# Description: you are given a bitwise pattern and a string
# you need to find the number of times the pattern matches in the string
# any one liner or simple pythonic solution?

import random

def matchIt(yourString, yourPattern):
        """find the number of times yourPattern occurs in yourString"""

        count = 0
        matchTimes = 0

        # How can you simplify the for-if structures?

        for coin in yourString:
            #return to base
            if  count == len(pattern):
                    matchTimes = matchTimes + 1
                    count = 0

            #special case to return to 2, there could be more this type of conditions
            #so this type of if-conditionals are screaming for a havoc
            if count == 2 and pattern[count] == 1:
                    count = count - 1

            #the work horse
            #it could be simpler by breaking the intial string of lenght 'l'
            #to blocks of pattern-length, the number of them is 'l - len(pattern)-1'
            if coin == pattern[count]:
                    count=count+1

        average = len(yourString)/matchTimes

        return [average, matchTimes]



# Generates the list
myString =[]
for x in range(10000):
    myString= myString + [int(random.random()*2)]

pattern = [1,0,0]
result = matchIt(myString, pattern)

print("The sample had "+str(result[1])+" matches and its size was "+str(len(myString))+".\n" +
        "So it took "+str(result[0])+" steps in average.\n" +
        "RESULT: "+str([a for a in "FAILURE" if result[0] != 8]))


# Sample Output
# 
# The sample had 1656 matches and its size was 10000.
# So it took 6 steps in average.
# RESULT: ['F', 'A', 'I', 'L', 'U', 'R', 'E']