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

• Difficulty Level : Expert
• Last Updated : 08 Dec, 2017

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

 // C++ program to implement wildcard // pattern matching algorithm #include using namespace std;    struct DP {     bool value;     char ch; };    // Function that matches input str with // given wildcard pattern bool strmatch(string str, string pattern,               int n, int m) {     // empty pattern can only match with     // empty string     if (m == 0)         return (n == 0);        // If first character of pattern is '+'     if (pattern == '+')         return false;        // lookup table for storing results of     // subproblems     struct DP lookup[m + 1][n + 1];        // initialize lookup table to false     for (int i = 0; i <= m; i++)         for (int j = 0; j <= n; j++) {             lookup[i][j].value = false;              lookup[i][j].ch = ' ';          }                   // empty pattern can match with     // empty string     lookup.value = true;        // Only '*' can match with empty string     for (int j = 1; j <= n; j++)         if (pattern[j - 1] == '*')             lookup[j].value =                     lookup[j - 1].value;        // fill the table in bottom-up fashion     for (int i = 1; i <= m; i++) {         for (int j = 1; j <= n; 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[i - 1] == '*') {                 lookup[i][j].value =                        lookup[i][j - 1].value ||                         lookup[i - 1][j].value;                 lookup[i][j].ch = str[j - 1];             }                // Current characters are considered as             // matching in two cases             // (a) current character of pattern is '?'             else if (pattern[i - 1] == '?') {                 lookup[i][j].value =                            lookup[i - 1][j - 1].value;                 lookup[i][j].ch = str[j - 1];             }                // (b) characters actually match             else if (str[j - 1] == pattern[i - 1])                 lookup[i][j].value =                        lookup[i - 1][j - 1].value;                // Current character match             else if (pattern[i - 1] == '+')                    // case 1: if previous character is                  // not symbol                 if (pattern[i - 2] != '+' ||                     pattern[i - 2] != '*' ||                     pattern[i - 2] != '?')                     if (pattern[i - 2] == str[j - 1]) {                         lookup[i][j].value =                             lookup[i - 1][j - 1].value;                         lookup[i][j].ch = str[j - 1];                     }                        // case 2 : if previous character                      // is symbol (+, *, ? ) then we                      // compare current text character                      // with the character that is used by                     // the symbol  at that point. we                      // access it by lookup[i-1][j-1]                     else if (str[j-1] == lookup[i-1][j-1].ch) {                         lookup[i][j].value =                               lookup[i - 1][j - 1].value;                         lookup[i][j].ch =                               lookup[i - 1][j - 1].ch;                     }                        // If characters don't match                     else                         lookup[i][j].value = false;         }     }        return lookup[m][n].value; }    // Driver code int main() {     string str = "baaabaaa";     string pattern = "*****+ba***+";        if (strmatch(str, pattern, str.length(),                        pattern.length()))         cout << "Yes" << endl;     else         cout << "No" << endl;        return 0; }

Output:

Yes

Time Complexity : O(n*m)

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