Longest Common Subsequence
"""
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()
About this Algorithm
Problem Statement
Given two strings S and T, find the length of the longest common subsequence (LCS).
Approach
Let the dp[i][j] be the length of the longest common subsequence of prefixes S[1..i] and T[1..j]. Our answer (the length of LCS) is dp[|S|][|T|] since the prefix of the length of string is the string itself.
Both dp[0][i] and dp[i][0] are 0 for any i since the LCS of empty prefix and anything else is an empty string.
Now let's try to calculate dp[i][j] for any i, j. Let's say S[1..i] = *A and T[1..j] = *B where * stands for any sequence of letters (could be different for S and T), A stands for any letter and B stands for any letter different from A. Since A != B, our LCS doesn't include A or B as a last character. So we could try to throw away A or B character. If we throw A, our LCS length will be dp[i - 1][j] (since we have prefixes S[1..i - 1] and T[1..j]). If we try to throw B character, we will have prefixes S[1..i] and T[1..j - 1] so the length of LCS will be dp[i][j - 1]. As we are looking for the Longest common subsequence, we will pick the maximum value from dp[i][j - 1] and dp[i - 1][j].
But what if S[1..i] = *A and T[1..j] = *A? We could say that the LCS of our prefixes is LCS of prefixes S[1..i - 1] and T[1..j - 1] plus the letter A. So dp[i][j] = dp[i - 1][j - 1] + 1 if S[i] = T[j].
We could see that we can fill our dp table row by row, column by column. So our algorithm will be like:
- Let's say that we have strings
Sof the length N andTof the length M (numbered from 1). Let's create the tabledpof size(N + 1) x (M + 1)numbered from 0. - Let's fill the 0th row and the 0th column of
dpwith 0. - Then, we follow the algorithm:
for i in range(1..N):
for j in range(1..M):
if(S[i] == T[j])
dp[i][j] = dp[i - 1][j - 1] + 1
else
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
Time Complexity
O(N * M) In any case
Space Complexity
O(N * M) - simple implementation
O(min {N, M}) - two-layers implementation (as dp[i][j] depends on only i-th and i-th layers, we coudld store only two layers).
Example
Let's say we have strings ABCB and BBCB. We will build such a table:
# # A B C B
# 0 0 0 0 0
B 0 ? ? ? ?
B 0 ? ? ? ?
C 0 ? ? ? ?
B 0 ? ? ? ?
Now we will start to fill our table from 1st row. Since S[1] = A and T[1] = B, the dp[1][1] will be the maximal value of dp[0][1] = 0 and dp[1][0] = 0. So dp[1][1] = 0. But now S[2] = B = T[1], so dp[1][2] = dp[0][1] + 1 = 1. dp[1][3] is 1 since A != C and we pick max{dp[1][2], dp[0][3]}. And dp[1][4] = dp[0][3] + 1 = 1.
# # A B C B
# 0 0 0 0 0
B 0 0 1 1 1
B 0 ? ? ? ?
C 0 ? ? ? ?
B 0 ? ? ? ?
Now let's fill the other part of the table:
# # A B C B
# 0 0 0 0 0
B 0 0 1 1 1
B 0 0 1 1 2
C 0 0 1 2 2
B 0 0 1 2 3
So the length of LCS is dp[4][4] = 3.
Code Implementation Links
Video Explanation
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