Write a program that counts the number of occurrences of "11" in an input sequence of zeros and ones. The input of the program is just the sequence and it should return a single number, which is the number of occurrences of "11".

coin-flip-next.py

def count11(seq):
   # define this function and return the number of occurrences as a number
   
   # define int for occurences of num 
   occurences = 0  

   # loop through each element 

   # get length of seq 
   seq_length = len(seq)

   for idx, val in enumerate(seq):
        # next iteration in the index 
        next_index = idx + 1

        # check if next index is last 
        if next_index >= seq_length:
            break

        # get the current and the next iteration in the index 
        current_int = val 
        next_int = seq[next_index]

        # join the the two ints to be a string 
        joined_string = str(current_int) + str(next_int)

        # compare to a "11" string
        # if the joined int string is equal to "11"
        #    then iterate occurences one forward ++
        #    else do nothing
        if joined_string == "11":
            occurences += 1

   return occurences

print(count11([0, 0, 1, 1, 1, 0])) # this should print 2