Add 0010-regular-expression-matching.go

Author: simhonchourasia

go/0010-regular-expression-matching.go

@@ -0,0 +1,41 @@
+func isMatch(s string, p string) bool {
+    // dp[i][j] is true if p[:i] matches s[:j]
+    dp := make([][]bool, len(p) + 1)
+    for i := 0; i < len(p) + 1; i++ {
+        dp[i] = make([]bool, len(s) + 1)
+        for j := 0; j < len(s) + 1; j++ {
+            dp[i][j] = false
+        }
+    }
+
+    dp[0][0] = true
+
+    // Base case for i = 0 is already set up, as empty pattern can only match empty string
+    // But a nonempty pattern can match an empty string, so we do base cases for j = 0
+    for i := 1; i < len(p); i++ {
+        if p[i] == '*' {
+            dp[i + 1][0] = dp[i - 1][0]
+        }
+    }
+    
+    // Now for the general case
+    for i := 0; i < len(p); i++ {
+        for j := 0; j < len(s); j++ {
+            if p[i] == '.' || p[i] == s[j] {
+                // Single character matches
+                dp[i + 1][j + 1] = dp[i][j]
+            } else if p[i] == '*' {
+                // Wildcard - check that the character matches
+                // or if we can have 0 repetitions of the previous char
+                dp[i + 1][j + 1] = dp[i - 1][j + 1] || dp[i][j + 1]
+                if p[i - 1] == '.' || p[i - 1] == s[j] {
+                    if dp[i + 1][j] {
+                        dp[i + 1][j + 1] = dp[i + 1][j]
+                    }
+                }
+            }
+        }
+    }
+
+    return dp[len(p)][len(s)]
+}