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‎.dev/logs.json

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}
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"design_add_and_search_words_data_structure.go": "f035897eeefd8eff6ecae6c7ee62b41ad404b70eef8d62ce0993eee71786bee6"
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}
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}
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‎README.md

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[**_Rejection Sampling_**][rejection sampling]  • 
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[**_Biconnected Component_**][biconnected component]
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### **Total Problems Solved: _177_**
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### **Total Problems Solved: _185_**
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---
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‎TOPICWISE.md

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‎docs/index.md

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[**_Rejection Sampling_**][rejection sampling]  • 
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[**_Biconnected Component_**][biconnected component]
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### **Total Problems Solved: _177_**
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### **Total Problems Solved: _185_**
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## Contributors
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‎src/0101-0200/106 - Construct Binary Tree from Inorder and Postorder Traversal/README.md

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# 106. Construct Binary Tree from Inorder and Postorder Traversal [![share]](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)
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![][medium]
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## Problem Statement
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<p>Given two integer arrays <code>inorder</code> and <code>postorder</code> where <code>inorder</code> is the inorder traversal of a binary tree and <code>postorder</code> is the postorder traversal of the same tree, construct and return <em>the binary tree</em>.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree.jpg" style="width: 277px; height: 302px;"/>
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```
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Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
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Output: [3,9,20,null,null,15,7]
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```
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<p><strong class="example">Example 2:</strong></p>
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```
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Input: inorder = [-1], postorder = [-1]
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Output: [-1]
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```
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= inorder.length &lt;= 3000</code></li>
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<li><code>postorder.length == inorder.length</code></li>
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<li><code>-3000 &lt;= inorder[i], postorder[i] &lt;= 3000</code></li>
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<li><code>inorder</code> and <code>postorder</code> consist of <strong>unique</strong> values.</li>
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<li>Each value of <code>postorder</code> also appears in <code>inorder</code>.</li>
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<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
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<li><code>postorder</code> is <strong>guaranteed</strong> to be the postorder traversal of the tree.</li>
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</ul>
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## Solutions
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### [_Go_](construct_binary_tree_from_inorder_and_postorder_traversal.go)
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```go [Go]
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package main
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// Definition for a binary tree node.
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type TreeNode struct {
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Val int
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Left *TreeNode
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Right *TreeNode
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}
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func buildTree(inorder []int, postorder []int) *TreeNode {
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mapInOrder := make(map[int]int)
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for i, v := range inorder {
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mapInOrder[v] = i
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}
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var helper func(int, int) *TreeNode
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helper = func(left, right int) *TreeNode {
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if left > right {
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return nil
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}
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pop := postorder[len(postorder)-1]
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postorder = postorder[:len(postorder)-1]
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root := &TreeNode{Val: pop}
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mid := mapInOrder[pop]
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root.Right = helper(mid+1, right)
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root.Left = helper(left, mid-1)
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return root
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}
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return helper(0, len(inorder)-1)
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}
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```
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### [_..._]()
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```
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```
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<!----------------------------------{ link }--------------------------------->
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[share]: https://graph.org/file/3ea5234dda646b71c574a.png
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[easy]: https://img.shields.io/badge/Difficulty-Easy-bright.svg
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[medium]: https://img.shields.io/badge/Difficulty-Medium-yellow.svg
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[hard]: https://img.shields.io/badge/Difficulty-Hard-red.svg

‎src/0201-0300/211 - Design Add and Search Words Data Structure/README.md

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# 211. Design Add and Search Words Data Structure [![share]](https://leetcode.com/problems/design-add-and-search-words-data-structure/)
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![][medium]
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## Problem Statement
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<p>Design a data structure that supports adding new words and finding if a string matches any previously added string.</p>
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<p>Implement the <code>WordDictionary</code> class:</p>
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<ul>
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<li><code>WordDictionary()</code> Initializes the object.</li>
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<li><code>void addWord(word)</code> Adds <code>word</code> to the data structure, it can be matched later.</li>
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<li><code>bool search(word)</code> Returns <code>true</code> if there is any string in the data structure that matches <code>word</code> or <code>false</code> otherwise. <code>word</code> may contain dots <code>'.'</code> where dots can be matched with any letter.</li>
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</ul>
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<p> </p>
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<p><strong class="example">Example:</strong></p>
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```
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Input
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["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
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[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
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Output
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[null,null,null,null,false,true,true,true]
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Explanation
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WordDictionary wordDictionary = new WordDictionary();
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wordDictionary.addWord("bad");
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wordDictionary.addWord("dad");
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wordDictionary.addWord("mad");
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wordDictionary.search("pad"); // return False
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wordDictionary.search("bad"); // return True
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wordDictionary.search(".ad"); // return True
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wordDictionary.search("b.."); // return True
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```
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= word.length &lt;= 25</code></li>
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<li><code>word</code> in <code>addWord</code> consists of lowercase English letters.</li>
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<li><code>word</code> in <code>search</code> consist of <code>'.'</code> or lowercase English letters.</li>
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<li>There will be at most <code>2</code> dots in <code>word</code> for <code>search</code> queries.</li>
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<li>At most <code>10<sup>4</sup></code> calls will be made to <code>addWord</code> and <code>search</code>.</li>
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</ul>
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<details>
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<summary>
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#### _Click to open Hints_
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</summary>
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- You should be familiar with how a Trie works. If not, please work on this problem: <a href="https://leetcode.com/problems/implement-trie-prefix-tree/">Implement Trie (Prefix Tree)</a> first.
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</details>
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## Solutions
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### [_Go_](design_add_and_search_words_data_structure.go)
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```go [Go]
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package main
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type WordDictionary struct {
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children [26]* WordDictionary
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isWord bool
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}
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func Constructor() WordDictionary {
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return WordDictionary{}
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}
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func (this *WordDictionary) AddWord(word string) {
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root := this
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for _,item := range word{
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char := item - 'a'
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if root.children[char] == nil {
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root.children[char] = &WordDictionary{}
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}
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root = root.children[char]
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}
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root.isWord = true
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}
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func (this *WordDictionary) Search(word string) bool {
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root := this
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for i,item := range word{
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if word[i] == '.'{
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for _,item := range root.children{
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if item != nil && item.Search(word[i+1:]) {
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return true
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}
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}
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return false
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}
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char := item - 'a'
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if root.children[char] == nil{
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return false
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}
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root = root.children[char]
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}
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return root.isWord
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}
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```
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### [_..._]()
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```
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```
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<!----------------------------------{ link }--------------------------------->
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[share]: https://graph.org/file/3ea5234dda646b71c574a.png
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[easy]: https://img.shields.io/badge/Difficulty-Easy-bright.svg
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[medium]: https://img.shields.io/badge/Difficulty-Medium-yellow.svg
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[hard]: https://img.shields.io/badge/Difficulty-Hard-red.svg

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