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Original file line number Diff line number Diff line change
@@ -0,0 +1,62 @@
"""
Find All Distinct Longest Common Subsequences

Complexity:
Time: Exponential in the number of LCSs
Space: O(m * n) + O(number of LCSs)
"""


def print_all_longest_common_subsequence(s: str, t: str) -> list[str]:
"""Compute all distinct longest common subsequences of two strings.

Args:
s: First string.
t: Second string.

Returns:
A list containing all distinct longest common subsequences.

Examples:
>>> print_all_longest_common_subsequence('rajroy', 'rxoyxraxj')
['raj', 'roy']
"""

n, m = len(s), len(t)
dp = [[0] * (m + 1) for _ in range(n + 1)]

for i in range(1, n + 1):
for j in range(1, m + 1):
if s[i - 1] == t[j - 1]:
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

res = set()
memo = {}

def solve(i: int, j: int, current: str) -> None:
if i < 1 or j < 1:
res.add(current[::-1])
return

key = (i, j, current)
if key in memo:
res.update(memo[key])
return

if s[i - 1] == t[j - 1]:
solve(i - 1, j - 1, current + s[i - 1])
else:
if dp[i - 1][j] > dp[i][j - 1]:
solve(i - 1, j, current)
elif dp[i][j - 1] > dp[i - 1][j]:
solve(i, j - 1, current)
else:
solve(i - 1, j, current)
solve(i, j - 1, current)

memo[key] = set(res)

solve(n, m, "")
return list(res)
Original file line number Diff line number Diff line change
@@ -0,0 +1,52 @@
"""
Print any longest common subsequence

Complexity:
Time: O(n * m)
Space: O(n * m)
"""


def print_any_longest_common_subsequence(s1: str, s2: str) -> str:
"""Return any one longest common subsequence of two strings.

Args:
s1: First string.
s2: Second string.

Returns:
A string representing any one longest common subsequence.

Example:
>>> print_any_longest_common_subsequence('rajroy', 'rxoyxraxj')
"raj" or "roy"
"""

n = len(s1)
m = len(s2)
dp = [[0] * (m + 1) for _ in range(n + 1)]

for ind1 in range(1, n + 1):
for ind2 in range(1, m + 1):
if s1[ind1 - 1] == s2[ind2 - 1]:
dp[ind1][ind2] = 1 + dp[ind1 - 1][ind2 - 1]
else:
dp[ind1][ind2] = max(dp[ind1 - 1][ind2], dp[ind1][ind2 - 1])

length = dp[n][m]
i, j = n, m
lcs_str = [""] * length
index = length - 1

while i > 0 and j > 0:
if s1[i - 1] == s2[j - 1]:
lcs_str[index] = s1[i - 1]
index -= 1
i -= 1
j -= 1
elif dp[i - 1][j] > dp[i][j - 1]:
i -= 1
else:
j -= 1

return "".join(lcs_str)