Create Regular Expression Matching solution with Recursion

The provided Java code solves the Regular Expression Matching Problem by determining if a given input string s matches a specified pattern p. The pattern can include special characters:
1. " . "matches any single character.
2. " * " matches zero or more occurrences of the preceding character.
The implementation uses memoization to store results for string and pattern combinations, enhancing performance, while employing recursive logic to efficiently handle character matching and special characters like . and *.

Author: Bucke200

src/main/java/com/thealgorithms/Recursion/RegexMatcher.java

@@ -0,0 +1,36 @@
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+    // Memoization map to store results of subproblems
+    private Map<String, Boolean> memo = new HashMap<>();
+
+    public boolean isMatch(String s, String p) {
+        // Check if the result is already computed
+        String key = s + "," + p;
+        if (memo.containsKey(key)) {
+            return memo.get(key);
+        }
+
+        // Base case: if the pattern is empty, the string must also be empty
+        if (p.isEmpty()) {
+            return s.isEmpty();
+        }
+
+        // Check if the first characters match
+        boolean firstMatch = (!s.isEmpty() && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.'));
+
+        boolean result;
+        if (p.length() >= 2 && p.charAt(1) == '*') {
+            // Handle the '*' case (zero or more occurrences of the preceding character)
+            result = (isMatch(s, p.substring(2)) || (firstMatch && isMatch(s.substring(1), p)));
+        } else {
+            // Handle the normal case (no '*')
+            result = firstMatch && isMatch(s.substring(1), p.substring(1));
+        }
+
+        // Memoize the result
+        memo.put(key, result);
+        return result;
+    }
+}