[LeetCode] 208. Implement Trie (Prefix Tree)

Implement a trie with insert, search, and startsWith methods.
Note:
You may assume that all inputs are consist of lowercase letters a-z.

Thought process:
Insert: create a new node and add it to the map if current node does not contain current character.
Search: return false if current node does not contain current character, or if the node at the end of searching is not a word.
Startswith: return false if current node does not contain current character.

Solution 1 (Java):

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
class Trie {
    class TrieNode {
        TrieNode[] children;
        boolean isWord;
        
        TrieNode() {
            children = new TrieNode[26];
        }
    }
    
    private TrieNode root;

    /** Initialize your data structure here. */
    public Trie() {
        root = new TrieNode();
    }
    
    /** Inserts a word into the trie. */
    public void insert(String word) {
        TrieNode current = root;
        for (char c : word.toCharArray()) {
            int index = c - 'a';
            if (current.children[index] == null) {
                current.children[index] = new TrieNode();
            }
            current = current.children[index];
        }
        current.isWord = true;
    }
    
    /** Returns if the word is in the trie. */
    public boolean search(String word) {
        TrieNode current = root;
        for (char c : word.toCharArray()) {
            int index = c - 'a';
            if (current.children[index] == null) {
                return false;
            }
            current = current.children[index];
        }
        return current.isWord;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    public boolean startsWith(String prefix) {
        TrieNode current = root;
        for (char c : prefix.toCharArray()) {
            int index = c - 'a';
            if (current.children[index] == null) {
                return false;
            }
            current = current.children[index];
        }
        return true;
    }
}

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * boolean param_2 = obj.search(word);
 * boolean param_3 = obj.startsWith(prefix);
 */

Given all inputs consist of lowercase letters, we can simplify the hashmap in each trie node to an array of size 26. All three methods require iterating through the input word.

Solution 2 (C++):
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
class TrieNode {
public:
    TrieNode* children[26];
    bool isWord;
    
    TrieNode() {
        memset(children, 0, sizeof(children));
        isWord = false;
    }
};

class Trie {
private:
    TrieNode* root;

public:
    /** Initialize your data structure here. */
    Trie() {
        root = new TrieNode();
    }
    
    /** Inserts a word into the trie. */
    void insert(string word) {
        TrieNode* current = root;
        
        for (int i = 0; i < word.length(); i++) {
            int index = word[i] - 'a';
            if (current->children[index] == NULL) {
                current->children[index] = new TrieNode();
            }
            current = current->children[index];
        }
        current->isWord = true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(string word) {
        TrieNode* current = root;
        
        for (int i = 0; i < word.length(); i++) {
            int index = word[i] - 'a';
            if (current->children[index] == NULL) {
                return false;
            }
            current = current->children[index];
        }
        
        return current->isWord;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string prefix) {
        TrieNode* current = root;
        
        for (int i = 0; i < prefix.length(); i++) {
            int index = prefix[i] - 'a';
            if (current->children[index] == NULL) {
                return false;
            }
            current = current->children[index];
        }
        
        return true;
    }
};

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * bool param_2 = obj.search(word);
 * bool param_3 = obj.startsWith(prefix);
 */

Time complexity:
All three methods takes O(n), where n is the length of the input string.
Location: Philadelphia, PA, USA

Comments

Post a Comment

Popular posts from this blog

[LeetCode] 269. Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of  non-empty  words from the dictionary, where  words are sorted lexicographically by the rules of this new language . Derive the order of letters in this language. Example 1: Given the following words in dictionary, [ "wrt", "wrf", "er", "ett", "rftt" ] The correct order is:  "wertf" . Example 2: Given the following words in dictionary, [ "z", "x" ] The correct order is:  "zx" . Example 3: Given the following words in dictionary, [ "z", "x", "z" ] The order is invalid, so return  "" . Note: You may assume all letters are in lowercase. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary. If the order is invalid, return an empty string
Read more

[LeetCode] 631. Design Excel Sum Formula

Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions: Excel(int H, char W):  This is the constructor. The inputs represents the height and width of the Excel form.  H  is a positive integer, range from 1 to 26. It represents the height.  W  is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to  W . The Excel form content is represented by a height * width 2D integer array  C , it should be initialized to zero. You should assume that the first row of  C  starts from 1, and the first column of  C  starts from 'A'. void Set(int row, char column, int val):  Change the value at  C(row, column)  to be val. int Get(int row, char column):  Return the value at  C(row, column) . int Sum(int row, char column, List of Strings : numbers):  This function calculate and set the value at  C(row, column) , wh
Read more

[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee

Your are given an array of integers  prices , for which the  i -th element is the price of a given stock on day  i ; and a non-negative integer  fee  representing a transaction fee. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.) Return the maximum profit you can make. Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices[0] = 1 Selling at prices[3] = 8 Buying at prices[4] = 4 Selling at prices[5] = 9 The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8. Note: 0 < prices.length <= 50000 . 0 < prices[i] < 50000 . 0 <= fee < 50000 . Thought process: Each day there are two options, buy or sell. Use two variables buy and sell to keep track of the max profit. Solution: 1 2 3 4 5 6
Read more