Difficulty solving this with recursive code - recursion

Question

I need to find the length of the longest common subsequence.
s and t are Strings, and n and m are their lengths. I would like to write a recursive code.
This is what I did so far but I cant get any progress:
def lcs_len_v1(s, t):
n = len(s)
m = len(t)
return lcs_len_rec(s,n,t,m)
def lcs_len_rec(s,size_s,t,size_t):
cnt= 0
if size_s==0 or size_t==0:
return 0
elif s[0]==t[0]:
cnt= +1
return cnt, lcs_len_rec(s[1:], len(s[1:]), t[1:], len(t[1:]))

Answer 1

This works:
def lcs(xstr, ystr):
if not xstr or not ystr:
return ""
x, xs, y, ys = xstr[0], xstr[1:], ystr[0], ystr[1:]
if x == y:
return x + lcs(xs, ys)
else:
return max(lcs(xstr, ys), lcs(xs, ystr), key=len)
print(lcs("AAAABCC","AAAACCB"))
# AAAACC
You should know that a recursive approach will only work with relatively trivial string; the complexity increases very rapidly with longer strings.

Answer 2

this is my code, how can I use on it the memoization technique?
def lcs_len_v1(s, t):
n = len(s)
m = len(t)
return lcs_len_rec(s,n,t,m)
def lcs_len_rec(s,size_s,t,size_t):
if size_s==0 or size_t==0:
return 0
elif s[0]==t[0]:
cnt=0
cnt+= 1
return cnt+ lcs_len_rec(s[1:], size_s-1, t[1:], size_t-1)
else:
return max(lcs_len_rec(s[1:], size_s-1, t, size_t), lcs_len_rec(s, size_s, t[1:], size_t-1))

Answer 3

Using the memoization technique, you can run the algorithm also with a very long strings. Infact it is just O(n^2):
def recursiveLCS(table, s1, s2):
if(table[len(s1)][len(s2)] != False):
return table[len(s1)][len(s2)]
elif len(s1) == 0 or len(s2) == 0:
val = ""
elif s1[0] == s2[0]:
val = s1[0] + recursiveLCS(table, s1[1:], s2[1:])
else:
res1 = recursiveLCS(table, s1[1:], s2)
res2 = recursiveLCS(table, s1, s2[1:])
val = res2
if len(res1) > len(res2):
val = res1
table[len(s1)][len(s2)] = val
return val
def computeLCS(s1, s2):
table = [[False for col in range(len(s2) + 1)] for row in range(len(s1) + 1)]
return recursiveLCS(table, s1, s2)
print computeLCS("testistest", "this_is_a_long_testtest_for_testing_the_algorithm")
Output:
teststest