From 3300bb7622f16b040cd1be7aa21c919836f636a4 Mon Sep 17 00:00:00 2001
From: Priyanshu72 <72194113+Priyanshu72@users.noreply.github.com>
Date: Sat, 28 Oct 2023 07:51:59 +0530
Subject: [PATCH] Create WildCardMatching.java

---
 WildCardMatching.java | 57 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 57 insertions(+)
 create mode 100644 WildCardMatching.java

diff --git a/WildCardMatching.java b/WildCardMatching.java
new file mode 100644
index 0000000..0563097
--- /dev/null
+++ b/WildCardMatching.java
@@ -0,0 +1,57 @@
+class Solution {
+    public boolean isMatch(String s, String p) {
+        dp = new int[s.length()][p.length()];
+        return helper(s, p, 0, 0);
+    }
+
+    int[][] dp;
+    
+    private boolean helper(String s, String p, int iS, int iP) {
+        // base case one: retrun true if we match all chars to the end
+        if((iS == s.length() && iP == p.length()) 
+        || (iS == s.length() && isRestAllAsterisk(p, iP))) {
+            return true;
+        }
+        // base case two: return false if only one of them make it to the end
+        if(iS >= s.length() || iP >= p.length()) {
+            return false;
+        }
+        // base case three: return false if current p is not '?' or '*' and is nat a matched char to s
+        if(p.charAt(iP) != '?' && p.charAt(iP) != '*' && s.charAt(iS) != p.charAt(iP)) {
+            return false;
+        }
+        // if we already have an answer, just return the answer
+        if(dp[iS][iP] != 0) {
+            return dp[iS][iP] == 1;
+        }
+        boolean res = false;
+        switch(p.charAt(iP)) {
+            case '?':
+                // have to match one char
+                res = res || helper(s, p, iS + 1, iP + 1);
+                break;
+            case '*':
+                // tricky part
+                // we can either match multiple chars or nothing
+                res = res || helper(s, p, iS, iP + 1); // match nothing
+                res = res || helper(s, p, iS + 1, iP + 1); // match one char
+                res = res || helper(s, p, iS + 1, iP); // match one char with no cost
+                break;
+            default:
+                // same char, have to match one char
+                res = res || helper(s, p, iS + 1, iP + 1);
+        }
+        // memorize answer for later use
+        dp[iS][iP] = res ? 1 : -1;
+        return res;
+    }
+
+    private boolean isRestAllAsterisk(String p, int iP) {
+        for(int i = iP ; i < p.length() ; i++) {
+            if(p.charAt(i) != '*') {
+                return false;
+            }
+        }
+        return true;
+    }
+}