import re

pattern = r"(?=(1[01]{3}0))"
s = "011010010111001101101111011011010110110101110011010000110010010111000100100110110101101111011011110111011001101100011011010111011001101000011001001101100011100010010110110011111011001110001001011011"

cnt = dict()
for i in re.findall(pattern, s):
    # get the value by key, or return 0 if the key does not exist
    cnt[i] = cnt.get(i, 0) + 1

print(cnt)