Add a solution to Word Break II

Author: soapyigu

DFS/WordBreakII.swift

@@ -0,0 +1,36 @@
+/**
+ * Question Link: https://leetcode.com/problems/word-break-ii/
+ * Primary idea: DFS. Termination case is index hits the end of the string
+ * 
+ * Time Complexity: O(n), Space Complexity: O(n)
+ *
+ */
+
+class WordBreakII {
+    func wordBreak(_ s: String, _ wordDict: [String]) -> [String] {
+        var res = [String](), path = [String]()
+        
+        dfs(&res, &path, Array(s), Set(wordDict), 0)
+        
+        return res
+    }
+    
+    private func dfs(_ res: inout [String], _ path: inout [String], _ s: [Character], _ dict: Set<String>, _ idx: Int) {
+        if idx >= s.count {
+            res.append(path.joined(separator: " "))
+            return
+        }
+        
+        for i in idx..<s.count {
+            let currentWord = String(s[idx...i])
+            
+            guard dict.contains(currentWord) else {
+                continue
+            }
+            
+            path.append(currentWord)
+            dfs(&res, &path, s, dict, i + 1)
+            path.removeLast()
+        }
+    }
+}