Python Program for KMP Algorithm for Pattern Searching
Given a text txt[0..n-1] and a pattern pat[0..m-1], write a function search(char pat[], char txt[]) that prints all occurrences of pat[] in txt[]. You may assume that n > m.
Examples:
Input: txt[] = "THIS IS A TEST TEXT"
pat[] = "TEST"
Output: Pattern found at index 10
Input: txt[] = "AABAACAADAABAABA"
pat[] = "AABA"
Output: Pattern found at index 0
Pattern found at index 9
Pattern found at index 12
Pattern searching is an important problem in computer science. When we do search for a string in notepad/word file or browser or database, pattern searching algorithms are used to show the search results.
Python
# Python program for KMP Algorithmdef KMPSearch(pat, txt): M = len(pat) N = len(txt) # create lps[] that will hold the longest prefix suffix # values for pattern lps = [0]*M j = 0 # index for pat[] # Preprocess the pattern (calculate lps[] array) computeLPSArray(pat, M, lps) i = 0 # index for txt[] while i < N: if pat[j] == txt[i]: i += 1 j += 1 if j == M: print "Found pattern at index " + str(i-j) j = lps[j-1] # mismatch after j matches elif i < N and pat[j] != txt[i]: # Do not match lps[0..lps[j-1]] characters, # they will match anyway if j != 0: j = lps[j-1] else: i += 1 def computeLPSArray(pat, M, lps): len = 0 # length of the previous longest prefix suffix lps[0] # lps[0] is always 0 i = 1 # the loop calculates lps[i] for i = 1 to M-1 while i < M: if pat[i]== pat[len]: len += 1 lps[i] = len i += 1 else: # This is tricky. Consider the example. # AAACAAAA and i = 7. The idea is similar # to search step. if len != 0: len = lps[len-1] # Also, note that we do not increment i here else: lps[i] = 0 i += 1 txt = "ABABDABACDABABCABAB"pat = "ABABCABAB"KMPSearch(pat, txt) # This code is contributed by Bhavya Jain |
Output:
Found pattern at index 10
Please refer complete article on KMP Algorithm for Pattern Searching for more details!
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