Lognest Common Substring (code from Wikipedia)

def lcs(s1, s2):
   M = [[0] * (len(s2) + 1) for i in xrange(len(s1) + 1)] #create a matrix M with len(s1)+1 rows and len(s2)+1 columns.
							  #The extra row and column is for conveniece, to avoid "index out of range"
							  #later when we use M[x][y] = M[x-1][y-1] + 1 as the last step.
   longest, x_longest = 0, 0		#"longest" keeps track of the length of the lcs found so far.
					#"x_longest" keeps track of the index (in s1) of beggining of lcs found so far. 
   for x in xrange(1, len(s1) + 1):
       for y in xrange(1, len(s2) + 1):
           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]

import numpy as np 				#Needless to say, Using Numpy is not necessary

def lcs(s1, s2):
    m = np.zeros(2*(len(s2) + 1)).reshape(2,(len(s2) + 1)) 
    longest, x_longest = 0, 0  
    for x in xrange(1, len(s1) + 1):
        for y in xrange(1, len(s2) + 1):
            if s1[x - 1] == s2[y - 1]:
                m[1][y] = m[0][y - 1] + 1
                if m[1][y] > longest:
                    longest = m[1][y]
                    x_longest = x
            else:
                m[1][y] = 0
        m[0] = m[1] 				#Marking the "current row" as "last row"
        m[1] = np.zeros(len(s2) + 1)		#Creating a fresh "current row"
    return s1[x_longest - longest: x_longest]