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:
Example
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'))Output
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.