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Commit 28eafd7

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Knuth–Morris–Pratt algorithm
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‎README.md

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@@ -28,4 +28,6 @@ Data Structures and Algorithms Patterns implemented in Python.
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- [x] [Shell Sort](Sorting-Algo/shellsort.py)
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- [x] [Selection Sort](Sorting-Algo/selectionsort.py)
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- [x] [Bucket Sort](Sorting-Algo/bucketsort.py)
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- [x] [Strings](Strings)
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- [x] [KMP (Knuth Morris Pratt) Pattern Searching](Strings/KMP.py)
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‎Strings/KMP.py

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'''
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The Knuth-Morris-Pratt (KMP)Algorithm
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For a given string ‘S’, string matching algorithm determines whether a pattern ‘p’ occurs in the given string ‘S’.
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Input :
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String = "Welcome to CodeKMPPattern"
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Pattern = "Code"
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Output :Pattern found at index 11.
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Time Complexity - O(n)
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'''
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def KMPSearch(pat, txt):
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M = len(pat)
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N = len(txt)
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# create lps[] that will hold the longest prefix suffix
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# values for pattern
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lps = [0]*M
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j = 0 # index for pat[]
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# Preprocess the pattern (calculate lps[] array)
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compute(pat, M, lps)
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i = 0 # index for txt[]
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while i < N:
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if pat[j] == txt[i]:
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i += 1
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j += 1
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if j == M:
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print ("Found pattern at index", str(i-j))
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j = lps[j-1]
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# mismatch after j matches
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elif i < N and pat[j] != txt[i]:
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# Do not match lps[0..lps[j-1]] characters,
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# they will match anyway
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if j != 0:
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j = lps[j-1]
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else:
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i += 1
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def compute(pat, M, lps):
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len = 0 # length of the previous longest prefix suffix
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lps[0] # lps[0] is always 0
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i = 1
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# the loop calculates lps[i] for i = 1 to M-1
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while i < M:
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if pat[i]== pat[len]:
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len += 1
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lps[i] = len
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i += 1
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else:
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# This is tricky. Consider the example.
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# AAACAAAA and i = 7. The idea is similar
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# to search step.
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if len != 0:
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len = lps[len-1]
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# Also, note that we do not increment i here
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else:
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lps[i] = 0
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i += 1
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txt = "ABABDABACDABABCABAB"
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pat = "ABABCABAB"
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KMPSearch(pat, txt)
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#Found pattern at index 10

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