Problem Statement
leetcode problem link
My Solution [Accepted]
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
memo = {}
def dp(i, j):
if (i >= len(text1) or j >= len(text2)):
return 0
if (i, j) in memo:
return memo[(i, j)]
add_text1 = add_text2 = add_both = skip = 0
if text1[i] == text2[j]:
add_both = dp(i + 1, j + 1) + 1
add_text1 = dp(i + 1, j)
add_text2 = dp(i, j + 1)
skip = dp(i + 1, j + 1)
memo[(i, j)] = max(add_text1, add_text2, add_both, skip)
return memo[(i, j)]
return dp(0, 0)
Editorial
Approach 1: Memoization
from functools import lru_cache
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
@lru_cache(maxsize=None)
def memo_solve(p1, p2):
# Base case: If either string is now empty, we can't match
# up anymore characters.
if p1 == len(text1) or p2 == len(text2):
return 0
# Option 1: We don't include text1[p1] in the solution.
option_1 = memo_solve(p1 + 1, p2)
# Option 2: We include text1[p1] in the solution, as long as
# a match for it in text2 at or after p2 exists.
first_occurence = text2.find(text1[p1], p2)
option_2 = 0
if first_occurence != -1:
option_2 = 1 + memo_solve(p1 + 1, first_occurence + 1)
# Return the best option.
return max(option_1, option_2)
return memo_solve(0, 0)
Approach 2: Improved Memoization
from functools import lru_cache
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
@lru_cache(maxsize=None)
def memo_solve(p1, p2):
# Base case: If either string is now empty, we can't match
# up anymore characters.
if p1 == len(text1) or p2 == len(text2):
return 0
# Recursive case 1.
if text1[p1] == text2[p2]:
return 1 + memo_solve(p1 + 1, p2 + 1)
# Recursive case 2.
else:
return max(memo_solve(p1, p2 + 1), memo_solve(p1 + 1, p2))
return memo_solve(0, 0)
Approach 3: Dynamic Programming
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
# Make a grid of 0's with len(text2) + 1 columns
# and len(text1) + 1 rows.
dp_grid = [[0] * (len(text2) + 1) for _ in range(len(text1) + 1)]
# Iterate up each column, starting from the last one.
for col in reversed(range(len(text2))):
for row in reversed(range(len(text1))):
# If the corresponding characters for this cell are the same...
if text2[col] == text1[row]:
dp_grid[row][col] = 1 + dp_grid[row + 1][col + 1]
# Otherwise they must be different...
else:
dp_grid[row][col] = max(dp_grid[row + 1][col], dp_grid[row][col + 1])
# The original problem's answer is in dp_grid[0][0]. Return it.
return dp_grid[0][0]
Approach 4: Dynamic Programming with Space Optimization
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
# If text1 doesn't reference the shortest string, swap them.
if len(text2) < len(text1):
text1, text2 = text2, text1
# The previous column starts with all 0's and like before is 1
# more than the length of the first word.
previous = [0] * (len(text1) + 1)
# Iterate up each column, starting from the last one.
for col in reversed(range(len(text2))):
# Create a new array to represent the current column.
current = [0] * (len(text1) + 1)
for row in reversed(range(len(text1))):
if text2[col] == text1[row]:
current[row] = 1 + previous[row + 1]
else:
current[row] = max(previous[row], current[row + 1])
# The current column becomes the previous one.
previous = current
# The original problem's answer is in previous[0]. Return it.
return previous[0]
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
# If text1 doesn't reference the shortest string, swap them.
if len(text2) < len(text1):
text1, text2 = text2, text1
# The previous and current column starts with all 0's and like
# before is 1 more than the length of the first word.
previous = [0] * (len(text1) + 1)
current = [0] * (len(text1) + 1)
# Iterate up each column, starting from the last one.
for col in reversed(range(len(text2))):
for row in reversed(range(len(text1))):
if text2[col] == text1[row]:
current[row] = 1 + previous[row + 1]
else:
current[row] = max(previous[row], current[row + 1])
# The current column becomes the previous one, and vice versa.
previous, current = current, previous
# The original problem's answer is in previous[0]. Return it.
return previous[0]