# 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```
```

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