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 :



title-img




                        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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