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Common dynamic programming implementations for the Longest Common Substring algorithm runs in O(nm) time. The following is an implementation of the longest common substring algorithm:

def longest_common_substring(s1, s2): m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))] longest, x_longest = 0, 0 for x in xrange(1, 1 + len(s1)): for y in xrange(1, 1 + len(s2)): if s1[x - 1] == s2[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 s1[x_longest - longest: x_longest] print(longest_common_substring('wellbeing', 'welcome'))

wel

This is how it works

Initially, we initialized the counter array(m) all 0.

Starting from the 1st row, we will compare the first character of a string s1 with all characters in a string s2.

While we traverse the characters in s2, if it matches with the character in s1, we increment the counter. It will be saved m[i][j] which is at diagonally one lower position.

At the end we return the longest sub-string using indices we calculated in the loops.

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