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Wildcard Pattern Matching

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)

For example,

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

Each occurrence of ‘?’ character in wildcard pattern can be replaced with any other character and each occurrence of ‘*’ with a sequence of characters such that the wildcard pattern becomes identical to the input string after replacement.

Let’s consider any character in the pattern.



Case 1: The character is ‘*’ 
Here two cases arise  

  1. We can ignore ‘*’ character and move to next character in the Pattern.
  2. ‘*’ 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 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.
We can use Dynamic Programming to solve this problem – 
Let T[i][j] is true if first i characters in given string matches the first j characters of pattern. 

DP Initialization: 

// both text and pattern are null
T[0][0] = true; 

// pattern is null
T[i][0] = false; 

// text is null
T[0][j] = T[0][j - 1] if pattern[j – 1] is '*'

DP relation : 

// If current characters match, result is same as 
// result for lengths minus one. Characters match
// in two cases:
// a) If pattern character is '?' then it matches  
//    with any character of text. 
// b) If current characters in both match
if ( pattern[j – 1] == ‘?’) || 
     (pattern[j – 1] == text[i - 1])
    T[i][j] = T[i-1][j-1]   
 
// If we encounter ‘*’, two choices are possible-
// a) We ignore ‘*’ character and move to next 
//    character in the pattern, i.e., ‘*’ 
//    indicates an empty sequence.
// b) '*' character matches with ith character in
//     input 
else if (pattern[j – 1] == ‘*’)
    T[i][j] = T[i][j-1] || T[i-1][j]  

else // if (pattern[j – 1] != text[i - 1])
    T[i][j]  = false

Below is the implementation of the above Dynamic Programming approach.

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// C++ program to implement wildcard
// pattern matching algorithm
#include <bits/stdc++.h>
using namespace std;
 
// Function that matches input str with
// given wildcard pattern
bool strmatch(char str[], char pattern[], int n, int m)
{
    // empty pattern can only match with
    // empty string
    if (m == 0)
        return (n == 0);
 
    // lookup table for storing results of
    // subproblems
    bool lookup[n + 1][m + 1];
 
    // initailze lookup table to false
    memset(lookup, false, sizeof(lookup));
 
    // empty pattern can match with empty string
    lookup[0][0] = true;
 
    // Only '*' can match with empty string
    for (int j = 1; j <= m; j++)
        if (pattern[j - 1] == '*')
            lookup[0][j] = lookup[0][j - 1];
 
    // fill the table in bottom-up fashion
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            // Two cases if we see a '*'
            // a) We ignore ‘*’ character and move
            //    to next  character in the pattern,
            //     i.e., ‘*’ indicates an empty sequence.
            // b) '*' character matches with ith
            //     character in input
            if (pattern[j - 1] == '*')
                lookup[i][j]
                    = lookup[i][j - 1] || lookup[i - 1][j];
 
            // Current characters are considered as
            // matching in two cases
            // (a) current character of pattern is '?'
            // (b) characters actually match
            else if (pattern[j - 1] == '?'
                     || str[i - 1] == pattern[j - 1])
                lookup[i][j] = lookup[i - 1][j - 1];
 
            // If characters don't match
            else
                lookup[i][j] = false;
        }
    }
 
    return lookup[n][m];
}
 
int main()
{
    char str[] = "baaabab";
    char pattern[] = "*****ba*****ab";
    // char pattern[] = "ba*****ab";
    // char pattern[] = "ba*ab";
    // char pattern[] = "a*ab";
    // char pattern[] = "a*****ab";
    // char pattern[] = "*a*****ab";
    // char pattern[] = "ba*ab****";
    // char pattern[] = "****";
    // char pattern[] = "*";
    // char pattern[] = "aa?ab";
    // char pattern[] = "b*b";
    // char pattern[] = "a*a";
    // char pattern[] = "baaabab";
    // char pattern[] = "?baaabab";
    // char pattern[] = "*baaaba*";
 
    if (strmatch(str, pattern, strlen(str),
                 strlen(pattern)))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
 
    return 0;
}
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// Java program to implement wildcard


// pattern matching algorithm



import java.util.Arrays;




public class GFG {



 



    // Function that matches input str with




    // given wildcard pattern




    static boolean strmatch(String str, String pattern,




                            int n, int m)




    {




        // empty pattern can only match with




        // empty string




        if (m == 0)




            return (n == 0);



 



        // lookup table for storing results of




        // subproblems




        boolean[][] lookup = new boolean[n + 1][m + 1];



 



        // initailze lookup table to false




        for (int i = 0; i < n + 1; i++)




            Arrays.fill(lookup[i], false);



 



        // empty pattern can match with empty string




        lookup[0][0] = true;



 



        // Only '*' can match with empty string




        for (int j = 1; j <= m; j++)




            if (pattern.charAt(j - 1) == '*')




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



 



        // fill the table in bottom-up fashion




        for (int i = 1; i <= n; i++)




        {




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




            {




                // Two cases if we see a '*'




                // a) We ignore '*'' character and move




                //    to next  character in the pattern,




                //     i.e., '*' indicates an empty




                //     sequence.




                // b) '*' character matches with ith




                //     character in input




                if (pattern.charAt(j - 1) == '*')




                    lookup[i][j] = lookup[i][j - 1]




                                   || lookup[i - 1][j];



 



                // Current characters are considered as




                // matching in two cases




                // (a) current character of pattern is '?'




                // (b) characters actually match




                else if (pattern.charAt(j - 1) == '?'




                         || str.charAt(i - 1)




                                == pattern.charAt(j - 1))




                    lookup[i][j] = lookup[i - 1][j - 1];



 



                // If characters don't match




                else




                    lookup[i][j] = false;




            }




        }



 



        return lookup[n][m];




    }



 



   



    // Driver code




    public static void main(String args[])




    {




        String str = "baaabab";




        String pattern = "*****ba*****ab";




        // String pattern = "ba*****ab";




        // String pattern = "ba*ab";




        // String pattern = "a*ab";




        // String pattern = "a*****ab";




        // String pattern = "*a*****ab";




        // String pattern = "ba*ab****";




        // String pattern = "****";




        // String pattern = "*";




        // String pattern = "aa?ab";




        // String pattern = "b*b";




        // String pattern = "a*a";




        // String pattern = "baaabab";




        // String pattern = "?baaabab";




        // String pattern = "*baaaba*";



 



        if (strmatch(str, pattern, str.length(),




                     pattern.length()))




            System.out.println("Yes");




        else




            System.out.println("No");




    }



}


// This code is contributed by Sumit Ghosh


















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# Python program to implement wildcard


# pattern matching algorithm


 


# Function that matches input strr with


# given wildcard pattern


 


 



def strrmatch(strr, pattern, n, m):



 



    # empty pattern can only match with




    # empty strring




    if (m == 0):




        return (n == 0)



 



    # lookup table for storing results of




    # subproblems




    lookup = [[False for i in range(m + 1)] for j in range(n + 1)]



 



    # empty pattern can match with empty strring




    lookup[0][0] = True



 



    # Only '*' can match with empty strring




    for j in range(1, m + 1):




        if (pattern[j - 1] == '*'):




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



 



    # fill the table in bottom-up fashion




    for i in range(1, n + 1):




        for j in range(1, m + 1):



 



            # Two cases if we see a '*'




            # a) We ignore ‘*’ character and move




            # to next character in the pattern,




            # i.e., ‘*’ indicates an empty sequence.




            # b) '*' character matches with ith




            # character in input




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




                lookup[i][j] = lookup[i][j - 1] or lookup[i - 1][j]



 



            # Current characters are considered as




            # matching in two cases




            # (a) current character of pattern is '?'




            # (b) characters actually match




            elif (pattern[j - 1] == '?' or strr[i - 1] == pattern[j - 1]):




                lookup[i][j] = lookup[i - 1][j - 1]



 



            # If characters don't match




            else:




                lookup[i][j] = False



 



    return lookup[n][m]



 


# Driver code


 


 



strr = "baaabab"




pattern = "*****ba*****ab"



# char pattern[] = "ba*****ab"


# char pattern[] = "ba*ab"


# char pattern[] = "a*ab"


# char pattern[] = "a*****ab"


# char pattern[] = "*a*****ab"


# char pattern[] = "ba*ab****"


# char pattern[] = "****"


# char pattern[] = "*"


# char pattern[] = "aa?ab"


# char pattern[] = "b*b"


# char pattern[] = "a*a"


# char pattern[] = "baaabab"


# char pattern[] = "?baaabab"


# char pattern[] = "*baaaba*"


 



if (strrmatch(strr, pattern, len(strr), len(pattern))):




    print("Yes")




else:




    print("No")



 


# This code is contributed by shubhamsingh10


















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// C# program to implement wildcard


// pattern matching algorithm



using System;



 



class GFG {



 



    // Function that matches input str with




    // given wildcard pattern




    static Boolean strmatch(String str, 




                            String pattern,




                            int n, int m)




    {




        // empty pattern can only match with




        // empty string




        if (m == 0)




            return (n == 0);



 



        // lookup table for storing results of




        // subproblems




        Boolean[, ] lookup = new Boolean[n + 1, m + 1];



 



        // initailze lookup table to false




        for (int i = 0; i < n + 1; i++)




            for (int j = 0; j < m + 1; j++)




                lookup[i, j] = false;



 



        // empty pattern can match with




        // empty string




        lookup[0, 0] = true;



 



        // Only '*' can match with empty string




        for (int j = 1; j <= m; j++)




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




                lookup[0, j] = lookup[0, j - 1];



 



        // fill the table in bottom-up fashion




        for (int i = 1; i <= n; i++) {




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




                // Two cases if we see a '*'




                // a) We ignore '*'' character and move




                // to next character in the pattern,




                //     i.e., '*' indicates an empty




                //     sequence.




                // b) '*' character matches with ith




                //     character in input




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




                    lookup[i, j] = lookup[i, j - 1]




                                   || lookup[i - 1, j];



 



                // Current characters are considered as




                // matching in two cases




                // (a) current character of pattern is '?'




                // (b) characters actually match




                else if (pattern[j - 1] == '?'




                         || str[i - 1] == pattern[j - 1])




                    lookup[i, j] = lookup[i - 1, j - 1];



 



                // If characters don't match




                else




                    lookup[i, j] = false;




            }




        }




        return lookup[n, m];




    }



 



    // Driver Code




    public static void Main(String[] args)




    {




        String str = "baaabab";




        String pattern = "*****ba*****ab";




        // String pattern = "ba*****ab";




        // String pattern = "ba*ab";




        // String pattern = "a*ab";




        // String pattern = "a*****ab";




        // String pattern = "*a*****ab";




        // String pattern = "ba*ab****";




        // String pattern = "****";




        // String pattern = "*";




        // String pattern = "aa?ab";




        // String pattern = "b*b";




        // String pattern = "a*a";




        // String pattern = "baaabab";




        // String pattern = "?baaabab";




        // String pattern = "*baaaba*";



 



        if (strmatch(str, pattern, str.Length,




                     pattern.Length))




            Console.WriteLine("Yes");




        else




            Console.WriteLine("No");




    }



}


 


// This code is contributed by Rajput-Ji


















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Output

Yes



Time complexity: O(m x n) 
Auxillary space: O(m x n)
DP Memoisation solution:-














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// C++ program to implement wildcard


// pattern matching algorithm


#include <bits/stdc++.h>



using namespace std;



 


// Function that matches input str with


// given wildcard pattern



vector<vector<int> > dp;




int finding(string& s, string& p, int n, int m)



{



    // return 1 if n and m are negative




    if (n < 0 && m < 0)




        return 1;




   



    // return 0 if m is negative




    if (m < 0)




        return 0;




   



    // return n if n is negative




    if (n < 0) 




    {




        // while m is positve




        while (m >= 0) 




        {




            if (p[m] != '*')




                return 0;




            m--;




        }




        return 1;




    }




    



    // if dp state is not visited




    if (dp[n][m] == -1) 




    {




        if (p[m] == '*')




        {




            return dp[n][m] = finding(s, p, n - 1, m)




                              || finding(s, p, n, m - 1);




        }




        else




        {




            if (p[m] != s[n] && p[m] != '?')




                return dp[n][m] = 0;




            else




                return dp[n][m]




                       = finding(s, p, n - 1, m - 1);




        }




    }




   



    // return dp[n][m] if dp state is previsited




    return dp[n][m];



}


 


 



bool isMatch(string s, string p)



{



    dp.clear();




     



    // resize the dp array




    dp.resize(s.size() + 1, vector<int>(p.size() + 1, -1));




    return dp[s.size()][p.size()]




           = finding(s, p, s.size() - 1, p.size() - 1);



}


 


// Driver code



int main()



{



    string str = "baaabab";




    string pattern = "*****ba*****ab";




    // char pattern[] = "ba*****ab";




    // char pattern[] = "ba*ab";




    // char pattern[] = "a*ab";




    // char pattern[] = "a*****ab";




    // char pattern[] = "*a*****ab";




    // char pattern[] = "ba*ab****";




    // char pattern[] = "****";




    // char pattern[] = "*";




    // char pattern[] = "aa?ab";




    // char pattern[] = "b*b";




    // char pattern[] = "a*a";




    // char pattern[] = "baaabab";




    // char pattern[] = "?baaabab";




    // char pattern[] = "*baaaba*";



 



    if (isMatch(str, pattern))




        cout << "Yes" << endl;




    else




        cout << "No" << endl;



 



    return 0;



}


















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Output

Yes



Time complexity: O(m x n). 
Auxiliary space:  O(m x n). 
Further Improvements: 
We can improve space complexity by making use of the fact that we only uses the result from last row. 
One more improvement is yo merge consecutive ‘*’ in the pattern to single ‘*’ as they mean the same thing. For example for pattern “*****ba*****ab”, if we merge consecutive stars, the resultant string will be “*ba*ab”. So, value of m is reduced from 14 to 6.
This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 
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