WildCard pattern matching having three symbols ( * , + , ? )

Given a text and a wildcard pattern, implement wildcard pattern matching algorithm that finds if wildcard pattern is matched with text. The matching should cover the entire text (not partial text). The wildcard pattern can include the characters ‘?’, ‘*’ and ‘+’.

‘?’ – matches any single character
‘*’ – Matches any sequence of characters
      (including the empty sequence)
'+' – Matches previous single character
      of pattern 

Examples:

Input :Text = "baaabaaa",
Pattern = “****+ba*****a+", output : true
Pattern = "baaa?ab", output : false 
Pattern = "ba*a?", output : true
Pattern = "+a*ab", output : false 

Input : Text = "aab"
Pattern = "*+"  output : false 
Pattern = "*+b" output : true    

• Case 1: The character is ‘*’ Here two cases arise: We can ignore ‘*’ character and move to next character in the pattern. ‘*’ character matches with one or more characters in text. Here we will move to next character in the string. 
• Case 2: The character is ‘?’ We can ignore current character in text and move to next character in the pattern and text. 
• Case 3: The character is ‘+’ Here two cases arise : We match current character of text with the previous character of pattern. If there is no previous character that mean ‘+’ is the first character of pattern so we print result as “text do not match”. If the previous character is either ‘+’, ‘?’ or ‘*’ we replace it with the last character used by them. 
• Case 4: The character is not a wildcard character If current character in text matches with current character in pattern, we move to next character in the pattern and text. If they do not match, wildcard pattern and text do not match. 

The process for “*”, “?” is similar to wildcard pattern Matching for two characters:

Here we use a dp table that will contain 
two fields 
Struct DP
{
   // value is true if match possible
   // for current indexes, else false.
   bool value; 

   // Stores the character used 
   // by the symbol that we used 
   // later for symbol '+'
   char ch; 
}  

Below c++ implementation of above idea. 

Yes

Time Complexity : O(n*m)
Auxiliary Space: O(n*m)