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def substring_exists(corpus, pattern):
"""
Return True if `pattern` appears in `corpus`.
This function runs in O(m) time where n is equal to the length of
`corpus`. To improve the efficiency of this algorithm, use a hashing
function the reduces the number of collisions, which will consequently
reduce the number of string-to-string, linear comparisons.
"""
m, n = len(corpus), len(pattern)
a = sum(ord(c) for c in corpus[0:n])
b = sum(ord(c) for c in pattern)
# (clumsily) prevent an off-by-one error...
if a == b and corpus[0:n] == pattern:
return True
for i in range(1, m - n):
# Update the hash of corpus by subtracting the hash of the character
# that is sliding out of view and adding the hash of the character that
# is sliding into view.
a = a - ord(corpus[i - 1]) + ord(corpus[i + n - 1])
# Integer comparison in O(0) time followed by string comparison in O(m)
# time.
if a == b and corpus[i:i + n] == pattern:
return True
return False
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