Longest Common Subsequence of two strings

'''
LCS: select the maximum common subsequence of two strings.
if current_character is matching, then add 1 and pass on n-1,m-1
else may be the n-1th character is matching with mth character or nth might be matching with m-1th char so find the max of both cases.
dp table: rows tells all the chars of str1 and col denotes all the chars of the str2 where dp[i][j]=maximum length of subsequence of str1[0->ith character] and str2[0->jth character].

Base Case:
1) if any one string is over, means no character can be common so subsequence=0
2) Just keep in mind, when you make dp table len(str2) will go for cols and len(str1) will go for the rows.
'''

def fun(str1,str2, n,m,dp):
    if n<0 or m<0:
        return 0
    if dp[n][m]!=-1:
        return dp[n][m]
    take=0
    if str1[n]==str2[m]:
        dp[n][m]=1+fun(str1,str2,n-1,m-1,dp)
    else:
        dp[n][m]=max(fun(str1,str2,n-1,m,dp), fun(str1,str2,n,m-1,dp))
    return dp[n][m]

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp=[[-1]*len(text2) for _ in range(len(text1))]
        return fun(text1,text2,len(text1)-1, len(text2)-1, dp)