feat: word-break

Author: minji-go

word-break/minji-go.java

@@ -0,0 +1,53 @@
+/**
+ * <a href="https://leetcode.com/problems/word-break/">week05-4.word-break</a>
+ * <li>Description: return true if s can be segmented into a space-separated sequence of one or more dictionary words.</li>
+ * <li>Topics: Array, Hash Table, String, Dynamic Programming, Trie, Memoization</li>
+ * <li>Time Complexity: O(N³), Runtime 7ms        </li>
+ * <li>Space Complexity: O(N+W), Memory 44.59MB     </li>
+ */
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        Trie trie = new Trie();
+        for (String str : wordDict) {
+            trie.insert(str);
+        }
+
+        List<Integer> dp = new ArrayList<>();
+        dp.add(-1);
+        for (int i = 0; i < s.length(); i++) {
+            for (int j = dp.size() - 1; j >= 0; j--) {
+                int startIndex = dp.get(j);
+                if (trie.search(s.substring(startIndex + 1, i + 1))) {
+                    dp.add(i);
+                    break;
+                }
+            }
+        }
+
+        return dp.contains(s.length() - 1);
+    }
+
+    class Trie {
+        private boolean end;
+        private final Map<Character, Trie> children = new HashMap<>();
+
+        public void insert(String word) {
+            Trie node = this;
+            for (Character c : word.toCharArray()) {
+                node = node.children.computeIfAbsent(c, k -> new Trie());
+            }
+            node.end = true;
+        }
+
+        public boolean search(String word) {
+            Trie node = this;
+            for (Character c : word.toCharArray()) {
+                node = node.children.get(c);
+                if (node == null) {
+                    return false;
+                }
+            }
+            return node.end;
+        }
+    }
+}