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Regular Expression Matching using DP

Certification: Advanced Level Accuracy: 57.14% Submissions: 7 Points: 15

Write a Java program to implement regular expression matching with support for '.' and '*'. The matching should cover the entire input string, not just a portion of it.

  • '.' Matches any single character.
  • '*' Matches zero or more of the preceding element.
Example 1
    
  
  • Input: s = "aa", p = "a*"
    • 
  
  • Output: true
    • 
  
  • Explanation: 
    • The pattern "a*" means zero or more occurrences of 'a'. 
    • The first 'a' in "aa" matches the 'a' in the pattern, and the second 'a' can be matched by the '*', which allows 'a' to appear multiple times.
    •
Example 2
    
  
  • Input: s = "mississippi", p = "mis*i.*pi"
    • 
  
  • Output: true
    • 
  
  • Explanation: 
    • The first three characters "mis" match directly. The pattern "s*" matches the single 's'. 
    • The 'i' matches the next 'i' in the string. The pattern ".*" matches any sequence of characters, which in this case is "ssi". 
    • The remaining "pi" in the pattern matches the final "pi" in the string.
    •
Constraints
    
  
  • 1 <= s.length <= 20
    • 
  
  • 1 <= p.length <= 30
    • 
  
  • s contains only lowercase English letters
    • 
  
  • p contains only lowercase English letters, '.', and '*'
    • 
  
  • It is guaranteed for each appearance of the character '*', there will be a previous valid character to match
    • 
  
  • Time Complexity: O(m*n) where m and n are the lengths of the string and pattern respectively
    • 
  
  • Space Complexity: O(m*n)
    •
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to solve this problem
    • 
  
  • Create a 2D boolean array dp[m+1][n+1] where dp[i][j] indicates if the first i characters of s match the first j characters of p
    • 
  
  • Handle the base cases properly: empty string and empty pattern
    • 
  
  • When you encounter a '*' in the pattern, you need to consider both zero occurrence and multiple occurrences of the preceding character
    • 
  
  • When you encounter a '.', it matches any single character in the string
    • 
  
  • Be careful with edge cases, especially those involving '*' at the beginning of patterns
    •

Steps to solve by this approach:

 Step 1: Create a 2D boolean array dp[m+1][n+1] where dp[i][j] indicates if the first i characters of s match the first j characters of p.

 Step 2: Initialize the base case: empty string matches empty pattern, so dp[0][0] = true.

 Step 3: Handle patterns like a*, a*b*, etc. for matching an empty string by checking if dp[0][j-2] is true when p[j-1] is '*'.

 Step 4: For each character in s and p, check if they match (either directly or through '.').

 Step 5: If the current pattern character is '*', consider both zero occurrence (dp[i][j-2]) and multiple occurrences (dp[i-1][j] if the preceding pattern character matches current string character).

 Step 6: Update the dp table accordingly for each character comparison.

 Step 7: The result is stored in dp[m][n], which indicates if the entire string matches the entire pattern.

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