Wildcard Pattern Matching
- Difficulty Level : Hard
- Last Updated : 03 Mar, 2022
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)
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
- 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 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.
// both text and pattern are null T = true; // pattern is null T[i] = false; // text is null T[j] = T[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.
Time complexity: O(m x n)
Auxiliary space: O(m x n)
DP Memoization solution:-
Time complexity: O(m x n).
Auxiliary space: O(m x n).
We can improve space complexity by making use of the fact that we only uses the result from last row.
One more improvement is you can 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.
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