# Trie - Amazon Top Interview Questions

### Problem Statement :

```Implement a trie data structure with the following methods:

Trie() constructs a new instance of a trie
exists(String word) returns whether word exists in the trie
startswith(String p) returns whether there's some word whose prefix is p
Constraints

n ≤ 100,000 where n is the number of calls to add, exists and startswith

Example 1

Input

arguments = [[], ["dog"], ["document"], ["dog"], ["doc"], ["doge"]]`

Output

[None, None, None, True, True, False]

Explanation

We create a Trie
We add the word "dog" to the trie
We add the word "document" to the trie
We check whether "dog" exists in the trie which it does
We check whether there is some word whose prefix is "doc" which there is ("document)
We check whether "doge" exists in the trie which it does not```

### Solution :

```                        ```Solution in C++ :

class Trie {  // Time: O(L) for every operation, where L is the length of the input string
private:
bool is_word;
Trie* children[26];

public:
Trie() {  // Constructor
is_word = false;
for (int i = 0; i < 26; i++) children[i] = nullptr;
}

~Trie() {  // Destructor
for (int i = 0; i < 26; i++)
if (children[i]) delete children[i];  // Release children nodes
}

Trie* search(string& word) {
Trie* node = this;
for (char c : word) {
if (node->children[c - 'a'] == nullptr) return nullptr;

node = node->children[c - 'a'];
}

return node;
}

Trie* node = this;
for (char c : s) {
if (node->children[c - 'a'] == nullptr) node->children[c - 'a'] = new Trie();

node = node->children[c - 'a'];
}

node->is_word = true;
}

bool exists(string word) {
Trie* node = search(word);
return (node != nullptr) && (node->is_word);
}

bool startswith(string p) {
Trie* node = search(p);
return node != nullptr;
}
};```
```

```                        ```Solution in Python :

class Node:
def __init__(self):
self.children = defaultdict(Node)
self.end = False

class Trie:
def __init__(self):
self.root = Node()

"""We simply iterate through the word and add each character
one by one. since we are using a defaultdict, we don't need
to worry about creating a new node if a child doesn't exist

In the end we set the `end` flag of the current node to true
"""
cur = self.root
for c in s:
cur = cur.children[c]
cur.end = True

def exists(self, word):
"""
We go by each character and see if the current node has a child
mapped to that character. if it doesn't, return false. else keep
going down. In the end check if the node's end flag is set to true
"""
cur = self.root
for c in word:
if c not in cur.children:
return False
cur = cur.children[c]
return cur.end

def startswith(self, p):
"""
We go by each character and see if the current node has a child
mapped to that character. if it doesn't, return false. else keep
going down. we just return true without checking the end flag
"""
cur = self.root
for c in p:
if c not in cur.children:
return False
cur = cur.children[c]
return True```
```

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty