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How to find the longest common substring from more than two strings in Python?
In this article, we are going to find the longest common substring from more than two strings in Python.
The first and only approach for finding the longest common substring is by using the Longest common substring Algorithm. We will define a function that accepts 2 strings, then we will iterate over both the strings and find for common subsequence and calculate the length.
We will check entire string by iterating over and calculating the length of the sub-sequences and we will return the longest common subsequence. We will get an empty string as output if there are not any common sub-sequences between two given strings.
Example
In the example given below, we are taking 2 strings as input and we are showing the implementation of the Longest Common Subsequence Algorithm and finding the Longest Common Subsequence −
def longest_common_substring(str1, sub): m = [[0] * (1 + len(sub)) for i in range(1 + len(str1))] longest, x_longest = 0, 0 for x in range(1, 1 + len(str1)): for y in range(1, 1 + len(sub)): if str1[x - 1] == sub[y - 1]: m[x][y] = m[x - 1][y - 1] + 1 if m[x][y] > longest: longest = m[x][y] x_longest = x else: m[x][y] = 0 return str1[x_longest - longest: x_longest] str1 = "Welcome to Tutorialspoint" print("The given string is ") print(str1) sub1 = "Tutorial" print("The first sub-string is") print(sub1) print("The longest common subsequence between'",str1,"'and '",sub1,"' is") print(longest_common_substring(str1, sub1)) sub2 = "12345" print("The second sub-string is") print(sub2) print("The longest common subsequence between'",str1,"'and '",sub2,"' is") print(longest_common_substring(str1, sub2))
Output
Following is an output of the above code −
The given string is Welcome to Tutorialspoint The first sub-string is Tutorial The longest common subsequence between' Welcome to Tutorialspoint 'and ' Tutorial ' is Tutorial 1 The second sub-string is 12345 The longest common subsequence between' Welcome to Tutorialspoint 'and ' 12345 ' is
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