simple approach

def count_substring(s, sub): count = 0

#   indiaisiajsndiasd   ia
k = 0
for i in range(0, len(s)):
    p = s.find(sub, k, len(s))
    if i == p:
        count = count + 1
        if p + len(sub) < len(s):
            k = p + 1

return count