Implement Trie (Prefix Tree)
Problem Statement :
A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.
Implement the Trie class:
Trie() Initializes the trie object.
void insert(String word) Inserts the string word into the trie.
boolean search(String word) Returns true if the string word is in the trie (i.e., was inserted before), and false otherwise.
boolean startsWith(String prefix) Returns true if there is a previously inserted string word that has the prefix prefix, and false otherwise.
Example 1:
Input
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
Output
[null, null, true, false, true, null, true]
Explanation
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // return True
trie.search("app"); // return False
trie.startsWith("app"); // return True
trie.insert("app");
trie.search("app"); // return True
Constraints:
1 <= word.length, prefix.length <= 2000
word and prefix consist only of lowercase English letters.
At most 3 * 104 calls in total will be made to insert, search, and startsWith.
Solution :
Solution in C :
#define N 26
typedef struct Trie{
struct Trie* children[N];
bool is_leaf;
}Trie;
Trie* trieCreate() {
Trie* node = malloc(sizeof(Trie));
for(int i = 0; i < N; i++){
node->children[i] = NULL;
}
node->is_leaf = false;
return node;
}
void trieInsert(Trie* obj, char * word) {
Trie* tmp = obj;
for(int i = 0; word[i] != '\0'; i++){
int idx = word[i] - 'a';
if(tmp->children[idx] == NULL){
tmp->children[idx] = trieCreate();
}
tmp = tmp->children[idx];
}
tmp->is_leaf = true;
}
bool trieSearch(Trie* obj, char * word) {
Trie* tmp = obj;
for(int i = 0; word[i] != '\0'; i++){
int idx = word[i] - 'a';
if(tmp->children[idx] == NULL)
return false;
tmp = tmp->children[idx];
}
return tmp->is_leaf ;
}
bool trieStartsWith(Trie* obj, char * prefix) {
Trie* tmp = obj;
for(int i = 0; prefix[i] != '\0'; i++){
int idx = prefix[i] - 'a';
if(tmp->children[idx] == NULL)
return false;
tmp = tmp->children[idx];
}
return true ;
}
void trieFree(Trie* obj) {
Trie* tmp = obj;
for(int i = 0; i < N; i++){
if(tmp->children[i] != NULL)
trieFree(tmp->children[i]);
}
free(obj);
}
Solution in C++ :
class TrieNode {
public:
TrieNode *child[26];
bool isWord;
TrieNode() {
isWord = false;
for (auto &a : child) a = nullptr;
}
};
class Trie {
TrieNode* root;
public:
Trie() {
root = new TrieNode();
}
void insert(string s) {
TrieNode *p = root;
for (auto &a : s) {
int i = a - 'a';
if (!p->child[i]) p->child[i] = new TrieNode();
p = p->child[i];
}
p->isWord = true;
}
bool search(string key, bool prefix=false) {
TrieNode *p = root;
for (auto &a : key) {
int i = a - 'a';
if (!p->child[i]) return false;
p = p->child[i];
}
if (prefix==false) return p->isWord;
return true;
}
bool startsWith(string prefix) {
return search(prefix, true);
}
};
Solution in Java :
class Trie {
Node root;
public Trie() {
root = new Node();
}
public void insert(String word) {
root.insert(word, 0);
}
public boolean search(String word) {
return root.search(word, 0);
}
public boolean startsWith(String prefix) {
return root.startsWith(prefix, 0);
}
class Node {
Node[] nodes;
boolean isEnd;
Node() {
nodes = new Node[26];
}
private void insert(String word, int idx) {
if (idx >= word.length()) return;
int i = word.charAt(idx) - 'a';
if (nodes[i] == null) {
nodes[i] = new Node();
}
if (idx == word.length()-1) nodes[i].isEnd = true;
nodes[i].insert(word, idx+1);
}
private boolean search(String word, int idx) {
if (idx >= word.length()) return false;
Node node = nodes[word.charAt(idx) - 'a'];
if (node == null) return false;
if (idx == word.length() - 1 && node.isEnd) return true;
return node.search(word, idx+1);
}
private boolean startsWith(String prefix, int idx) {
if (idx >= prefix.length()) return false;
Node node = nodes[prefix.charAt(idx) - 'a'];
if (node == null) return false;
if (idx == prefix.length() - 1) return true;
return node.startsWith(prefix, idx+1);
}
}
}
Solution in Python :
class Trie:
def __init__(self):
self.root={}
def insert(self, word: str) -> None:
cur=self.root
for letter in word:
if letter not in cur:
cur[letter]={}
cur=cur[letter]
cur['*']=''
def search(self, word: str) -> bool:
cur=self.root
for letter in word:
if letter not in cur:
return False
cur=cur[letter]
return '*' in cur
def startsWith(self, prefix: str) -> bool:
cur=self.root
for letter in prefix:
if letter not in cur:
return False
cur=cur[letter]
return True
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