Palindromic Partitioning

def isPalindrome(s,i,j,palin):
    if i>=j:
        return True

    if palin[i][j]!=-1:
        return palin[i][j]

    palin[i][j]=(s[i]==s[j] and isPalindrome(s,i+1,j-1,palin))

    return palin[i][j]

def fun(s,i,dp,palin):
    # Returning -1 avoids counting an extra cut after the last partition.
    # Example: "aa|b" already used 1 cut.
    # After processing last partition, recursion reaches end of string.
    # If we return 0 here, final answer becomes 2 instead of 1.
    if i>=len(s): #it must need to take a cut(and before cut it should be palindrome) for calling the fun on further so suppose aa is there so for calling after aa, it must consider 1 cut.
        return -1

    if dp[i]!=-1:
        return dp[i]

    temp=float('inf')

    for j in range(i,len(s)):

        if isPalindrome(s,i,j,palin):
            temp=min(temp,1+fun(s,j+1,dp,palin))

    dp[i]=temp

    return dp[i]

class Solution:
    def palPartition(self,s):
        n=len(s)

        #dp[i] means minimum palindrome cuts needed for s[i:]
        dp=[-1]*n

        palin=[[-1]*n for _ in range(n)]

        return fun(s,0,dp,palin)