I can't think of a better solution than this. Getting run time error

#1
#2

'*' can match with empty string

for ( int j = 1; j <= m; j++)

if (pattern[j - 1] == '*' )

`lookup[0][j] = lookup[0][j-1]

#3

You are not checking this case and if you want to look a simpler solution Google wildcard matching. YouTube and various other sites have a really good tutorial.

#4

I have changed my code. It’s passing all test cases except now except tc 2 where is says runtime error. I submitted my code on leetcode, it is getting accepted there.
Please edit my code if necessary
https://ide.codingblocks.com/s/243885

#5

In line no.47 if value of j= 1 and you are trying to access J-2 then you will definitely get a runtime error

#6

Yeah you are right. That was the problem. It got submitted but I still have a doubt. Is “*a” a valid pattern for input?

#7

It is a right input.

#8

Actually in my accepted code, I am not checking for the case where the first character is ’ * ’ that’s why I am confused. All I can make out is that a ’ * ’ as the first character of the string is same as an empty string.


Accepted code

#9

Oh maybe the test cases are not strong. Doesn’t matter good job. Atleast you observe these things.

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