feat: solve word break

Author: GangBean

word-break/GangBean.java

@@ -0,0 +1,34 @@
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        /**
+        1. understanding
+        - check if s's segments are all in wordDict.
+        - wordDict can be used multiple times
+        2. strategy
+        a) dynamic programming 
+        - dp[k]: substring(0,k+1) can be constructed by wordDict.
+        - dp[0]: false
+        - dp[1]: substring(0, 2) in wordDict
+        - dp[k+1] = substring(0, k+2) in wordDict || Any(dp[k] && substring(k+1, k+2) in wordDict)
+        - return dp[s.length()]
+        3. complexity
+        - time: O(N^2 * S), where N is the length of s, S is search time each segment in wordDict. You can calculate S in O(1) time, when change wordDict to Set. so O(N) is final time complexity.
+        - space: O(N + W), W is the size of wordDict
+        */
+        Set<String> bow = new HashSet<>(wordDict);
+        boolean[] dp = new boolean[s.length() + 1];
+
+        for (int i = 1; i < dp.length; i++) { // O(N)
+            String segment = s.substring(0, i);
+            dp[i] = bow.contains(segment);
+            for (int j = 0; j < i; j++) { // O(N)
+                if (dp[i]) break;
+                segment = s.substring(j, i);
+                dp[i] = dp[i] || (dp[j] && bow.contains(segment)); // O(1)
+            }
+        }
+
+        return dp[s.length()];
+    }
+}
+