They have 5 length-2 common subsequences: (AB), (AC), (ad), (BD), and (CD); 2 length-3 common subsequences: (ABD) and (ACD); and no longer common subsequences. It differs from the longest common substring problem: unlike substrings, subsequences are not need to occupy consecutive positions within the original sequences.

COMING SOON!

```
"""
LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them.
A subsequence is a sequence that appears in the same relative order, but not necessarily continuous.
Example:"abc", "abg" are subsequences of "abcdefgh".
"""
def longest_common_subsequence(x: str, y: str):
"""
Finds the longest common subsequence between two strings. Also returns the
The subsequence found
Parameters
----------
x: str, one of the strings
y: str, the other string
Returns
-------
L[m][n]: int, the length of the longest subsequence. Also equal to len(seq)
Seq: str, the subsequence found
>>> longest_common_subsequence("programming", "gaming")
(6, 'gaming')
>>> longest_common_subsequence("physics", "smartphone")
(2, 'ph')
>>> longest_common_subsequence("computer", "food")
(1, 'o')
"""
# find the length of strings
assert x is not None
assert y is not None
m = len(x)
n = len(y)
# declaring the array for storing the dp values
L = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if x[i - 1] == y[j - 1]:
match = 1
else:
match = 0
L[i][j] = max(L[i - 1][j], L[i][j - 1], L[i - 1][j - 1] + match)
seq = ""
i, j = m, n
while i > 0 and j > 0:
if x[i - 1] == y[j - 1]:
match = 1
else:
match = 0
if L[i][j] == L[i - 1][j - 1] + match:
if match == 1:
seq = x[i - 1] + seq
i -= 1
j -= 1
elif L[i][j] == L[i - 1][j]:
i -= 1
else:
j -= 1
return L[m][n], seq
if __name__ == "__main__":
a = "AGGTAB"
b = "GXTXAYB"
expected_ln = 4
expected_subseq = "GTAB"
ln, subseq = longest_common_subsequence(a, b)
print("len =", ln, ", sub-sequence =", subseq)
import doctest
doctest.testmod()
```