# A Naive recursive Python implementation of LCS problem
def lcs(X, Y, m, n):
if m == 0 or n == 0:
return 0;
elif X[m-1] == Y[n-1]:
return 1 + lcs(X, Y, m-1, n-1);
else:
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
# Driver program to test the above function
X = "AGGTAB"
Y = "GXTXAYB"
print ("Length of LCS is ", lcs(X, Y, len(X), len(Y)))
Output:
Length of LCS is 4
Following is a tabulated implementation for the LCS problem.
# Dynamic Programming implementation of LCS problem
def lcs(X, Y):
# find the length of the strings
m = len(X)
n = len(Y)
# declaring the array for storing the dp values
L = [[None]*(n + 1) for i in range(m + 1)]
"""Following steps build L[m + 1][n + 1] in bottom up fashion
Note: L[i][j] contains length of LCS of X[0..i-1]
and Y[0..j-1]"""
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0 :
L[i][j] = 0
elif X[i-1] == Y[j-1]:
L[i][j] = L[i-1][j-1]+1
else:
L[i][j] = max(L[i-1][j], L[i][j-1])
# L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1]
return L[m][n]
# end of function lcs
# Driver program to test the above function
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of LCS is ", lcs(X, Y))