hello there, I tried to solve a problem which is directly based on the Longest common substring. I solved it quite easily with the bottom-up (DP) approach because it seemed more intuitive to me then top-down since I just needed to extend to the idea of Lcs(longest common subsequence).
below, how is implemented bottom-up approach:
def findLength(A,B):
m,n=len(A),len(B)
dp=[0]*(n+1)
ans=0
for i in range(m):
for j in range(n,0,-1):
if A[i]==B[j-1]:
dp[j]=dp[j-1]+1
ans=max(ans,dp[j])
else:
dp[j]=0
return ans
I tried to implement the same using top-down approach but did not succeed. I would like to share my thought that how I landed on my recursive relation. let A, B are two strings are given and def lcs(i,j): # function. dp[i][j] represents LC_Substring after considering ith elements and jth elements of two string respectively. Now if A[i]==B[j] then dp[i][j] = 1+lcs(i-1,j-1) else: dp[i][j]=0, also we need check at lcs(i-1,j) and lcs(i,j-1) irrespective of if-else,to clarify my points i would take an example ,A= 'aab' and B='aabb' ,recursive call start with i=3 and j=4, since A[3]==B[4] , dp[3][4]=1+lcs(2,3) but the required output is achieved when we consider lcs(3,3) .
below is the code I submitted which passed 31/54 test cases and throws TLE .
m,n=len(A),len(B)
dp={}
ans=0
def lcs(i,j):
if i==0 or j==0:
return 0
elif dp.get((i,j)):
return dp[(i,j)]
if A[i-1]==B[j-1]:
dp[(i,j)]=1+lcs(i-1,j-1)
ans=max(ans,dp[(i,j)])
else:
dp[(i,j)]=0
lcs(i-1,j)
lcs(i,j-1)
return dp[(i,j)]
lcs(m,n)
return ans
my queries :
1. is the time complexity of my top-down approach is O(m*n)?
2. is my recursive relation for top-down right?
3. it is considered good practice for a beginner to solve first top-down then move to bottom-up , but in this case, is it not case that bottom-up seems more intuitive?
4. where my thought went process wrong?
5. what would be right or efficient way to interpret these types of questions and approach them?
Thanks a million in advance. happy coding