[Dynamic Programming] Wildcard Matching - Hard

Wildcard matching is a tricky problem in computer science. It helps us see if a string fits a certain pattern with wildcard characters. The pattern can have the wildcard character '*'. This character matches any number of characters, even no characters at all. The '?' character matches just one character. We can solve the wildcard matching problem quickly using dynamic programming. This way, we do not get the slow performance that comes with a simple recursive method.

In this article, we will look into the details of the wildcard matching problem with dynamic programming methods. We will give a clear overview of the problem. We will explain the dynamic programming approach step by step. We will also show how to implement it in Java, Python, and C++. We will talk about how to make space usage better. We will compare different methods. We will look at common edge cases and answer questions that people often ask about wildcard matching.

  • Dynamic Programming Wildcard Matching Problem Overview
  • Understanding the Problem Statement
  • Dynamic Programming Approach Explained
  • Java Implementation of Wildcard Matching
  • Python Implementation of Wildcard Matching
  • C++ Implementation of Wildcard Matching
  • Optimizing Space Complexity in Wildcard Matching
  • Comparative Analysis of Different Approaches
  • Common Edge Cases in Wildcard Matching
  • Frequently Asked Questions

Understanding the Problem Statement

The Wildcard Matching problem is about checking if a string s matches a pattern p. The pattern can have some special characters called wildcards.

  • ? can match any single character.
  • * can match any group of characters. This includes an empty group.

Problem Definition

We have two strings: - s: This is the string we want to match. - p: This is the pattern string that may have wildcards.

Our task is to create a function that checks if s matches p based on the wildcard rules.

Example Cases

  1. Example 1:
    • Input: s = "aa", p = "a"
    • Output: false
    • Explanation: The pattern a does not match the string aa.
  2. Example 2:
    • Input: s = "aa", p = "*"
    • Output: true
    • Explanation: The pattern * matches any string.
  3. Example 3:
    • Input: s = "cb", p = "?a"
    • Output: false
    • Explanation: The pattern ?a needs a character before a, but cb does not fit.
  4. Example 4:
    • Input: s = "adceb", p = "*a*b"
    • Output: true
    • Explanation: The pattern *a*b can match adceb by matching dce.
  5. Example 5:
    • Input: s = "acdcb", p = "a*c?b"
    • Output: false
    • Explanation: The pattern a*c?b does not match the input string acdcb.

Constraints

  • The length of s and p can be up to 100.
  • The allowed characters in s and p are lowercase English letters.

We can solve this problem fast using dynamic programming methods. We will talk about that in the next sections.

Dynamic Programming Approach Explained

The Wildcard Matching problem is about checking if a string matches a pattern with wildcard characters. The wildcards are *, which can match any group of characters, and ?, which can match just one character. We can solve this problem well using dynamic programming.

Dynamic Programming Table

  1. Define the DP Table:
    • We use dp[i][j]. It is True if the first i characters of the string match the first j characters of the pattern. If not, it is False.
  2. Initialization:
    • We set dp[0][0] = True. This means an empty string matches an empty pattern.
    • If the pattern starts with *, we set dp[0][j] = dp[0][j-1] if the j-th character is *. This means * can match an empty string.
  3. Filling the Table:
    • We go through each character of the string and pattern:
      • If characters match or the pattern has a ?, we take the match from dp[i-1][j-1].
      • If the pattern has a *, we look at two cases:
        • Treat * as matching no characters: dp[i][j] = dp[i][j-1].
        • Treat * as matching one or more characters: dp[i][j] = dp[i-1][j].

Transition Formula

  • If s[i-1] == p[j-1] or p[j-1] == '?':

    dp[i][j] = dp[i-1][j-1]

  • If p[j-1] == '*':

    dp[i][j] = dp[i][j-1] || dp[i-1][j]

Complexity

  • Time Complexity: O(m * n). Here, m is the string length and n is the pattern length.
  • Space Complexity: O(m * n) for the DP table.

Example Code Snippet (Java)

public boolean isMatch(String s, String p) {
    int m = s.length(), n = p.length();
    boolean[][] dp = new boolean[m + 1][n + 1];
    dp[0][0] = true;

    for (int j = 1; j <= n; j++) {
        if (p.charAt(j - 1) == '*') {
            dp[0][j] = dp[0][j - 1];
        }
    }

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (p.charAt(j - 1) == s.charAt(i - 1) || p.charAt(j - 1) == '?') {
                dp[i][j] = dp[i - 1][j - 1];
            } else if (p.charAt(j - 1) == '*') {
                dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
            }
        }
    }
    return dp[m][n];
}

With this dynamic programming method, we can solve the Wildcard Matching problem quickly. It works well for cases where we need to match patterns, like in search algorithms or text processing tasks.

Java Implementation of Wildcard Matching

The wildcard matching problem is about checking if a string matches a pattern with wildcard characters. The wildcard characters are * that can match any group of characters and ? that can match any single character.

Java Code Implementation

Here is a simple Java code for the wildcard matching problem using dynamic programming:

public class WildcardMatching {
    public boolean isMatch(String s, String p) {
        int sLen = s.length(), pLen = p.length();
        boolean[][] dp = new boolean[sLen + 1][pLen + 1];
        
        // Empty pattern matches empty string
        dp[0][0] = true;

        // Handle patterns with leading '*'
        for (int j = 1; j <= pLen; j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 1];
            }
        }

        for (int i = 1; i <= sLen; i++) {
            for (int j = 1; j <= pLen; j++) {
                if (p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                } else if (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }
        
        return dp[sLen][pLen];
    }

    public static void main(String[] args) {
        WildcardMatching matcher = new WildcardMatching();
        String s = "adceb";
        String p = "*a*b";
        boolean result = matcher.isMatch(s, p);
        System.out.println("Does the string match the pattern? " + result);
    }
}

Explanation of the Code

  • DP Table Initialization: We create a 2D boolean array dp. Each dp[i][j] shows if the first i characters of the string s match the first j characters of the pattern p.
  • Filling the DP Table:
    • We start with the base case where both string and pattern are empty.
    • We fill the first row to handle cases where the pattern starts with *.
    • The main logic checks three conditions:
      • If the current character in the pattern is *, it can match zero characters (left) or one or more characters (up).
      • If the current character is ? or it matches the character in the string, we take the value from the diagonal cell.
  • Result: The final answer is in dp[sLen][pLen].

This method is efficient and runs in O(m * n) time. Here, m is the length of the string and n is the length of the pattern. This makes it good for large inputs. If you want to read more about dynamic programming techniques, you can check this article on the Edit Distance problem.

Python Implementation of Wildcard Matching

The wildcard matching problem is about checking if a string matches a pattern that can have wildcards. The wildcards are *, which can match any group of characters, even nothing, and ?, which can match one character.

Python Code Implementation

Here is a Python code that uses dynamic programming to solve the wildcard matching problem:

def isMatch(s: str, p: str) -> bool:
    m, n = len(s), len(p)
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    
    dp[0][0] = True  # Both string and pattern are empty

    # Handle patterns with '*' at the start
    for j in range(1, n + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 1]
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[j - 1] == '*':
                dp[i][j] = dp[i][j - 1] or dp[i - 1][j]
            elif p[j - 1] == '?' or p[j - 1] == s[i - 1]:
                dp[i][j] = dp[i - 1][j - 1]

    return dp[m][n]

# Example usage
s = "adceb"
p = "*a*b"
print(isMatch(s, p))  # Output: True

Explanation of the Code

  • Matrix Initialization: We create a 2D list dp. The dp[i][j] shows if the first i characters of s match the first j characters of p.
  • Base Case: We set dp[0][0] to True because two empty strings match.
  • First Row Initialization: We fill the first row based on the pattern. It allows initial * characters to match an empty string.
  • Filling the DP Table: We fill the table based on two cases:
    • If the current character in the pattern is *, it can match no characters or some characters.
    • If the current character is ? or matches the character in s, we take the value from the diagonal cell.
  • Final Result: The result is in dp[m][n], showing if the whole string s matches the whole pattern p.

This dynamic programming method checks for wildcard matches in O(m * n) time, where m is the length of the string and n is the length of the pattern. For more about dynamic programming, we can check the Dynamic Programming: Regular Expression Matching article.

C++ Implementation of Wildcard Matching

In C++, we can solve the wildcard matching problem using dynamic programming. This problem is about matching a string s with a pattern p. The pattern can have wildcard characters. The * can match any sequence of characters. The ? can match any single character.

Code Implementation

Here is a simple implementation of the wildcard matching algorithm in C++:

#include <iostream>
#include <vector>
#include <string>

using namespace std;

bool isMatch(string s, string p) {
    int sLen = s.length(), pLen = p.length();
    vector<vector<bool>> dp(sLen + 1, vector<bool>(pLen + 1, false));
    
    dp[0][0] = true;

    for (int j = 1; j <= pLen; j++) {
        if (p[j - 1] == '*') {
            dp[0][j] = dp[0][j - 1];
        }
    }

    for (int i = 1; i <= sLen; i++) {
        for (int j = 1; j <= pLen; j++) {
            if (p[j - 1] == '*') {
                dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
            } else if (p[j - 1] == '?' || s[i - 1] == p[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            }
        }
    }

    return dp[sLen][pLen];
}

int main() {
    string s = "adceb";
    string p = "*a*b";
    cout << (isMatch(s, p) ? "Match" : "No Match") << endl;
    return 0;
}

Explanation of the Code

  • Dynamic Programming Table: We create a 2D vector dp with size (sLen + 1) x (pLen + 1). The dp[i][j] tells us if the first i characters of s match the first j characters of p.
  • Initialization: We set dp[0][0] to true. This means an empty pattern matches an empty string. Patterns that start with * can match an empty string too.
  • Filling the DP Table:
    • For every character in s and p, we check the conditions:
      • If p[j-1] is *, it can match zero characters (so dp[i][j-1]) or one character (so dp[i-1][j]).
      • If p[j-1] is ? or matches s[i-1], then it depends on the previous characters (dp[i-1][j-1]).
  • Final Result: The value at dp[sLen][pLen] gives the final match result.

This C++ code gives a clear and good solution to the wildcard matching problem using dynamic programming. For more reading on dynamic programming, we can check topics like Edit Distance and Regular Expression Matching.

Optimizing Space Complexity in Wildcard Matching

In wildcard matching, we want to check if a string (s) fits a pattern (p). This pattern can have special characters like * which matches any string and ? which matches a single character. A common way to solve this problem is using dynamic programming with a 2D table. But this takes a lot of memory, especially with long strings.

Space Optimization Technique

To save space, we can change the 2D table to a 1D array. We notice that the current state only relies on the previous state. So we only need two arrays. One is for the current row and the other is for the previous row.

Implementation Details

  1. Use a 1D DP Array: We create a boolean array dp. Here, dp[j] shows if s up to index i matches p up to index j.

  2. Iterate through Characters: We loop through each character in the pattern and string. We update the dp array based on current characters and wildcards.

  3. Initialization: Set dp[0] to true if p is empty. Also, we check if p starts with *.

Java Implementation

public boolean isMatch(String s, String p) {
    boolean[] dp = new boolean[p.length() + 1];
    dp[0] = true; // Empty pattern matches empty string

    // Handle patterns like "*", "**", etc.
    for (int j = 1; j <= p.length(); j++) {
        if (p.charAt(j - 1) == '*') {
            dp[j] = dp[j - 1];
        }
    }

    for (int i = 1; i <= s.length(); i++) {
        boolean prev = dp[0]; // Store dp[i-1][j-1]
        dp[0] = false; // Empty pattern can't match non-empty string
        for (int j = 1; j <= p.length(); j++) {
            boolean temp = dp[j]; // Store dp[i-1][j]
            if (p.charAt(j - 1) == '*') {
                dp[j] = dp[j - 1] || dp[j];
            } else if (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1)) {
                dp[j] = prev;
            } else {
                dp[j] = false;
            }
            prev = temp; // Update prev for next iteration
        }
    }
    return dp[p.length()];
}

Python Implementation

def isMatch(s: str, p: str) -> bool:
    dp = [False] * (len(p) + 1)
    dp[0] = True  # Empty pattern matches empty string

    # Handle patterns like "*", "**", etc.
    for j in range(1, len(p) + 1):
        if p[j - 1] == '*':
            dp[j] = dp[j - 1]

    for i in range(1, len(s) + 1):
        prev = dp[0]  # Store dp[i-1][j-1]
        dp[0] = False  # Empty pattern can't match non-empty string
        for j in range(1, len(p) + 1):
            temp = dp[j]  # Store dp[i-1][j]
            if p[j - 1] == '*':
                dp[j] = dp[j - 1] or dp[j]
            elif p[j - 1] == '?' or s[i - 1] == p[j - 1]:
                dp[j] = prev
            else:
                dp[j] = False
            prev = temp  # Update prev for next iteration
    return dp[len(p)]

C++ Implementation

bool isMatch(string s, string p) {
    vector<bool> dp(p.size() + 1, false);
    dp[0] = true; // Empty pattern matches empty string

    // Handle patterns like "*", "**", etc.
    for (int j = 1; j <= p.size(); j++) {
        if (p[j - 1] == '*') {
            dp[j] = dp[j - 1];
        }
    }

    for (int i = 1; i <= s.size(); i++) {
        bool prev = dp[0]; // Store dp[i-1][j-1]
        dp[0] = false; // Empty pattern can't match non-empty string
        for (int j = 1; j <= p.size(); j++) {
            bool temp = dp[j]; // Store dp[i-1][j]
            if (p[j - 1] == '*') {
                dp[j] = dp[j - 1] || dp[j];
            } else if (p[j - 1] == '?' || s[i - 1] == p[j - 1]) {
                dp[j] = prev;
            } else {
                dp[j] = false;
            }
            prev = temp; // Update prev for next iteration
        }
    }
    return dp[p.size()];
}

Complexity Analysis

  • Time Complexity: O(m * n). Here, m is the length of the string and n is the length of the pattern.
  • Space Complexity: O(n). We reduce the space from O(m*n) to O(n) using the 1D array.

This method helps us save memory while still matching well in the wildcard matching problem. For more on dynamic programming, we can check articles like Dynamic Programming Regular Expression Matching - Hard.

Comparative Analysis of Different Approaches

When we look at the wildcard matching problem, we can use different methods. Each method has its good and bad sides. We will look at three main methods: brute force, dynamic programming, and regular expressions.

1. Brute Force Approach

  • Description: The brute force method tries to create all possible strings that fit the pattern. Then, it checks each one against the input string.
  • Time Complexity: O(2^(m + n)), where m is the pattern length and n is the input string length.
  • Space Complexity: O(m + n) for the recursion stack.
  • Pros: It is simple to use for small inputs.
  • Cons: It does not work well for larger strings because of its slow speed.

2. Dynamic Programming Approach

  • Description: This method uses a 2D table to keep results of smaller problems. This helps to calculate things quicker.
  • Time Complexity: O(m * n), where m is the pattern length and n is the input string length.
  • Space Complexity: O(m * n) for the DP table. We can also make it O(n) by using only two rows.
  • Pros: It works well for larger inputs. It also gives a clear way to solve the problem.
  • Cons: It needs more memory for the DP table.

Dynamic Programming Table Construction:

boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true; // Empty pattern matches empty string

for (int j = 1; j <= m; j++) {
    if (pattern.charAt(j - 1) == '*') {
        dp[j][0] = dp[j - 1][0]; // '*' can match empty string
    }
}

for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
        if (pattern.charAt(j - 1) == '*') {
            dp[j][i] = dp[j][i - 1] || dp[j - 1][i];
        } else if (pattern.charAt(j - 1) == '?' || pattern.charAt(j - 1) == str.charAt(i - 1)) {
            dp[j][i] = dp[j - 1][i - 1];
        }
    }
}

3. Regular Expression Approach

  • Description: This method changes the wildcard pattern into a regular expression. We then use regex tools to match.
  • Time Complexity: It changes based on the regex engine but can be O(m * n) in the worst case.
  • Space Complexity: O(m + n) for regex processing.
  • Pros: It is small and often works fast thanks to the regex engine.
  • Cons: Sometimes, it can be hard to read and may have speed issues with complicated patterns.

Summary of Comparison

Approach Time Complexity Space Complexity Pros Cons
Brute Force O(2^(m + n)) O(m + n) Simple to use Not good for large inputs
Dynamic Programming O(m * n) O(m * n) (O(n) optimized) Works well for larger inputs Needs more memory
Regular Expression O(m * n) O(m + n) Small and fast Can be hard to read

In the end, we choose a method based on the problem needs and the input size limits. For most cases, we like the dynamic programming approach because it is efficient and clear. If we have more complex matching needs, we may find the regular expression method to be the best choice.

Common Edge Cases in Wildcard Matching

When we make a wildcard matching algorithm with dynamic programming, we can face some edge cases. These cases can change how our solution works and what it gives us as an output. It is important to look at these situations to make sure our matching is strong and correct. Here are some common edge cases we should think about:

  • Empty String and Pattern:
    • Input: s = "", p = ""
    • Output: True (both string and pattern are empty).
  • Empty String with Non-empty Pattern:
    • Input: s = "", p = "a*"
    • Output: True (the pattern can match an empty string).
  • Non-empty String with Empty Pattern:
    • Input: s = "abc", p = ""
    • Output: False (a non-empty string cannot match an empty pattern).
  • Leading Wildcards:
    • Input: s = "abc", p = "*"
    • Output: True (a wildcard can match any string).
  • Multiple Consecutive Wildcards:
    • Input: s = "abc", p = "a**b*"
    • Output: True (consecutive wildcards are like one wildcard).
  • Wildcard at the End of the Pattern:
    • Input: s = "abc", p = "abc*"
    • Output: True (the pattern matches the string).
  • Wildcard in Middle of the Pattern:
    • Input: s = "abc", p = "a*c"
    • Output: True (the wildcard can match characters in between).
  • No Matching Characters:
    • Input: s = "abc", p = "a*d"
    • Output: False (no way to match the characters).
  • Multiple Wildcards Interspersed:
    • Input: s = "abcde", p = "*a*b*c*e*"
    • Output: True (wildcards can match characters in between).
  • Complex Patterns:
    • Input: s = "ab", p = "a?b"
    • Output: True (the ‘?’ can match any single character).

Testing these edge cases help us to make sure that our wildcard matching works in all cases. The code below shows how to do wildcard matching using dynamic programming in Python, while considering these edge cases:

def isMatch(s: str, p: str) -> bool:
    dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
    dp[0][0] = True

    for j in range(1, len(p) + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 1]

    for i in range(1, len(s) + 1):
        for j in range(1, len(p) + 1):
            if p[j - 1] == '*':
                dp[i][j] = dp[i][j - 1] or dp[i - 1][j]
            elif p[j - 1] == '?' or s[i - 1] == p[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]

    return dp[len(s)][len(p)]

This code looks at the different edge cases we talked about. It gives us a strong base for wildcard matching.

Frequently Asked Questions

1. What is the time complexity of the dynamic programming approach to wildcard matching?

The time complexity for the dynamic programming method in wildcard matching is O(m * n). Here, m means the length of the input string and n means the length of the pattern. We fill a 2D table to show the states of the matching process. If you want to learn more about dynamic programming problems, check this article on Dynamic Programming: Fibonacci Number.

2. How do wildcards work in string matching?

Wildcards are special symbols in string matching. They can stand for one or more characters in a string. The most used wildcards are the asterisk (*) that matches zero or more characters and the question mark (?) that matches exactly one character. This makes pattern matching more flexible. For a better understanding of this, look at the Dynamic Programming: Regular Expression Matching.

3. Can wildcard matching be solved using recursion?

Yes, we can solve wildcard matching with recursion. But this method is not the best because it has exponential time complexity. We can improve the recursive solution by using memoization. Memoization helps to store results and change it into a dynamic programming solution. For more on recursion in dynamic programming, see Dynamic Programming: Edit Distance.

4. How can space complexity be optimized in the wildcard matching problem?

We can optimize space complexity in the wildcard matching problem from O(m * n) to O(n) by using a rolling array. Instead of keeping a full 2D table, we only need two rows. Each calculation only depends on the current row and the previous one. This method helps to save memory but keeps the same time complexity. For more tips on optimizing space complexity, check Dynamic Programming: Minimum Cost Climbing Stairs.

5. What are common edge cases in wildcard matching?

Common edge cases in wildcard matching are when the pattern or string is empty. It also includes when the pattern has two or more asterisks together or when both the string and pattern do not match at all. We need to handle these cases well to make sure the algorithm works correctly. For more examples of edge cases in dynamic programming, you might find this article on Dynamic Programming: Longest Common Subsequence useful.