Add a solution to Word Break

Author: soapyigu

DP/WordBreak.swift

@@ -0,0 +1,26 @@
+/**
+ * Question Link: https://leetcode.com/problems/word-break/
+ * Primary idea: DP. dp[i] = dp[j] && dict.contains(s[j..<i]). Break the operation if current index is already true.
+ * Time Complexity: O(n^3), Space Complexity: O(n)
+ */
+
+class WordBreak {
+    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
+        let dict = Set(wordDict), s = Array(s)
+        var dp = Array(repeating: false, count: s.count + 1)
+        dp[0] = true
+        
+        for i in 1...s.count {
+            for j in 0..<i {
+                let currentStr = String(s[j..<i])
+                
+                if dp[j] && wordDict.contains(currentStr) {
+                    dp[i] = true
+                    break
+                }   
+            }
+        }
+        
+        return dp[s.count]
+    }
+}