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| 1 | +"""Author Anurag Kumar(mailto:anuragkumarak95@gmail.com) |
| 2 | + The Knuth-Morris-Pratt Algorithm for finding a pattern within a piece of te$ |
| 3 | + with complexity O(n + m) |
| 4 | + 1) Preprocess pattern to identify any suffixes that are identical to prefix$ |
| 5 | + This tells us where to continue from if we get a mismatch between a cha$ |
| 6 | + and the text. |
| 7 | + 2) Step through the text one character at a time and compare it to a charac$ |
| 8 | + updating our location within the pattern if necessary |
| 9 | +""" |
| 10 | +def kmp(pattern, text, len_p=None, len_t=None): |
| 11 | + # 1) Construct the failure array |
| 12 | + failure = [0] |
| 13 | + i = 0 |
| 14 | + for index, char in enumerate(pattern[1:]): |
| 15 | + if pattern[i] == char: |
| 16 | + i += 1 |
| 17 | + else: |
| 18 | + i = 0 |
| 19 | + failure.append(i) |
| 20 | + |
| 21 | + # 2) Step through text searching for pattern |
| 22 | + i, j = 0, 0 # index into text, pattern |
| 23 | + while i < len(text): |
| 24 | + if pattern[j] == text[i]: |
| 25 | + if j == (len(pattern) - 1): |
| 26 | + return True |
| 27 | + i += 1 |
| 28 | + j += 1 |
| 29 | + |
| 30 | + # if this is a prefix in our pattern |
| 31 | + # just go back far enough to continue |
| 32 | + elif failure[j] > 0: |
| 33 | + j = failure[j] - 1 |
| 34 | + else: |
| 35 | + i += 1 |
| 36 | + return False |
| 37 | + |
| 38 | + |
| 39 | +if __name__ == '__main__': |
| 40 | + |
| 41 | + # Test 1) |
| 42 | + pattern = "abc1abc12" |
| 43 | + text1 = "alskfjaldsabc1abc1abc12k23adsfabcabc" |
| 44 | + text2 = "alskfjaldsk23adsfabcabc" |
| 45 | + assert kmp(pattern, text1) and not kmp(pattern, text2) |
| 46 | + |
| 47 | + # Test 2) |
| 48 | + pattern = "ABABX" |
| 49 | + text = "ABABZABABYABABX" |
| 50 | + assert kmp(pattern, text) |
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