**Trie - Amazon Top Interview Questions**

### Problem Statement :

Implement a trie data structure with the following methods: Trie() constructs a new instance of a trie add(String word) adds a word into the 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 methods = ["constructor", "add", "add", "exists", "startswith", "exists"] 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;
}
void add(string s) {
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()
def add(self, s):
"""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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