Corrects sequences of parentesis can be defined recursively:
The empty string "" is a correct sequence.
If "X" and "Y" are correct sequences, then "XY" (the concatenation of
X and Y) is a correct sequence.
If "X" is a correct sequence, then "(X)" is a correct sequence.
Each correct parentheses sequence can be derived using the above
rules.
Given two strings s1 and s2. Each character in these strings is a parenthesis, but the strings themselves are not necessarily correct sequences of parentheses.
You would like to interleave the two sequences so that they will form a correct parentheses sequence. Note that sometimes two different ways of interleaving the two sequences will produce the same final sequence of characters. Even if that happens, we count each of the ways separately.
Compute and return the number of different ways to produce a correct parentheses sequence, modulo 10^9 + 7.
Example s1 = (() and s2 = ())
corrects sequences of parentheses, s1 (red) and s2(blue)
I don't understand the recursive algorithm, what does X and Y mean? And modulo 10^9 + 7?
First, I tried defining all permutations of s1 and s2 and then calculate the number of balanced parentheses. But that way is wrong, isn't it?
class InterleavingParenthesis:
def countWays(self, s1, s2):
sequences = list(self.__exchange(list(s1 + s2)))
corrects = 0
for sequence in sequences:
if self.__isCorrect(sequence):
corrects += 1
def __isCorrect(self, sequence):
s = Stack()
balanced = True
i = 0
while i < len(sequence) and balanced:
if '(' == sequence[i]:
s.stack(sequence[i])
elif s.isEmpty():
balanced = False
else: s.remove()
i += 1
if s.isEmpty() and balanced: return True
else: return False
def __exchange(self, s):
if len(s) <= 0: yield s
else:
for i in range(len(s)):
for p in self.__exchange(s[:i] + s[i + 1:]):
yield [s[i]] + p
class Stack:
def __init__(self):
self.items = []
def stack(self, data):
self.items.append(data)
def remove(self):
self.items.pop()
def isEmpty(self):
return self.items == []
Here's an example that shows how this recursive property works:
Start with:
X = "()()(())"
Through property 2, we break this into further X and Y:
X = "()" ; Y = "()(())"
For X, we can look at the insides with property 3.
X = ""
Because of property 1, we know this is valid.
For Y, we use property 2 again:
X = "()"
Y = "(())"
Using the same recursion as before (property 2, then property 1) we know that X is valid. Note that in code, you usually have to go through the same process, I'm just saving time for humans. For Y, you use property 3:
X = "()"
And again.. :
X = ""
And with property 1, you know this is valid.
Because all sub-parts of "()()(())" are valid, "()()(())" is valid. That's an example of recursion: You keep breaking things down into smaller problems until they are solvable. In code, you would have the function call itself with regards to a smaller part of it, in your case, X and Y.
As for the question you were given, there is a bit that doesn't make sense to me. I don't get how there is any room for doubt in any string of parentheses, like in the image you linked. In "((()())())" for example, there is no way these two parentheses do not match up: "((()())())". Therefore my answer would be that there is only one permutation for every valid string of parentheses, but this obviously is wrong somehow.
Could you or anyone else expand on this?
Related
I am trying to write a function that returns true or false if a given string has exactly 6 consecutive characters with the same value. If the string has more or less than 6, it will return false:
I am not allowed to use lists, sets or import any packages. I am only restricted to while loops, for loops, and utilizing basic mathematical operations
Two example runs are shown below:
Enter a string: 367777776
True
Enter a string: 3677777777776
False
Note that although I entered numbers, it is actually a string within the function argument for example: consecutive('3777776')
I tried to convert the string into an ASCII table and then try and filter out the numbers there. However, I
def consecutive(x):
storage= ' '
acc=0
count=0
for s in x:
storage+= str(ord(s)) + ' '
acc+=ord(s)
if acc == acc:
count+=1
for s in x-1:
return count
My intention is to compare the previous character's ASCII code to the current character's ASCII code in the string. If the ASCII doesnt match, I will add an accumulator for it. The accumulator will list the number of duplicates. From there, I will implement an if-else statement to see if it is greater or less than 6 However, I have a hard time translating my thoughts into python code.
Can anyone assist me?
That's a pretty good start!
A few comments:
Variables storage and acc play the same role, and are a little more complicated than they have to be. All you want to know when you arrive at character s is whether or not s is identical to the previous character. So, you only need to store the previously seen character.
Condition acc == acc is always going to be True. I think you meant acc == s?
When you encounter an identical character, you correctly increase the count with count += 1. However, when we change characters, you should reset the count.
With these comments in mind, I fixed your code, then blanked out a few parts for you to fill. I've also renamed storage and acc to previous_char which I think is more explicit.
def has_6_consecutive(x):
previous_char = None
count = 0
for s in x:
if s == previous_char:
???
elif count == 6:
???
else:
???
previous_char = ???
???
You could use recursion. Loop over all the characters and for each one check to see of the next 6 are identical. If so, return true. If you get to the end of the array (or even within 6 characters of the end), return false.
For more info on recursion, check this out: https://www.programiz.com/python-programming/recursion
would something like this be allowed?
def consecF(n):
consec = 1
prev = n[0]
for i in n:
if i==prev:
consec+=1
else:
consec=1
if consec == 6:
return True
prev = i
return False
n = "12111123333221"
print(consecF(n))
You can try a two pointer approach, where the left pointer is fixed at the first instance of some digit and the right one is shifted as long as the digit is seen.
def consecutive(x):
left = 0
while left != len(x):
right = left
while right < len(x) and x[right] == x[left]:
right += 1
length = (right - 1) - left + 1 # from left to right - 1 inclusive, x[left] repeated
if length == 6: # found desired length
return True
left = right
return False # no segment found
tests = [
'3677777777776',
'367777776'
]
for test in tests:
print(f"{test}: {consecutive(test)}")
Output
3677777777776: False
367777776: True
You should store the current sequence of repeated chars.
def consecutive(x):
sequencechar = ' '
repetitions = 0
for ch in x:
if ch != sequencechar:
if repetitions == 6:
break
sequencechar = ch
repetitions = 1
else:
repetitions += 1
return repetitions == 6
If I could, I would not have given the entire solution, but this still is a simple problem. However one has to take care of some points.
As you see the current sequence is stored, and when the sequence is ended and a new starts, on having found a correct sequence it breaks out of the for loop.
Also after the for loop ends normally, the last sequence is checked (which was not done in the loop).
I'm working on the problem stated in the question statement. I know my solution is correct (ran the program) but I'm curious as to whether or not I'm analyzing my code (below) correctly.
def parens(num)
return ["()"] if num == 1
paren_arr = []
parens(num-1).each do |paren|
paren_arr << paren + "()" unless "()#{paren}" == "#{paren}()"
paren_arr << "()#{paren}"
paren_arr << "(#{paren})"
end
paren_arr
end
parens(3), as an example, will output the following:
["()()()", "(()())", "(())()", "()(())", "((()))"]
Here's my analysis:
Every f(n) value is roughly 3 times as many elements as f(n-1). So:
f(n) = 3 * f(n-1) = 3 * 3 * f(n-2) ~ (3^n) time cost.
By a similar analysis, the stack will be occupied by f(1)..f(n) and so the space complexity should be 3^n.
I'm not sure if this analysis for either time or space is correct. On the one hand, the stack only holds n function calls, but each of these calls returns an array ~3 times as big as the call before it. Does this factor into space cost? And is my time analysis correct?
As others have mentioned, your solution is not correct.
My favourite solution to this problem generates all the valid combinations by repeatedly incrementing the current string to the lexically next valid combination.
"Lexically next" breaks down into a few rules that make it pretty easy:
The first difference in the string changes a '(' to a ')'. Otherwise the next string would be lexically before the current one.
The first difference is as far to the right as possible. Otherwise there would be smaller increments.
The part after the first difference is lexically minimal, again because otherwise there would be smaller increments. In this case that means that all the '('s come before all the ')'.
So all you have to do is find the rightmost '(' that can be changed to a ')', flip it, and then append the correct number of '('s and ')'s.
I don't know Ruby, but in Python it looks like this:
current="(((())))"
while True:
print(current)
opens=0
closes=0
pos=0
for i in range(len(current)-1,-1,-1):
if current[i]==')':
closes+=1
else:
opens+=1
if closes > opens:
pos=i
break
if pos<1:
break
current = current[:pos]+ ")" + "("*opens + ")"*(closes-1)
Output:
(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())(())
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()
Solutions like this turn out to be easy and fast for many types of "generate all the combinations" problems.
Recursive reasoning makes a simple solution. If the number of left parens remaining to emit is positive, emit one and recur. If the number of right parens remaining to emit is greater than the number of left, emit and recur. The base case is when all parens, both left and right, have been emitted. Print.
def parens(l, r = l, s = "")
if l > 0 then parens(l - 1, r, s + "(") end
if r > l then parens(l, r - 1, s + ")") end
if l + r == 0 then print "#{s}\n" end
end
As others have said, the Catalan numbers give the number of strings that will be printed.
While this Ruby implementation doesn't achieve it, a lower level language (like C) would make it easy to use a single string buffer: O(n) space. Due to substring copies, this one is O(n^2). But since the run time and output length are O(n!), O(n) space inefficiency doesn't mean much.
I found Tom Davis' article, "Catalan Numbers," very helpful in explaining one recursive method for defining the Catalan Numbers. I'll try to explain it myself (in part, to see how much of it I've understood) as it may be applied to finding the set of all unique arrangements of N matched parentheses (e.g., 1 (); 2 ()(), (()); etc. ).
For N > 1 let (A)B represent one arrangement of N matched parentheses, where A and B each have only balanced sets of parentheses. Then we know that if A contains k matched sets, B must have the other N - k - 1, where 0 <= k <= N - 1.
In the following example, a dot means the group has zero sets of parentheses:
C_0 => .
C_1 => (.)
To enumerate C_2, we arrange C_1 as AB in all ways and place the second parentheses around A:
. () = AB = C_0C_1 => (.)()
() . = AB = C_1C_0 => (()) .
Now for C_3, we have three partitions for N - 1, each with its own combinations: C_0C_2, C_1C_1, C_2C_0
C_0C_2 = AB = . ()() and . (()) => ()()(), ()(())
C_1C_1 = AB = ()() => (())()
C_2C_0 = AB = ()() . and (()) . => (()()), ((()))
We can code this method by keeping a set for each N and iterating over the combinations for each partition. We'll keep the individual arrangements as bits: 0 for left and 1 for right (this appears backwards when cast as a binary string).
def catalan
Enumerator.new do |y|
# the zero here represents none rather than left
s = [[0],[2]]
y << [0]
y << [2]
i = 2
while true
s[i] = []
(0..i - 1).each do |k|
as = s[k]
bs = s[i - k - 1]
as.each do |a|
bs.each do |b|
if a != 0
s[i] << ((b << (2*k + 2)) | (1 << (2*k + 1)) | (a << 1))
else
s[i] << (2 | (b << 2))
end
end # bs
end # as
end # k
y.yield(s[i])
i = i + 1
end # i
end # enumerator
end
catalan.take(4)
# => [[0], [2], [10, 12], [42, 50, 44, 52, 56]]
The yielder is lazy: although the list is infinite, we can generate as little as we like (using .take for example):
catalan.take(4).last.map{|x| x.to_s(2)}
# => ["101010", "110010", "101100", "110100", "111000"]
The former generation obliges us to keep all previous sets in order to issue the next. Alternatively, we can build a requested set through a more organic type, meandering recursion. This next version yields each arrangement to the block, so we can type:
catalan(4){
|x| (0..7).reduce(""){
|y,i| if x[i] == 0 then y + "(" else y + ")" end
}
}.take(14)
# => ["(((())))", "((()()))", "((())())", "((()))()", "(()(()))", "(()()())",
# "(()())()", "(())(())", "(())()()", "()((()))", "()(()())", "()(())()",
# "()()(())", "()()()()"]
Direct generation:
def catalan(n)
Enumerator.new do |y|
s = [[0,0,0]]
until s.empty?
left,right,result = s.pop
if left + right == 2 * n
y << yield(result)
end
if right < left
s << [left, right + 1, result | 1 << (left + right)]
end
if left < n
s << [left + 1, right, result]
end
end
end
end
Given a number n of x digits. How to remove y digits in a way the remaining digits results in the greater possible number?
Examples:
1)x=7 y=3
n=7816295
-8-6-95
=8695
2)x=4 y=2
n=4213
4--3
=43
3)x=3 y=1
n=888
=88
Just to state: x > y > 0.
For each digit to remove: iterate through the digits left to right; if you find a digit that's less than the one to its right, remove it and stop, otherwise remove the last digit.
If the number of digits x is greater than the actual length of the number, it means there are leading zeros. Since those will be the first to go, you can simply reduce the count y by a corresponding amount.
Here's a working version in Python:
def remove_digits(n, x, y):
s = str(n)
if len(s) > x:
raise ValueError
elif len(s) < x:
y -= x - len(s)
if y <= 0:
return n
for r in range(y):
for i in range(len(s)):
if s[i] < s[i+1:i+2]:
break
s = s[:i] + s[i+1:]
return int(s)
>>> remove_digits(7816295, 7, 3)
8695
>>> remove_digits(4213, 4, 2)
43
>>> remove_digits(888, 3, 1)
88
I hesitated to submit this, because it seems too simple. But I wasn't able to think of a case where it wouldn't work.
if x = y we have to remove all the digits.
Otherwise, you need to find maximum digit in first y + 1 digits. Then remove all the y0 elements before this maximum digit. Then you need to add that maximum to the answer and then repeat that task again, but you need now to remove y - y0 elements now.
Straight forward implementation will work in O(x^2) time in the worst case.
But finding maximum in the given range can be done effectively using Segment Tree data structure. Time complexity will be O(x * log(x)) in the worst case.
P. S. I just realized, that it possible to solve in O(x) also, using the fact, that exists only 10 digits (but the algorithm maybe a little bit complicated). We need to find the minimum in the given range [L, R], but the ranges in this task will "change" from left to the right (L and R always increase). And we just need to store 10 pointers to the digits (1 per digit) to the first position in the number such that position >= L. Then to find the minimum, we need to check only 10 pointers. To update the pointers, we will try to move them right.
So the time complexity will be O(10 * x) = O(x)
Here's an O(x) solution. It builds an index that maps (i, d) to j, the smallest number > i such that the j'th digit of n is d. With this index, one can easily find the largest possible next digit in the solution in O(1) time.
def index(digits):
next = [len(digits)+1] * 10
for i in xrange(len(digits), 0, -1):
next[ord(digits[i-1])-ord('0')] = i-1
yield next[::-1]
def minseq(n, y):
n = str(n)
idx = list(index(n))[::-1]
i, r = 0, []
for ry in xrange(len(n)-y):
i = next(j for j in idx[i] if j <= y+ry) + 1
r.append(n[i - 1])
return ''.join(r)
print minseq(7816295, 3)
print minseq(4213, 2)
Pseudocode:
Number.toDigits().filter (sortedSet (Number.toDigits()). take (y))
Imho you don't need to know x.
For efficiency, Number.toDigits () could be precalculated
digits = Number.toDigits()
digits.filter (sortedSet (digits).take (y))
Depending on language and context, you either output the digits and are done or have to convert the result into a number again.
Working Scala-Code for example:
def toDigits (l: Long) : List [Long] = if (l < 10) l :: Nil else (toDigits (l /10)) :+ (l % 10)
val num = 734529L
val dig = toDigits (num)
dig.filter (_ > ((dig.sorted).take(2).last))
A sorted set is a set which is sorted, which means, every element is only contained once and then the resulting collection is sorted by some criteria, for example numerical ascending. => 234579.
We take two of them (23) and from that subset the last (3) and filter the number by the criteria, that the digits have to be greater than that value (3).
Your question does not explicitly say, that each digit is only contained once in the original number, but since you didn't give a criterion, which one to remove in doubt, I took it as an implicit assumption.
Other languages may of course have other expressions (x.sorted, x.toSortedSet, new SortedSet (num), ...) or lack certain classes, functions, which you would have to build on your own.
You might need to write your own filter method, which takes a pedicate P, and a collection C, and returns a new collection of all elements which satisfy P, P being a Method which takes one T and returns a Boolean. Very useful stuff.
I have a two-fold homework problem, Implement Karp-Rabin and run it on a test file and the second part:
For the hash values modulo q, explain why it is a bad idea to use q as a power of 2. Can you construct a terrible example e.g. for q=64
and n=15?
This is my implementation of the algorithm:
def karp_rabin(text, pattern):
# setup
alphabet = 'ACGT'
d = len(alphabet)
n = len(pattern)
d_n = d**n
q = 2**32-1
m = {char:i for i,char in enumerate(alphabet)}
positions = []
def kr_hash(s):
return sum(d**(n-i-1) * m[s[i]] for i in range(n))
def update_hash():
return d*text_hash + m[text[i+n-1]] - d_n * m[text[i-1]]
pattern_hash = kr_hash(pattern)
for i in range(0, len(text) - n + 1):
text_hash = update_hash() if i else kr_hash(text[i:n])
if pattern_hash % q == text_hash % q and pattern == text[i:i+n]:
positions.append(i)
return ' '.join(map(str, positions))
...The second part of the question is referring to this part of the code/algo:
pattern_hash = kr_hash(pattern)
for i in range(0, len(text) - n + 1):
text_hash = update_hash() if i else kr_hash(text[i:n])
# the modulo q used to check if the hashes are congruent
if pattern_hash % q == text_hash % q and pattern == text[i:i+n]:
positions.append(i)
I don't understand why it would be a bad idea to use q as a power of 2. I've tried running the algorithm on the test file provided(which is the genome of ecoli) and there's no discernible difference.
I tried looking at the formula for how the hash is derived (I'm not good at math) trying to find some common factors that would be really bad for powers of two but found nothing. I feel like if q is a power of 2 it should cause a lot of clashes for the hashes so you'd need to compare strings a lot more but I didn't find anything along those lines either.
I'd really appreciate help on this since I'm stumped. If someone wants to point out what I can do better in the first part (code efficiency, readability, correctness etc.) I'd also be thrilled to hear your input on that.
There is a problem if q divides some power of d, because then only a few characters contribute to the hash. For example in your code d=4, if you take q=64 only the last three characters determine the hash (d**3 = 64).
I don't really see a problem if q is a power of 2 but gcd(d,q) = 1.
Your implementation looks a bit strange because instead of
if pattern_hash % q == text_hash % q and pattern == text[i:i+n]:
you could also use
if pattern_hash == text_hash and pattern == text[i:i+n]:
which would be better because you get fewer collisions.
The Thue–Morse sequence has among its properties that its polynomial hash quickly becomes zero when a power of 2 is the hash module, for whatever polynomial base (d). So if you will try to search a short Thue-Morse sequence in a longer one, you will have a great lot of hash collisions.
For example, your code, slightly adapted:
def karp_rabin(text, pattern):
# setup
alphabet = '01'
d = 15
n = len(pattern)
d_n = d**n
q = 32
m = {char:i for i,char in enumerate(alphabet)}
positions = []
def kr_hash(s):
return sum(d**(n-i-1) * m[s[i]] for i in range(n))
def update_hash():
return d*text_hash + m[text[i+n-1]] - d_n * m[text[i-1]]
pattern_hash = kr_hash(pattern)
for i in range(0, len(text) - n + 1):
text_hash = update_hash() if i else kr_hash(text[i:n])
if pattern_hash % q == text_hash % q : #and pattern == text[i:i+n]:
positions.append(i)
return ' '.join(map(str, positions))
print(karp_rabin('0110100110010110100101100110100110010110011010010110100110010110', '0110100110010110'))
outputs a lot of positions, although only three of then are proper matches.
Note that I have dropped the and pattern == text[i:i+n] check. Obviously if you restore it, the result will be correct, but also it is obvious that the algorithm will do much more work checking this additional condition than for other q. In fact, because there are so many collisions, the whole idea of algorithm becomes not working: you could almost as effectively wrote a simple algorithm that checks every position for a match.
Also note that your implementation is quite strange. The whole idea of polynomial hashing is to take the modulo operation each time you compute the hash. Otherwise your pattern_hash and text_hash are very big numbers. In other languages this might mean arithmetic overflow, but in Python this will invoke big integer arithmetic, which is slow and once again loses the whole idea of the algorithm.
I'm taking my first steps into recursion and dynamic programming and have a question about forming subproblems to model the recursion.
Problem:
How many different ways are there to
flip a fair coin 5 times and not have
three or more heads in a row?
If some could put up some heavily commented code (Ruby preferred but not essential) to help me get there. I am not a student if that matters, this is a modification of a Project Euler problem to make it very simple for me to grasp. I just need to get the hang of writing recursion formulas.
If you would like to abstract the problem into how many different ways are there to flip a fair coin Y times and not have Z or more heads in a row, that may be beneficial as well. Thanks again, this website rocks.
You can simply create a formula for that:
The number of ways to flip a coin 5 times without having 3 heads in a row is equal to the number of combinations of 5 coin flips minus the combinations with at least three heads in a row. In this case:
HHH-- (4 combinations)
THHH- (2 combinations)
TTHHH (1 combination)
The total number of combinations = 2^5 = 32. And 32 - 7 = 25.
If we flip a coin N times without Q heads in a row, the total amount is 2^N and the amount with at least Q heads is 2^(N-Q+1)-1. So the general answer is:
Flip(N,Q) = 2^N - 2^(N-Q+1) +1
Of course you can use recursion to simulate the total amount:
flipme: N x N -> N
flipme(flipsleft, maxhead) = flip(flipsleft, maxhead, 0)
flip: N x N x N -> N
flip(flipsleft, maxhead, headcount) ==
if flipsleft <= 0 then 0
else if maxhead<=headcount then 0
else
flip(flipsleft - 1, maxhead, headcount+1) + // head
flip(flipsleft - 1, maxhead, maxhead) // tail
Here's my solution in Ruby
def combination(length=5)
return [[]] if length == 0
combination(length-1).collect {|c| [:h] + c if c[0..1]!= [:h,:h]}.compact +
combination(length-1).collect {|c| [:t] + c }
end
puts "There are #{combination.length} ways"
All recursive methods start with an early out for the end case.
return [[]] if length == 0
This returns an array of combinations, where the only combination of zero length is []
The next bit (simplified) is...
combination(length-1).collect {|c| [:h] + c } +
combination(length-1).collect {|c| [:t] + c }
So.. this says.. I want all combinations that are one shorter than the desired length with a :head added to each of them... plus all the combinations that are one shorter with a tail added to them.
The way to think about recursion is.. "assuming I had a method to do the n-1 case.. what would I have to add to make it cover the n case". To me it feels like proof by induction.
This code would generate all combinations of heads and tails up to the given length.
We don't want ones that have :h :h :h. That can only happen where we have :h :h and we are adding a :h. So... I put an if c[0..1] != [:h,:h] on the adding of the :h so it will return nil instead of an array when it was about to make an invalid combination.
I then had to compact the result to ignore all results that are just nil
Isn't this a matter of taking all possible 5 bit sequences and removing the cases where there are three sequential 1 bits (assuming 1 = heads, 0 = tails)?
Here's one way to do it in Python:
#This will hold all possible combinations of flipping the coins.
flips = [[]]
for i in range(5):
#Loop through the existing permutations, and add either 'h' or 't'
#to the end.
for j in range(len(flips)):
f = flips[j]
tails = list(f)
tails.append('t')
flips.append(tails)
f.append('h')
#Now count how many of the permutations match our criteria.
fewEnoughHeadsCount = 0
for flip in flips:
hCount = 0
hasTooManyHeads = False
for c in flip:
if c == 'h': hCount += 1
else: hCount = 0
if hCount >= 3: hasTooManyHeads = True
if not hasTooManyHeads: fewEnoughHeadsCount += 1
print 'There are %s ways.' % fewEnoughHeadsCount
This breaks down to:
How many ways are there to flip a fair coin four times when the first flip was heads + when the first flip was tails:
So in python:
HEADS = "1"
TAILS = "0"
def threeOrMoreHeadsInARow(bits):
return "111" in bits
def flip(n = 5, flips = ""):
if threeOrMoreHeadsInARow(flips):
return 0
if n == 0:
return 1
return flip(n - 1, flips + HEADS) + flip(n - 1, flips + TAILS)
Here's a recursive combination function using Ruby yield statements:
def combinations(values, n)
if n.zero?
yield []
else
combinations(values, n - 1) do |combo_tail|
values.each do |value|
yield [value] + combo_tail
end
end
end
end
And you could use regular expressions to parse out three heads in a row:
def three_heads_in_a_row(s)
([/hhh../, /.hhh./, /..hhh/].collect {|pat| pat.match(s)}).any?
end
Finally, you would get the answer using something like this:
total_count = 0
filter_count = 0
combinations(["h", "t"], 5) do |combo|
count += 1
unless three_heads_in_a_row(combo.join)
filter_count += 1
end
end
puts "TOTAL: #{ total_count }"
puts "FILTERED: #{ filter_count }"
So that's how I would do it :)