Implementation of Longest Common Subsequence So I implemented the LCS algorithm once using a 2-D matrix to store values while the other one used a python dictionary. I found the dictionary implementation was easier to implement and was a more natural and intuitive way of solving the problem.

I just wanted to make sure if it's a correct way of implementing the LCS algorithm and how can I improve on it.

Using 2-D matrix

def LowestCommonSubstring(s1, s2):
    LCS = [[0 for x in range(len(s2) + 1)] for x in range(len(s1) + 1)]
    for i in range(1, len(s1)+1):
        for j in range(1, len(s2)+1):
            if s1[i - 1] == s2[j - 1]:
                LCS[i][j] = 1 + LCS[i-1][j-1]
            else:
                LCS[i][j] = max(LCS[i-1][j], LCS[i][j-1])
    return LCS[i][j]


using dictionary

cache = {}

def lcs(s1, s2):
  global cache
  if len(s1) == 0 or len(s2) == 0:
      return 0
  if (s1, s2) in cache:
      return cache[(s1, s2)]
  else:
      if s1[-1] == s2[-1]:
          cache[(s1, s2)] = 1 + lcs(s1[:-1], s2[:-1])
      else:
          cache[(s1, s2)] = max(lcs(s1[:-1], s2), lcs(s1, s2[:-1]))
  return cache[(s1, s2)]


This is the problem I'm trying to implement and for now my solution only calculates the length of the longest common substring.
You could use an explicit memoizer rather than incorporating caching into your function.
This will make your function easier to understand.
You can use functools lru_cache if you are using Python3.
You should evaluate based on truth, rather than length. len(s1) == 0 is 'bad'.
You can use a turnery operator to reduce the repetition of your code.
Your else can be removed, to prevent the arrow anti-pattern.
And reduce the amount of indentation.




def memoize(fn)
    def inner(*args):
        try:
            return cache[args]
        except IndexError:
            cache[args] = fn(*args)
            return cache[args]
    return inner

# Natural abstracted algorithm.
@memoize
def lcs(s1, s2):
    if not s1 or not s2:
        return 0

    return (
        1 + lcs(s1[:-1], s2[:-1])
        if s1[-1] == s2[-1] else
        max(lcs(s1[:-1], s2), lcs(s1, s2[:-1]))
    )