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