From 6989c3a91a99d18fbe527fd453e2f1f9a5a1c1af Mon Sep 17 00:00:00 2001 From: William Carroll Date: Mon, 16 Nov 2020 17:14:08 +0000 Subject: [PATCH] Implement the Rabin Karp string matching algorithm This algorithm is pretty interesting because it runs in linear time with respect to the length of the `corpus` string. It does this by using a sliding window hash. This hash -- because it's a sliding window -- runs in constant time for each iteration; we're only adding and subtracting one character each time and not re-hashing the whole "window". When our hashes match, only then do we compare the "window" to the `pattern`. String comparisons are linear because they compare each character to each character one at a time. But because we only compare strings when are hashes match (a check which runs in constant time), this spares us the performance hit. --- scratch/facebook/rabin-karp.py | 27 +++++++++++++++++++++++++++ 1 file changed, 27 insertions(+) create mode 100644 scratch/facebook/rabin-karp.py diff --git a/scratch/facebook/rabin-karp.py b/scratch/facebook/rabin-karp.py new file mode 100644 index 000000000..53a47b278 --- /dev/null +++ b/scratch/facebook/rabin-karp.py @@ -0,0 +1,27 @@ +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