String matching with * (that matches with any) in any of the two strings

You are given two strings A and B. Strings also contains special character * . you can replace * with any alphabetic character. Finally, you have to tell whether it is possible to make both string same or not.

Examples:

Input : A = "gee*sforgeeks"
 B = "geeksforgeeks"
Output :Yes
Input :A = "abs*"
 B = "abds"
Output :No

Explanation: How we can solve above problem, Basically we three cases, 

• Case 1: Both strings contain * at a particular position, at that time we can replace both * with any character to make the string equal at that position. 
• Case 2: If one string has character and the other has * at that position. So, we can replace * with the same character in another string. 
• Case 3: If both strings has a character at that position, then they must be same, otherwise we can't make them equal. 

Implementation:

1

Time Complexity: O(N)

Auxiliary Space: O(1)

Two pointer Approach:

Steps:

• If the lengths of strings A != B, return No because it's not possible to make them the same.
• Initialize two pointers, i=0 and j= 0, pointing to the first characters of strings A and B, respectively.
• Iterate while both pointers are within the string bounds:
  1. If the characters at pointers i and j are equal or either of them is , move both pointers to the next character.
  2. If the character at pointer i is , move pointer j until the characters at pointers i and j are equal or until pointer j reaches the end of string B.
  3. If pointer j reaches the end of string B and pointer i is not pointing to , return No because it's not possible to make them the same.
  4. If the characters at pointers i and j are not equal and pointer i is not , return No because it's not possible to make them the same.
• If both pointers have reached the end of their respective strings, return Yes because it's possible to make them the same.
• If either pointer i or j has not reached the end of its string, return No because it's not possible to make them the same.

Below is the implementation of above approach:

No

Time Complexity: O(N), where N is the length of the input strings A and B.

Auxiliary Space: O(1)