Trie

Author: awangdev

Java/Implement Trie (Prefix Tree).java

@@ -53,11 +53,10 @@
 TrieNode: contains the single character, and a list of children.
 Note: we will use hashmap<child character, child TrieNode>, because child access is O(1)
 */
-
 class Trie {
     class TrieNode {
         public boolean isEnd;
-        public Map<Character, TrieNode> children;
+        public Map<Character, TrieNode> children; // Map is more applicable to all chars, not limited to 256 ASCII
         public TrieNode() {
             this.isEnd = false;
             this.children = new HashMap<>();
@@ -72,33 +71,22 @@ public Trie() {
 
     /** Inserts a word into the trie. */
     public void insert(String word) {
-        if (word == null || word.length() == 0) {
-            return;
-        }
-        if (search(word)) {
-            return;
-        }
+        if (word == null || word.length() == 0 || search(word)) return;
+
         TrieNode node = root;
-        for (int i = 0; i < word.length(); i++) {
-            char c = word.charAt(i);
-            if (!node.children.containsKey(c)) {
-                node.children.put(c, new TrieNode());
-            }
+        for (char c : word.toCharArray()) {
+            node.children.putIfAbsent(c, new TrieNode());
             node = node.children.get(c);
-            if (i == word.length() - 1) {
-                node.isEnd = true;
-            }
         }
+        node.isEnd = true; // can set word to node as well, if needed
     }
 
-    /** Returns if the word is in the trie. */
+    /** Returns if the word is in the trie: correct path + isEnd */
     public boolean search(String word) {
-        if (word == null || word.length() == 0) {
-            return false;
-        }
+        if (word == null || word.length() == 0) return false;
+        
         TrieNode node = root;
-        for (int i = 0; i < word.length(); i++) {
-            char c = word.charAt(i);
+        for (char c : word.toCharArray()) {
             if (!node.children.containsKey(c)) {
                 return false;
             }
@@ -109,12 +97,10 @@ public boolean search(String word) {
 
     /** Returns if there is any word in the trie that starts with the given prefix. */
     public boolean startsWith(String prefix) {
-        if (prefix == null || prefix.length() == 0) {
-            return false;
-        }
+        if (prefix == null || prefix.length() == 0) return false;
+        
         TrieNode node = root;
-        for (int i = 0; i < prefix.length(); i++) {
-            char c = prefix.charAt(i);
+        for (char c : prefix.toCharArray()) {
             if (!node.children.containsKey(c)) {
                 return false;
             }

Java/Read N Characters Given Read4.java

@@ -0,0 +1,59 @@
+E
+1533408053
+tags: String
+
+Read4 题目. 理解题目: 是有个input object buff, 会被populated with data.
+
+#### String in char[] format
+- 理解题目: 其实就是track `可以read多少bytes by read4() response`
+- 另外一个有用的function `System.arraycopy(src, srcIndex, dest, destIndex, length)`
+
+```
+/*
+The API: int read4(char *buf) reads 4 characters at a time from a file.
+
+The return value is the actual number of characters read. For example, 
+it returns 3 if there is only 3 characters left in the file.
+
+By using the read4 API, implement the function int read(char *buf, int n) 
+that reads n characters from the file.
+
+Example 1:
+
+Input: buf = "abc", n = 4
+Output: "abc"
+Explanation: The actual number of characters read is 3, which is "abc".
+Example 2:
+
+Input: buf = "abcde", n = 5 
+Output: "abcde"
+Note:
+The read function will only be called once for each test case.
+*/
+
+/*
+- use temp to store all items read, and System.copy into buf.
+- track overall progress: i 
+- if read4 returns < 4, mark end and return.
+*/
+public class Solution extends Reader4 {
+    /**
+     * @param buf Destination buffer
+     * @param n   Maximum number of characters to read
+     * @return    The number of characters read
+     */
+    public int read(char[] buf, int n) {
+        if (buf == null || n <= 0) return 0;
+        int i = 0;
+        while (i < n) {
+            char[] temp = new char[4];
+            int count = read4(temp);
+            int range =  i + 3 < n ? count : Math.min(n - i, count);
+            System.arraycopy(temp, 0, buf, i, range);
+            i += range;
+            if (count < 4) break;
+        }
+        return i;
+    }
+}
+```

Java/Two Sum IV - Input is a BST.java

@@ -0,0 +1,50 @@
+E
+1533407180
+tags: Tree
+
+HashSet to store visited items. Same old 2 sum trick.
+
+```
+/*
+Given a Binary Search Tree and a target number, 
+return true if there exist two elements in the BST such that their sum is equal to the given target.
+
+Example 1:
+Input: 
+    5
+   / \
+  3   6
+ / \   \
+2   4   7
+
+Target = 9
+
+Output: True
+Example 2:
+Input: 
+    5
+   / \
+  3   6
+ / \   \
+2   4   7
+
+Target = 28
+
+Output: False
+*/
+
+// hashmap, in-order traverse for smaller items, recursively
+class Solution {
+    Set<Integer> set = new HashSet<>();
+    
+    public boolean findTarget(TreeNode root, int k) {
+        if (root == null) return false;
+        if (findTarget(root.left, k)) return true;
+        if (set.contains(k - root.val)) return true;
+        set.add(root.val);
+        if (findTarget(root.right, k)) return true;
+        
+        return false;
+    }
+}
+```

LevelReviewPage.md

@@ -1,7 +1,7 @@
 
 
 
-## Easy (146)
+## Easy (148)
 **0. [Group Shifted Strings.java](https://github.com/awangdev/LintCode/blob/master/Java/Group%20Shifted%20Strings.java)**      Level: Easy      Tags: []
 
 相同shift规则的string, 能被推算到同一个零起始点，就是共同减去一个char,最后就相等。以此作为key，用HashMap。一目了然。
@@ -2271,6 +2271,28 @@ space: O(1)
 
 ---
 
+**146. [Two Sum IV - Input is a BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Two%20Sum%20IV%20-%20Input%20is%20a%20BST.java)**      Level: Easy      Tags: [Tree]
+      
+
+HashSet to store visited items. Same old 2 sum trick.
+
+
+
+---
+
+**147. [Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)**      Level: Easy      Tags: [String]
+      
+
+Read4 题目. 理解题目: 是有个input object buff, 会被populated with data.
+
+#### String in char[] format
+- 理解题目: 其实就是track `可以read多少bytes by read4() response`
+- 另外一个有用的function `System.arraycopy(src, srcIndex, dest, destIndex, length)`
+
+
+
+---
+
 
 
 

README.md

@@ -485,3 +485,5 @@
 |464|[Sliding Window Median.java](https://github.com/awangdev/LintCode/blob/master/Java/Sliding%20Window%20Median.java)|Hard|Java|[Design, Heap, MaxHeap, MinHeap]||
 |465|[Inorder Successor in BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Inorder%20Successor%20in%20BST.java)|Medium|Java|[BST, Tree]||
 |466|[Subtree of Another Tree.java](https://github.com/awangdev/LintCode/blob/master/Java/Subtree%20of%20Another%20Tree.java)|Easy|Java|[DFS, Divide and Conquer, Tree]||
+|467|[Two Sum IV - Input is a BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Two%20Sum%20IV%20-%20Input%20is%20a%20BST.java)|Easy|Java|[Tree]||
+|468|[Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)|Easy|Java|[String]||

ReviewPage.md

@@ -9229,5 +9229,27 @@ Data Stream Median 的同理题目: 不只是不断增加的Sequence, 而且要r
 
 
 
+---
+
+**467. [Two Sum IV - Input is a BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Two%20Sum%20IV%20-%20Input%20is%20a%20BST.java)**      Level: Easy      Tags: [Tree]
+      
+
+HashSet to store visited items. Same old 2 sum trick.
+
+
+
+---
+
+**468. [Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)**      Level: Easy      Tags: [String]
+      
+
+Read4 题目. 理解题目: 是有个input object buff, 会被populated with data.
+
+#### String in char[] format
+- 理解题目: 其实就是track `可以read多少bytes by read4() response`
+- 另外一个有用的function `System.arraycopy(src, srcIndex, dest, destIndex, length)`
+
+
+
 ---
 

TagREADME.md

@@ -6,8 +6,8 @@ Table of Contents
 * [DFS (85)](#dfs-85)
 * [DP (82)](#dp-82)
 * [Hash Table (57)](#hash-table-57)
-* [Tree (51)](#tree-51)
-* [String (50)](#string-50)
+* [Tree (52)](#tree-52)
+* [String (51)](#string-51)
 * [Binary Search (39)](#binary-search-39)
 * [Math (34)](#math-34)
 * [Divide and Conquer (34)](#divide-and-conquer-34)
@@ -445,7 +445,7 @@ Table of Contents
 
 
 
-## Tree (51)
+## Tree (52)
 | Squence | Problem       | Level  | Language  | Tags | Video Tutorial|
 |:-------:|:--------------|:------:|:---------:|:----:|:-------------:|
 |0|[Construct Binary Tree from Inorder and Preorder Traversal.java](https://github.com/awangdev/LintCode/blob/master/Java/Construct%20Binary%20Tree%20from%20Inorder%20and%20Preorder%20Traversal.java)|Medium|Java|[Array, DFS, Divide and Conquer, Hash Table, Tree]||
@@ -499,13 +499,14 @@ Table of Contents
 |48|[Convert Binary Search Tree to Sorted Doubly Linked List (extra space).java](https://github.com/awangdev/LintCode/blob/master/Java/Convert%20Binary%20Search%20Tree%20to%20Sorted%20Doubly%20Linked%20List%20(extra%20space).java)|Medium|Java|[Linked List, Stack, Tree]||
 |49|[Maximum Binary Tree.java](https://github.com/awangdev/LintCode/blob/master/Java/Maximum%20Binary%20Tree.java)|Medium|Java|[Stack, Tree]||
 |50|[Binary Tree Postorder Traversal.java](https://github.com/awangdev/LintCode/blob/master/Java/Binary%20Tree%20Postorder%20Traversal.java)|Medium|Java|[Stack, Tree, Two Stacks]||
+|51|[Two Sum IV - Input is a BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Two%20Sum%20IV%20-%20Input%20is%20a%20BST.java)|Easy|Java|[Tree]||
 
 
 
 
 
 
-## String (50)
+## String (51)
 | Squence | Problem       | Level  | Language  | Tags | Video Tutorial|
 |:-------:|:--------------|:------:|:---------:|:----:|:-------------:|
 |0|[Unique Characters.java](https://github.com/awangdev/LintCode/blob/master/Java/Unique%20Characters.java)|Easy|Java|[Array, String]||
@@ -555,9 +556,10 @@ Table of Contents
 |44|[Reverse Words in a String.java](https://github.com/awangdev/LintCode/blob/master/Java/Reverse%20Words%20in%20a%20String.java)|Medium|Java|[String]||
 |45|[Reverse Words in a String II.java](https://github.com/awangdev/LintCode/blob/master/Java/Reverse%20Words%20in%20a%20String%20II.java)|Medium|Java|[String]||
 |46|[Reverse Words in a String III.java](https://github.com/awangdev/LintCode/blob/master/Java/Reverse%20Words%20in%20a%20String%20III.java)|Easy|Java|[String]||
-|47|[Reverse Vowels of a String.java](https://github.com/awangdev/LintCode/blob/master/Java/Reverse%20Vowels%20of%20a%20String.java)|Easy|Java|[String, Two Pointers]||
-|48|[Implement strStr().java](https://github.com/awangdev/LintCode/blob/master/Java/Implement%20strStr().java)|Easy|Java|[String, Two Pointers]||
-|49|[Valid Palindrome.java](https://github.com/awangdev/LintCode/blob/master/Java/Valid%20Palindrome.java)|Easy|Java|[String, Two Pointers]||
+|47|[Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)|Easy|Java|[String]||
+|48|[Reverse Vowels of a String.java](https://github.com/awangdev/LintCode/blob/master/Java/Reverse%20Vowels%20of%20a%20String.java)|Easy|Java|[String, Two Pointers]||
+|49|[Implement strStr().java](https://github.com/awangdev/LintCode/blob/master/Java/Implement%20strStr().java)|Easy|Java|[String, Two Pointers]||
+|50|[Valid Palindrome.java](https://github.com/awangdev/LintCode/blob/master/Java/Valid%20Palindrome.java)|Easy|Java|[String, Two Pointers]||
 
 
 

TagReviewPage.md

@@ -331,7 +331,7 @@ SegmentTree大集合. Methods: `build, query, modify`. 不难。只是要都记
 
 
 
-## String (50)
+## String (51)
 **0. [Space Replacement.java](https://github.com/awangdev/LintCode/blob/master/Java/Space%20Replacement.java)**      Level: Medium      Tags: [String]
 
 
@@ -1227,6 +1227,19 @@ time: O(n)
 
 ---
 
+**50. [Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)**      Level: Easy      Tags: [String]
+      
+
+Read4 题目. 理解题目: 是有个input object buff, 会被populated with data.
+
+#### String in char[] format
+- 理解题目: 其实就是track `可以read多少bytes by read4() response`
+- 另外一个有用的function `System.arraycopy(src, srcIndex, dest, destIndex, length)`
+
+
+
+---
+
 
 
 
@@ -10218,7 +10231,7 @@ space: O(n)
 
 
 
-## Tree (51)
+## Tree (52)
 **0. [Unique Binary Search Tree.java](https://github.com/awangdev/LintCode/blob/master/Java/Unique%20Binary%20Search%20Tree.java)**      Level: Medium      Tags: [BST, DP, Tree]
 
 
@@ -11288,6 +11301,15 @@ space: O(1)
 
 ---
 
+**51. [Two Sum IV - Input is a BST.java](https://github.com/awangdev/LintCode/blob/master/Java/Two%20Sum%20IV%20-%20Input%20is%20a%20BST.java)**      Level: Easy      Tags: [Tree]
+      
+
+HashSet to store visited items. Same old 2 sum trick.
+
+
+
+---
+
 
 
 

review/String.md

@@ -1,7 +1,7 @@
 
 
 
-## String (50)
+## String (51)
 **0. [Space Replacement.java](https://github.com/awangdev/LintCode/blob/master/Java/Space%20Replacement.java)**      Level: Medium      Tags: [String]
 
 
@@ -895,5 +895,18 @@ time: O(n)
 
 
 
+---
+
+**50. [Read N Characters Given Read4.java](https://github.com/awangdev/LintCode/blob/master/Java/Read%20N%20Characters%20Given%20Read4.java)**      Level: Easy      Tags: [String]
+      
+
+Read4 题目. 理解题目: 是有个input object buff, 会被populated with data.
+
+#### String in char[] format
+- 理解题目: 其实就是track `可以read多少bytes by read4() response`
+- 另外一个有用的function `System.arraycopy(src, srcIndex, dest, destIndex, length)`
+
+
+
 ---