0208. Implement Trie (Prefix Tree)
Medium | Design + String | 112 ms (99.40%), 29.6 MB (93.89%)
Source: LeetCode - Implement Trie (Prefix Tree) GitHub: Solution / Performance
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 stringwordinto the trie.boolean search(String word)Returnstrueif the stringwordis in the trie (i.e., was inserted before), andfalseotherwise.boolean startsWith(String prefix)Returnstrueif there is a previously inserted stringwordthat has the prefixprefix, andfalseotherwise.
Constraints:
1 <= word.length, prefix.length <= 2000wordandprefixconsist only of lowercase English letters.At most
3 * 104calls in total will be made toinsert,search, andstartsWith.
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 TrueInsert
Iterate each char in the word and insert each of them into the hash table. Note that the symbol that represents the end of the word also needs to be added. (Here I use #)
Search
To search for a certain word, we iterate each char in the word and check the existence of each of them. Besides, if the search operation is done for each char, check whether the symbol # exists in the current node or not.
StartWith
Almost the same as the search operation. Except that we don't need to check the symbol # since the input string is only a prefix.
class Trie:
def __init__(self):
self.root = {}
def insert(self, word: str) -> None:
# time : O(n), n is the lenght of str
# space : O(n)
cur = self.root
for char in word:
if char not in cur: cur[char] = {}
cur = cur[char]
cur['#'] = True
def search(self, word: str) -> bool:
# time : O(n)
# space : O(1)
cur = self.root
for char in word:
if char not in cur: return False
cur = cur[char]
return '#' in cur
def startsWith(self, prefix: str) -> bool:
# time : O(n)
# space : O(1)
cur = self.root
for char in prefix:
if char not in cur: return False
cur = cur[char]
return True
class Trie {
Trie[] arr;
boolean wordEnd;
/* Initialize your data structure here */
public Trie() {
arr = new Trie[26];
}
/* Inserts a word into the trie */
public void insert(String word) {
insert( word, 0 );
}
private void insert(String word, int i) {
if( i == word.length() ) {
wordEnd = true;
return;
}
int index = word.charAt(i) - 'a';
if( arr[index] == null ) arr[index] = new Trie();
arr[index].insert(word, i + 1);
}
/* Returns if the word is in the trie */
public boolean search(String word) {
return search( word, 0 );
}
private boolean search(String word, int i) {
if( i == word.length() ) return wordEnd;
int index = word.charAt(i) - 'a';
if( arr[index] == null ) return false;
else return arr[index].search( word, i + 1 );
}
/* Returns if there is any word in the trie that starts with the given prefix */
public boolean startsWith(String prefix) {
return startsWith( prefix, 0 );
}
public boolean startsWith(String prefix, int i) {
if( i == prefix.length() ) return true;
int index = prefix.charAt(i) - 'a';
if( arr[index] == null ) return false;
else return arr[index].startsWith( prefix, i + 1 );
}
}Last updated
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