [[3, 1, 2], [4, 3], 1]
I have to create a program in ruby on rails so that it will take less time to solve the particular condition. Now i am to getting the less response time for k=4 but response time is more in case of k>5
Problem:
Problem is response time.
When value of k is more than 5 (k>5) response time is too late for given below equation.
Input: K, N (where 0 < N < ∞, 0 < K < ∞, and K <= N)
Output: Number of possible equations of K numbers whose sum is N.
Example Input:
N=10 K=3
Example Output:
Total unique equations = 8
1 + 1 + 8 = 10
1 + 2 + 7 = 10
1 + 3 + 6 = 10
1 + 4 + 5 = 10
2 + 2 + 6 = 10
2 + 3 + 5 = 10
2 + 4 + 4 = 10
3 + 3 + 4 = 10
For reference, N=100, K=3 should have a result of 833 unique sets
Here is my ruby code
module Combination
module Pairs
class Equation
def initialize(params)
#arr=[]
#n = params[:n]
#k = params[:k]
end
#To create possible equations
def create_equations
return "Please Enter value of n and k" if #k.blank? && #n.blank?
begin
Integer(#k)
rescue
return "Error: Please enter any +ve integer value of k"
end
begin
Integer(#n)
rescue
return "Error: Please enter any +ve integer value of n"
end
return "Please enter k < n" if #n < #k
create_equations_sum
end
def create_equations_sum
aar = []
#arr = []
#list_elements=(1..#n).to_a
(1..#k-1).each do |i|
aar << [*0..#n-1]
end
traverse([], aar, 0)
return #arr.uniq #return result
end
#To check sum
def generate_sum(*args)
new_elements = []
total= 0
args.flatten.each do |arg|
total += #list_elements[arg]
new_elements << #list_elements[arg]
end
if total < #n
new_elements << #n - total
#arr << new_elements.sort
else
return
end
end
def innerloop(arrayOfCurrentValues)
generate_sum(arrayOfCurrentValues)
end
#Recursive method to create dynamic nested loops.
def traverse(accumulated,params, index)
if (index==params.size)
return innerloop(accumulated)
end
currentParam = params[index]
currentParam.each do |currentElementOfCurrentParam|
traverse(accumulated+[currentElementOfCurrentParam],params, index+1)
end
end
end
end
end
run the code using
params = {:n =>100, :k =>4}
c = Combination::Pairs::Equation.new(params)
c.create_equations
Here are two ways to compute your answer. The first is simple but not very efficient; the second, which relies on an optimization technique, is much faster, but requires considerably more code.
Compact but Inefficient
This is a compact way to do the calculation, making use of the method Array#repeated_combination:
Code
def combos(n,k)
[*(1..n-k+1)].repeated_combination(3).select { |a| a.reduce(:+) == n }
end
Examples
combos(10,3)
#=> [[1, 1, 8], [1, 2, 7], [1, 3, 6], [1, 4, 5],
# [2, 2, 6], [2, 3, 5], [2, 4, 4], [3, 3, 4]]
combos(100,4).size
#=> 832
combos(1000,3).size
#=> 83333
Comment
The first two calculations take well under one second, but the third took a couple of minutes.
More efficient, but increased complexity
Code
def combos(n,k)
return nil if k.zero?
return [n] if k==1
return [1]*k if k==n
h = (1..k-1).each_with_object({}) { |i,h| h[i]=[[1]*i] }
(2..n-k+1).each do |i|
g = (1..[n/i,k].min).each_with_object(Hash.new {|h,k| h[k]=[]}) do |m,f|
im = [i]*m
mxi = m*i
if m==k
f[mxi].concat(im) if mxi==n
else
f[mxi] << im if mxi + (k-m)*(i+1) <= n
(1..[(i-1)*(k-m), n-mxi].min).each do |j|
h[j].each do |a|
f[mxi+j].concat([a+im]) if
((a.size==k-m && mxi+j==n) ||
(a.size<k-m && (mxi+j+(k-m-a.size)*(i+1))<=n))
end
end
end
end
g.update({ n=>[[i]*k] }) if i*k == n
h.update(g) { |k,ov,nv| ov+nv }
end
h[n]
end
Examples
p combos(10,3)
#=> [[3, 3, 4], [2, 4, 4], [2, 3, 5], [1, 4, 5],
# [2, 2, 6], [1, 3, 6], [1, 2, 7], [1, 1, 8]]
p combos(10,4)
#=> [[2, 2, 3, 3], [1, 3, 3, 3], [2, 2, 2, 4], [1, 2, 3, 4], [1, 1, 4, 4],
# [1, 2, 2, 5], [1, 1, 3, 5], [1, 1, 2, 6], [1, 1, 1, 7]]
puts "size=#{combos(100 ,3).size}" #=> 833
puts "size=#{combos(100 ,5).size}" #=> 38224
puts "size=#{combos(1000,3).size}" #=> 83333
Comment
The calculation combos(1000,3).size took about five seconds, the others were all well under one second.
Explanation
This method employs dynamic programming to compute a solution. The state variable is the largest positive integer used to compute arrays with sizes no more than k whose elements sum to no more than n. Begin with the largest integer equal to one. The next step is compute all combinations of k or fewer elements that include the numbers 1 and 2, then 1, 2 and 3, and so on, until we have all combinations of k or fewer elements that include the numbers 1 through n. We then select all combinations of k elements that sum to n from the last calculation.
Suppose
k => 3
n => 7
then
h = (1..k-1).each_with_object({}) { |i,h| h[i]=[[1]*i] }
#=> (1..2).each_with_object({}) { |i,h| h[i]=[[1]*i] }
#=> { 1=>[[1]], 2=>[[1,1]] }
This reads, using the only the number 1, [[1]] is the array of all arrays that sum to 1 and [[1,1]] is the array of all arrays that sum to 2.
Notice that this does not include the element 3=>[[1,1,1]]. That's because, already having k=3 elments, if cannot be combined with any other elements, and sums to 3 < 7.
We next execute:
enum = (2..n-k+1).each #=> #<Enumerator: 2..5:each>
We can convert this enumerator to an array to see what values it will pass into its block:
enum.to_a #=> [2, 3, 4, 5]
As n => 7 you may be wondering why this array ends at 5. That's because there are no arrays containing three positive integers, of which at least one is a 6 or a 7, whose elements sum to 7.
The first value enum passes into the block, which is represented by the block variable i, is 2. We will now compute a hash g that includes all arrays that sum to n => 7 or less, have at most k => 3 elements, include one or more 2's and zero or more 1's. (That's a bit of a mouthful, but it's still not precise, as I will explain.)
enum2 = (1..[n/i,k].min).each_with_object(Hash.new {|h,k| h[k]=[]})
#=> (1..[7/2,3].min).each_with_object(Hash.new {|h,k| h[k]=[]})
#=> (1..3).each_with_object(Hash.new {|h,k| h[k]=[]})
Enumerable#each_with_object creates an initially-empty hash that is represented by the block variable f. The default value of this hash is such that:
f[k] << o
is equivalent to
(f[k] |= []) << o
meaning that if f does not have a key k,
f[k] = []
is executed before
f[k] << o
is performed.
enum2 will pass the following elements into its block:
enum2.to_a #=> => [[1, {}], [2, {}], [3, {}]]
(though the hash may not be empty when elements after the first are passed into the block). The first element passed to the block is [1, {}], represented by the block variables:
m => 1
f => Hash.new {|h,k| h[k]=[]}
m => 1 means we will intially construct arrays that contain one (i=) 2.
im = [i]*m #=> [2]*1 => [2]
mxi = m*i #=> 2*1 => 2
As (m == k) #=> (1 == 3) => false, we next execute
f[mxi] << im if mxi + (k-m)*(i+1) <= n
#=> f[2] << [2] if 2 + (3-1)*(1+1) <= 7
#=> f[2] << [2] if 8 <= 7
This considers whether [2] should be added to f[2] without adding any integers j < i = 2. (We have yet to consider the combining of one 2 with integers less than 2 [i.e., 1].) As 8 <= 7, we do not add [2] to f[2]. The reason is that, for this to be part of an array of length k=3, it would be of the form [2,x,y], where x > 2 and y > 2, so 2+x+y >= 2+3+3 = 8 > n = 7. Clear as mud?
Next,
enum3 = (1..[(i-1)*(k-m), n-mxi].min).each
#=> = (1..[2,5].min).each
#=> = (1..2).each
#=> #<Enumerator: 1..2:each>
which passes the values
enum3.to_a #=> [1, 2]
into its block, represented by the block variable j, which is the key of the hash h. What we will be doing here is combine one 2 (m=1) with arrays of elements containing integers up to 1 (i.e., just 1) that sum to j, so the elements of the resulting array will sum to m * i + j => 1 * 2 + j => 2 + j.
The reason enum3 does not pass values of j greater than 2 into its block is that h[l] is empty for l > 2 (but its a little more complicated when i > 2).
For j => 1,
h[j] #=> [[1]]
enum4 = h[j].each #=> #<Enumerator: [[1]]:each>
enum4.to_a #=> [[1]]
a #=> [1]
so
f[mxi+j].concat([a+im]) if
((a.size==k-m && mxi+j==n) || (a.size<k-m && (mxi+j+(k-m-a.size)*(i+1))<=n))
#=> f[2+1].concat([[1]+[2]) if ((1==2 && 2+1==7) || (1<=3-1 && (2+1+(1)*(3)<=7))
#=> f[3].concat([1,2]) if ((false && false) || (1<=2 && (6<=7))
#=> f[3] = [] << [[1,2]] if (false || (true && true)
#=> f[3] = [[1,2]] if true
So the expression on the left is evaluated. Again, the conditional expressions are a little complex. Consider first:
a.size==k-m && mxi+j==n
which is equivalent to:
([2] + f[j]).size == k && ([2] + f[j]).reduce(:+) == n
That is, include the array [2] + f[j] if it has k elements that sum to n.
The second condition considers whether the array the arrays [2] + f[j] with fewer than k elements can be "completed" with integers l > i = 2 and have a sum of n or less.
Now, f #=> {3=>[[1, 2]]}.
We now increment j to 2 and consider arrays [2] + h[2], whose elements will total 4.
For j => 2,
h[j] #=> [[1, 1]]
enum4 = h[j].each #=> #<Enumerator: [[1, 1]]:each>
enum4.to_a #=> [[1, 1]]
a #=> [1, 1]
f[mxi+j].concat([a+im]) if
((a.size==k-m && mxi+j==n) || (a.size<k-m && (mxi+j+(k-m-a.size)*(i+1)<=n))
#=> f[4].concat([1, 1, 2]) if ((2==(3-1) && 2+2 == 7) || (2+2+(3-1-2)*(3)<=7))
#=> f[4].concat([1, 1, 2]) if (true && false) || (false && true))
#=> f[4].concat([1, 1, 2]) if false
so this operation is not performed (since [1,1,2].size => 3 = k and [1,1,2].reduce(:+) => 4 < 7 = n.
We now increment m to 2, meaning that we will construct arrays having two (i=) 2's. After doing so, we see that:
f={3=>[[1, 2]], 4=>[[2, 2]]}
and no other arrays are added when m => 3, so we have:
g #=> {3=>[[1, 2]], 4=>[[2, 2]]}
The statement
g.update({ n=>[i]*k }) if i*k == n
#=> g.update({ 7=>[2,2,2] }) if 6 == 7
adds the element 7=>[2,2,2] to the hash g if the sum of its elements equals n, which it does not.
We now fold g into h, using Hash#update (aka Hash#merge!):
h.update(g) { |k,ov,nv| ov+nv }
#=> {}.update({3=>[[1, 2]], 4=>[[2, 2]]} { |k,ov,nv| ov+nv }
#=> {1=>[[1]], 2=>[[1, 1]], 3=>[[1, 2]], 4=>[[2, 2]]}
Now h contains all the arrays (values) whose keys are the array totals, comprised of the integers 1 and 2, which have at most 3 elements and sum to at most 7, excluding those arrays with fewer than 3 elements which cannot sum to 7 when integers greater than two are added.
The operations performed are as follows:
i m j f
h #=> { 1=>[[1]], 2=>[[1,1]] }
2 1 1 {3=>[[1, 2]]}
2 1 2 {3=>[[1, 2]]}
2 2 1 {3=>[[1, 2]], 4=>[[2, 2]]}
{3=>[[1, 2]], 4=>[[2, 2]]}
3 1 1 {}
3 1 2 {}
3 1 3 {}
3 1 4 {7=>[[2, 2, 3]]}
3 2 1 {7=>[[2, 2, 3], [1, 3, 3]]}
g before g.update: {7=>[[2, 2, 3], [1, 3, 3]]}
g after g.update: {7=>[[2, 2, 3], [1, 3, 3]]}
h after h.update(g): {1=>[[1]],
2=>[[1, 1]],
3=>[[1, 2]],
4=>[[2, 2]],
7=>[[2, 2, 3], [1, 3, 3]]}
4 1 1 {}
4 1 2 {}
4 1 3 {7=>[[1, 2, 4]]}
g before g.update: {7=>[[1, 2, 4]]}
g after g.update: {7=>[[1, 2, 4]]}
h after h.update(g): {1=>[[1]],
2=>[[1, 1]],
3=>[[1, 2]],
4=>[[2, 2]],
7=>[[2, 2, 3], [1, 3, 3], [1, 2, 4]]}
5 1 1 {}
5 1 2 {7=>[[1, 1, 5]]}
g before g.update: {7=>[[1, 1, 5]]}
g after g.update: {7=>[[1, 1, 5]]}
h after h.update(g): {1=>[[1]],
2=>[[1, 1]],
3=>[[1, 2]],
4=>[[2, 2]],
7=>[[2, 2, 3], [1, 3, 3], [1, 2, 4], [1, 1, 5]]}
And lastly,
h[n].select { |a| a.size == k }
#=> h[7].select { |a| a.size == 3 }
#=> [[2, 2, 3], [1, 3, 3], [1, 2, 4], [1, 1, 5]]
#Cary's answer is very in-depth and impressive, but it appears to me that there is a much more naive solution, which proved to be much more efficient as well - good old recursion:
def combos(n,k)
if k == 1
return [n]
end
(1..n-1).flat_map do |i|
combos(n-i,k-1).map { |r| [i, *r].sort }
end.uniq
end
This solution simply reduces the problem each level by taking decreasing the target sum by each number between 1 and the previous target sum, while reducing k by one. Now make sure you don't have duplicates (by sort and uniq) - and you have your answer...
This is great for k < 5, and is much faster than Cary's solution, but as k gets larger, I found that it makes much too many iterations, sort and uniq took a very big toll on the calculation.
So I made sure that won't be needed, by making sure I get only sorted answers - each recursion should check only numbers larger than those already used:
def combos(n,k,min = 1)
if n < k || n < min
return []
end
if k == 1
return [n]
end
(min..n-1).flat_map do |i|
combos(n-i,k-1, i).map { |r| [i, *r] }
end
end
This solution is on par with Cary's on combos(100, 7):
user system total real
My Solution 2.570000 0.010000 2.580000 ( 2.695615)
Cary's 2.590000 0.000000 2.590000 ( 2.609374)
But we can do better: caching! This recursion does many calculations again and again, so caching stuff we already did will save us a lot of work when dealing with long sums:
def combos(n,k,min = 1, cache = {})
if n < k || n < min
return []
end
cache[[n,k,min]] ||= begin
if k == 1
return [n]
end
(min..n-1).flat_map do |i|
combos(n-i,k-1, i, cache).map { |r| [i, *r] }
end
end
end
This solution is mighty fast and passes Cary's solution for large n by light-years:
Benchmark.bm do |bm|
bm.report('Uri') { combos(1000, 3) }
bm.report('Cary') { combos_cary(1000, 3) }
end
user system total real
Uri 0.200000 0.000000 0.200000 ( 0.214080)
Cary 7.210000 0.000000 7.210000 ( 7.220085)
And is on par with k as high as 9, and I believe it is still less complicated than his solution.
You want the number of integer partitions of n into exactly k summands. There is a (computationally) somewhat ugly recurrence for that number.
The idea is this: let P(n,k) be the number of ways to partition n into k nonzero summands; then P(n,k) = P(n-1,k-1) + P(n-k,k). Proof: every partition either contains a 1 or it doesn't contain a 1 as one of the summands. The first case P(n-1,k-1) calculates the number of cases where there is a 1 in the sum; take that 1 away from the sum and partition the remaining n-1 into the now available k-1 summands. The second case P(n-k,k) considers the case where every summand is strictly greater than 1; to do that, reduce all of the k summands by 1 and recurse from there. Obviously, P(n,1) = 1 for all n > 0.
Here's a link that mentions that probably, no closed form is known for general k.
I have integers i.e. 9, 5, 4, 3, 1, 6, 7, 8. I want to return the index where a sequence of three descending or ascending integers exists. In the example above I would get indices 1 and 5. What is the ruby code for this?
def seq
array = [9,5,4,3,1,6,7,8]
array.each_with_index |val, index|
if (val < (array[index + 1]).val < (array[index + 1]).val)
puts "#{index}"
# Skip two indexes
end
end
I think the logic behind your solution is almost correct, but your syntax is pretty far off from valid Ruby.
Here are a pair of pretty verbose solutions that will (hopefully) be fairly obvious:
numbers = [9, 6, 5, 4, 3, 1, 6, 7, 8]
# Find non-overlapping sets
i = 0
until i > numbers.length - 2
a, b, c = numbers[i..i + 2]
if (a - b == b - c) && (a - b).abs == 1
puts "#{i} (#{a},#{b},#{c})"
# Skip next two indexes
i += 3
else
i += 1
end
end
# Find overlapping sets (same solution, but don't skip indexes)
(0...numbers.length - 2).each do |i|
a, b, c = numbers[i..i + 2]
if (a - b == b - c) && (a - b).abs == 1
puts "#{i} (#{a},#{b},#{c})"
end
end
Since the question is not clear enough. I will assume the question is about finding 3 ascending or descending continuous numbers. If the length of the satisfied sequence it longer than 3, e.g [2, 3, 4, 5], it returns 0 and 1.
Here is the algorithm, do list[index] - list[index - 1] for all elements, and repeat it for another time, the answer will be the index of 0 elements after the calculation.
Intuitively,
original 9, 5, 4, 3, 1, 6, 7, 8
first pass -4, -1, -1, -2, 5, 1, 1
2nd pass 3, 0, -1, 7, 4, 0 -> the answer will be the indexes of 0's, which is 1, 5
Algorithm:
lst = [9, 5, 4, 3, 1, 6, 7, 8]
lst1 = lst.each_cons(2).map{ |a, b| b-a }
lst2 = lst1.each_cons(2).map{ |a, b| b-a }
result = lst2.each_index.select{|i| lst2[i] == 0}
result = [1, 5]
Here’s a solution using each_cons(3).with_index:
[9,5,4,3,1,6,7,8].each_cons(3).with_index.select { |s, i| s[0] < s[1] && s[1] < s[2] }.map(&:last)