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String matching where one string contains wildcard characters

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Given two strings where first string may contain wild card characters and second string is a normal string. Write a function that returns true if the two strings match. The following are allowed wild card characters in first string. 

* --> Matches with 0 or more instances of any character or set of characters.
? --> Matches with any one character.

For example, “g*ks” matches with “geeks” match. And string “ge?ks*” matches with “geeksforgeeks” (note ‘*’ at the end of first string). But “g*k” doesn’t match with “gee” as character ‘k’ is not present in second string. 

Recommended Practice

C++




// A C program to match wild card characters
#include <stdbool.h>
#include <stdio.h>
  
// The main function that checks if two given strings
// match. The first string may contain wildcard characters
bool match(char* first, char* second)
{
    // If we reach at the end of both strings, we are done
    if (*first == '\0' && *second == '\0')
        return true;
  
    // Make sure to eliminate consecutive '*'
    if (*first == '*') {
        while (*(first + 1) == '*')
            first++;
    }
  
    // Make sure that the characters after '*' are present
    // in second string. This function assumes that the
    // first string will not contain two consecutive '*'
    if (*first == '*' && *(first + 1) != '\0'
        && *second == '\0')
        return false;
  
    // If the first string contains '?', or current
    // characters of both strings match
    if (*first == '?' || *first == *second)
        return match(first + 1, second + 1);
  
    // If there is *, then there are two possibilities
    // a) We consider current character of second string
    // b) We ignore current character of second string.
    if (*first == '*')
        return match(first + 1, second)
               || match(first, second + 1);
    return false;
}
  
// A function to run test cases
void test(char* first, char* second)
{
    match(first, second) ? puts("Yes") : puts("No");
}
  
// Driver program to test above functions
int main()
{
    test("g*ks", "geeks"); // Yes
    test("ge?ks*", "geeksforgeeks"); // Yes
    test("g*k", "gee"); // No because 'k' is not in second
    test("*pqrs",
         "pqrst"); // No because 't' is not in first
    test("abc*bcd", "abcdhghgbcd"); // Yes
    test("abc*c?d", "abcd"); // No because second must have
                             // 2 instances of 'c'
    test("*c*d", "abcd"); // Yes
    test("*?c*d", "abcd"); // Yes
    test("geeks**", "geeks"); // Yes
  
    return 0;
}


Java




// Java program to match wild card characters 
class GFG 
{
  
// The main function that checks if 
// two given strings match. The first string 
// may contain wildcard characters
static boolean match(String first, String second) 
{
  
    // If we reach at the end of both strings, 
    // we are done
    if (first.length() == 0 && second.length() == 0)
        return true;
    
  // Make sure to eliminate consecutive '*'
    if (first.length() > 1 &&first.charAt(0) == '*') {
      int i=0;
        while (i+1<first.length() && first.charAt(i+1) == '*')
          i++;
      first=first.substring(i);
    }
  
    // Make sure that the characters after '*' 
    // are present in second string. 
    // This function assumes that the first
    // string will not contain two consecutive '*'
    if (first.length() > 1 && first.charAt(0) == '*' && 
                              second.length() == 0)
        return false;
  
    // If the first string contains '?', 
    // or current characters of both strings match
    if ((first.length() > 1 && first.charAt(0) == '?') || 
        (first.length() != 0 && second.length() != 0 && 
         first.charAt(0) == second.charAt(0)))
        return match(first.substring(1), 
                     second.substring(1));
  
    // If there is *, then there are two possibilities
    // a) We consider current character of second string
    // b) We ignore current character of second string.
    if (first.length() > 0 && first.charAt(0) == '*')
        return match(first.substring(1), second) || 
               match(first, second.substring(1));
    return false;
}
  
// A function to run test cases
static void test(String first, String second)
{
    if (match(first, second))
        System.out.println("Yes");
    else
        System.out.println("No");
}
  
// Driver Code
public static void main(String[] args) 
{
    test("g*ks", "geeks"); // Yes
    test("ge?ks*", "geeksforgeeks"); // Yes
    test("g*k", "gee"); // No because 'k' is not in second
    test("*pqrs", "pqrst"); // No because 't' is not in first
    test("abc*bcd", "abcdhghgbcd"); // Yes
    test("abc*c?d", "abcd"); // No because second must have 2
                            // instances of 'c'
    test("*c*d", "abcd"); // Yes
    test("*?c*d", "abcd"); // Yes
      test("geeks**", "geeks"); // Yes
}
}
  
// This code is contributed by
// sanjeev2552


Python3




# Python program to match wild card characters
  
# The main function that checks if two given strings match.
# The first string may contain wildcard characters
  
  
def match(first, second):
  
    # If we reach at the end of both strings, we are done
    if len(first) == 0 and len(second) == 0:
        return True
  
    # Make sure to eliminate consecutive '*'
    if len(first) > 1 and first[0] == '*':
        i = 0
        while i+1 < len(first) and first[i+1] == '*':
            i = i+1
        first = first[i:]
  
    # Make sure that the characters after '*' are present
    # in second string. This function assumes that the first
    # string will not contain two consecutive '*'
    if len(first) > 1 and first[0] == '*' and len(second) == 0:
        return False
  
    # If the first string contains '?', or current characters
    # of both strings match
    if (len(first) > 1 and first[0] == '?') or (len(first) != 0
                                                and len(second) != 0 and first[0] == second[0]):
        return match(first[1:], second[1:])
  
    # If there is *, then there are two possibilities
    # a) We consider current character of second string
    # b) We ignore current character of second string.
    if len(first) != 0 and first[0] == '*':
        return match(first[1:], second) or match(first, second[1:])
  
    return False
  
# A function to run test cases
  
  
def test(first, second):
    if match(first, second):
        print("Yes")
    else:
        print("No")
  
  
# Driver program
test("g*ks", "geeks"# Yes
test("ge?ks*", "geeksforgeeks"# Yes
test("g*k", "gee"# No because 'k' is not in second
test("*pqrs", "pqrst"# No because 't' is not in first
test("abc*bcd", "abcdhghgbcd"# Yes
test("abc*c?d", "abcd"# No because second must have 2 instances of 'c'
test("*c*d", "abcd"# Yes
test("*?c*d", "abcd"# Yes
test("geeks**", "geeks"# Yes
  
# This code is contributed by BHAVYA JAIN and ROHIT SIKKA


C#




// C# program to match wild card characters
using System;
  
class GFG {
  
    // The main function that checks if
    // two given strings match. The first string
    // may contain wildcard characters
    static bool match(String first, String second)
    {
  
        // If we reach at the end of both strings,
        // we are done
        if (first.Length == 0 && second.Length == 0)
            return true;
  
        // Make sure to eliminate consecutive '*'
        if (first.Length > 1 && first[0] == '*') {
            int i = 0;
            while (i + 1 < first.Length
                   && first[i + 1] == '*')
                i++;
            first = first.Substring(i);
        }
  
        // Make sure that the characters after '*'
        // are present in second string.
        // This function assumes that the first
        // string will not contain two consecutive '*'
        if (first.Length > 1 && first[0] == '*'
            && second.Length == 0)
            return false;
  
        // If the first string contains '?',
        // or current characters of both strings match
        if ((first.Length > 1 && first[0] == '?')
            || (first.Length != 0 && second.Length != 0
                && first[0] == second[0]))
            return match(first.Substring(1),
                         second.Substring(1));
  
        // If there is *, then there are two possibilities
        // a) We consider current character of second string
        // b) We ignore current character of second string.
        if (first.Length > 0 && first[0] == '*')
            return match(first.Substring(1), second)
                || match(first, second.Substring(1));
        return false;
    }
  
    // A function to run test cases
    static void test(String first, String second)
    {
        if (match(first, second))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        test("g*ks", "geeks"); // Yes
        test("ge?ks*", "geeksforgeeks"); // Yes
        test("g*k",
             "gee"); // No because 'k' is not in second
        test("*pqrs",
             "pqrst"); // No because 't' is not in first
        test("abc*bcd", "abcdhghgbcd"); // Yes
        test("abc*c?d", "abcd"); // No because second must
                                 // have 2 instances of 'c'
        test("*c*d", "abcd"); // Yes
        test("*?c*d", "abcd"); // Yes
        test("geeks**", "geeks"); // Yes
    }
}
  
// This code is contributed by Rajput-Ji


Javascript




<script>
  
// Javascript program to match wild card characters
  
// The main function that checks if 
// two given strings match. The first string 
// may contain wildcard characters
function match(first, second) 
{
  
    // If we reach at the end of both strings, 
    // we are done
    if (first.length == 0 && second.length == 0)
        return true;
  
    // Make sure that the characters after '*' 
    // are present in second string. 
    // This function assumes that the first
    // string will not contain two consecutive '*'
    if (first.length > 1 && first[0] == '*' && 
        second.length == 0)
        return false;
  
    // If the first string contains '?', 
    // or current characters of both strings match
    if ((first.length > 1 && first[0] == '?') || 
        (first.length != 0 && second.length != 0 && 
         first[0] == second[0]))
        return match(first.substring(1), 
                     second.substring(1));
  
    // If there is *, then there are two possibilities
    // a) We consider current character of second string
    // b) We ignore current character of second string.
    if (first.length > 0 && first[0] == '*')
        return match(first.substring(1), second) || 
               match(first, second.substring(1));
                 
    return false;
}
  
// A function to run test cases
function test(first, second)
{
    if (match(first, second))
       document.write("Yes" + "<br>");
    else
       document.write("No" + "<br>");
}
  
// Driver code
test("g*ks", "geeks"); // Yes
test("ge?ks*", "geeksforgeeks"); // Yes
test("g*k", "gee"); // No because 'k' is not in second
test("*pqrs", "pqrst"); // No because 't' is not in first
test("abc*bcd", "abcdhghgbcd"); // Yes
test("abc*c?d", "abcd"); // No because second must have 2
                        // instances of 'c'
test("*c*d", "abcd"); // Yes
test("*?c*d", "abcd"); // Yes
  
// This code is contributed by SoumikMondal
  
</script>


Output: 

Yes
Yes
No
No
Yes
No
Yes
Yes
Yes

Time Complexity: O(n)

Auxiliary Space: O(1)

Exercise 
1) In the above solution, all non-wild characters of first string must be there is second string and all characters of second string must match with either a normal character or wildcard character of first string. Extend the above solution to work like other pattern searching solutions where the first string is pattern and second string is text and we should print all occurrences of first string in second. 
2) Write a pattern searching function where the meaning of ‘?’ is same, but ‘*’ means 0 or more occurrences of the character just before ‘*’. For example, if first string is ‘a*b’, then it matches with ‘aaab’, but doesn’t match with ‘abb’. 
This article is compiled by Vishal Chaudhary and reviewed by GeeksforGeeks team.
 



Last Updated : 11 Sep, 2023
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