Create RegularExpressionMatching.java

Author: kokatesaurabh

src/main/java/com/thealgorithms/dynamicprogramming/RegularExpressionMatching.java

@@ -0,0 +1,191 @@
+package com.thealgorithms.dynamicprogramming;
+
+/**
+ * Implements regular expression matching with support for '.' and '*'.
+ * 
+ * Problem: Given an input string s and a pattern p, implement regular expression
+ * matching with support for '.' and '*' where:
+ * - '.' Matches any single character
+ * - '*' Matches zero or more of the preceding element
+ * - The matching should cover the entire input string (not partial)
+ * 
+ * This solution uses dynamic programming with memoization for efficient computation.
+ * 
+ * Example:
+ * Input: s = "aa", p = "a" → Output: false
+ * Input: s = "aa", p = "a*" → Output: true  
+ * Input: s = "ab", p = ".*" → Output: true
+ * 
+ * Time Complexity: O(m * n) where m is length of s and n is length of p
+ * Space Complexity: O(m * n) for the memoization table
+ * 
+ * @author Your Name (replace with your GitHub username)
+ */
+public final class RegularExpressionMatching {
+    private RegularExpressionMatching() {
+        // Private constructor to prevent instantiation
+    }
+
+    /**
+     * Determines if the input string matches the given pattern.
+     *
+     * @param s the input string to match (contains only lowercase English letters)
+     * @param p the pattern (contains lowercase English letters, '.', and '*')
+     * @return true if the entire string matches the pattern, false otherwise
+     * @throws IllegalArgumentException if input strings are null or pattern is invalid
+     */
+    public static boolean isMatch(String s, String p) {
+        if (s == null || p == null) {
+            throw new IllegalArgumentException("Input strings cannot be null");
+        }
+        
+        if (!isValidPattern(p)) {
+            throw new IllegalArgumentException("Invalid pattern format");
+        }
+
+        // Create memoization table with Boolean wrapper for null checks
+        Boolean[][] memo = new Boolean[s.length() + 1][p.length() + 1];
+        return dp(0, 0, s, p, memo);
+    }
+
+    /**
+     * Helper method that performs the actual dynamic programming computation.
+     *
+     * @param i current index in string s
+     * @param j current index in pattern p  
+     * @param s the input string
+     * @param p the pattern
+     * @param memo memoization table storing computed results
+     * @return true if s[i:] matches p[j:], false otherwise
+     */
+    private static boolean dp(int i, int j, String s, String p, Boolean[][] memo) {
+        // Return cached result if available
+        if (memo[i][j] != null) {
+            return memo[i][j];
+        }
+
+        boolean result;
+        
+        // Base case: pattern is exhausted
+        if (j == p.length()) {
+            result = (i == s.length());
+        } else {
+            // Check if current characters match
+            boolean currentMatch = i < s.length() && 
+                                  (p.charAt(j) == '.' || p.charAt(j) == s.charAt(i));
+
+            // Handle '*' operator (lookahead)
+            if (j + 1 < p.length() && p.charAt(j + 1) == '*') {
+                // Two possibilities:
+                // 1. Use '*' as zero occurrences (skip current pattern character and '*')
+                // 2. Use '*' as one or more occurrences (if current characters match)
+                result = dp(i, j + 2, s, p, memo) || 
+                        (currentMatch && dp(i + 1, j, s, p, memo));
+            } else {
+                // No '*' operator, simply advance both pointers if current characters match
+                result = currentMatch && dp(i + 1, j + 1, s, p, memo);
+            }
+        }
+
+        // Cache the result
+        memo[i][j] = result;
+        return result;
+    }
+
+    /**
+     * Validates that the pattern follows the constraints:
+     * - Only contains lowercase English letters, '.', and '*'
+     * - '*' always follows a valid character (not at start and not after another '*')
+     *
+     * @param p the pattern to validate
+     * @return true if pattern is valid, false otherwise
+     */
+    private static boolean isValidPattern(String p) {
+        if (p.isEmpty()) {
+            return true;
+        }
+
+        // Check first character is not '*'
+        if (p.charAt(0) == '*') {
+            return false;
+        }
+
+        // Check all characters are valid and '*' always follows valid character
+        for (int i = 0; i < p.length(); i++) {
+            char c = p.charAt(i);
+            if (!isValidPatternChar(c)) {
+                return false;
+            }
+            
+            // Check if '*' is properly positioned
+            if (c == '*' && (i == 0 || p.charAt(i - 1) == '*')) {
+                return false;
+            }
+        }
+
+        return true;
+    }
+
+    /**
+     * Checks if a character is valid in a pattern (lowercase letter, '.', or '*')
+     */
+    private static boolean isValidPatternChar(char c) {
+        return (c >= 'a' && c <= 'z') || c == '.' || c == '*';
+    }
+
+    /**
+     * Alternative iterative DP solution (bottom-up approach)
+     * This version uses a 2D boolean array for tabulation.
+     *
+     * @param s the input string
+     * @param p the pattern
+     * @return true if string matches pattern, false otherwise
+     */
+    public static boolean isMatchIterative(String s, String p) {
+        if (s == null || p == null) {
+            throw new IllegalArgumentException("Input strings cannot be null");
+        }
+        
+        if (!isValidPattern(p)) {
+            throw new IllegalArgumentException("Invalid pattern format");
+        }
+
+        int m = s.length();
+        int n = p.length();
+        
+        // dp[i][j] means s[0..i-1] matches p[0..j-1]
+        boolean[][] dp = new boolean[m + 1][n + 1];
+        
+        // Empty string matches empty pattern
+        dp[0][0] = true;
+        
+        // Handle patterns like a*, a*b*, a*b*c* that can match empty string
+        for (int j = 2; j <= n; j++) {
+            if (p.charAt(j - 1) == '*') {
+                dp[0][j] = dp[0][j - 2];
+            }
+        }
+        
+        // Fill the DP table
+        for (int i = 1; i <= m; i++) {
+            for (int j = 1; j <= n; j++) {
+                char sc = s.charAt(i - 1);
+                char pc = p.charAt(j - 1);
+                
+                if (pc == '.' || pc == sc) {
+                    dp[i][j] = dp[i - 1][j - 1];
+                } else if (pc == '*') {
+                    char prev = p.charAt(j - 2);
+                    // Zero occurrences of previous character
+                    dp[i][j] = dp[i][j - 2];
+                    // One or more occurrences if previous character matches
+                    if (prev == '.' || prev == sc) {
+                        dp[i][j] = dp[i][j] || dp[i - 1][j];
+                    }
+                }
+            }
+        }
+        
+        return dp[m][n];
+    }
+}