[Leetcode] 211 & 208 Trie Tree

[Leetcode] 211 & 208 Trie Tree

Trie

In computer science, a trie, also called digital tree and sometimes radix tree or prefix tree (on the ground that they can be searched by prefixes), is a kind of search tree. Unlike binary tree, no node in the tree stores the key associated with that node; instead, its position in the tree defines the key with which it is associated. All the descendants of a node have a common prefix of the string associated with that node, and the root is associated with the empty string. Values are not necessarily associated with every node. Rather, values tend only to be associated with leaves, and with some inner nodes that correspond to keys of interest.

Problem description

211. Add and Search Word - Data structure design
208. Implement Trie (Prefix Tree)

As the above two question shown, it won’t be easy to solve it by brutal force because it will spend too much time. However, Trie is a effective data structure which can be used to solve such questions. Just think about that, I want to find a word if it has been inserted into the tree, then if I can find that there isn’t any word with the same prefix as the word, which I want to find, in the tree, then I can definitely make a conclusion the word doesn’t be inserted.

To construct Trie is easy and simple, just like the below figure shown. However, we have to accept that the trie is not a perfect solution as it sacrifices space to save time. For any node, it has to store (26 * 4 + 1) Bytes. Compared with the heavily time cost, it will be OK.

Q:208

This question is about implementing the function of Trie tree.
The data struct of tree node is important. As the below code shown, the node store a boolean value which represents if it is an end of word and an array of pointer of node which represents the successors.

struct TreeNodes {
    bool isWord;
    struct TreeNodes *next[MAX];
    TreeNodes() { 
        isWord = false;
        for (int i = 0; i < MAX; i++) {
            next[i] = nullptr;
        }
    }
};

The whole code:

#include <iostream>
#define MAX 26
using namespace std;

struct TreeNodes {
    bool isWord;
    struct TreeNodes *next[MAX];
    TreeNodes() { 
        isWord = false;
        for (int i = 0; i < MAX; i++) {
            next[i] = nullptr;
        }
    }
};

class Trie {
private:
    TreeNodes *root;
public:
    /** Initialize your data structure here. */
    Trie() {
        this->root = new TreeNodes();
    }

    /** Inserts a word into the trie. */
    void insert(string word) {
        TreeNodes* temp = this->root;
        for (auto ch : word) {
            if (temp->next[ch - 'a'] == nullptr) {
                temp->next[ch - 'a'] = new TreeNodes();
            }
            temp = temp->next[ch - 'a'];
        }
        temp->isWord = true;
    }

    /** Returns if the word is in the trie. */
    bool search(string word) {
        TreeNodes* temp = this->root;
        for (auto ch : word) {
            if (temp->next[ch - 'a'] == nullptr) {
                return false;
            } else {
                temp = temp->next[ch - 'a'];
            }
        }
        return temp->isWord;
    }

    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string prefix) {
        TreeNodes* temp = this->root;
        for (auto ch : prefix) {
            if (temp->next[ch - 'a'] == nullptr) {
                return false;
            } else {
                temp = temp->next[ch - 'a'];
            }
        }
        return temp != nullptr;
    }
};

int main() {
    Trie tree;
    tree.insert("abc");
    cout << tree.search("abc") << endl;
    cout << tree.startsWith("ab") << endl;
    return 0;
}

The Time Complexity: O(n) (for every function)

Q:211

This question is about how to use Trie tree to implement non-definitely search pattern. As the ‘.’ can represent any character in lower case. So it is necessary to change a bit. I think it is feasible to use recursion to search. If I encounter a char ‘.’, I just ignore that char and continue to search next char.

class WordDictionary {
private:
    TreeNodes *root;
    bool search(string word, TreeNodes* temp) {
        if (word.length() == 0) {
            return temp->isWord;
        } else {
            if (word[0] == '.') {
                bool isWord = false;
                for (int i = 0; i < MAX; i++) {
                    if (temp->next[i] != nullptr) {
                        isWord |= search(word.substr(1), temp->next[i]);
                    }
                }
                return isWord;
            } else {
                if (temp->next[word[0] - 'a'] != nullptr) {
                    return search(word.substr(1), temp->next[word[0] - 'a']);
                } else {
                    return false;
                }
            }
        }
    }
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new TreeNodes();
    }

    /** Adds a word into the data structure. */
    void addWord(string word) {
        TreeNodes *temp = root;
        for (auto ch : word) {
            if (temp->next[ch - 'a'] == nullptr) {
                temp->next[ch - 'a'] = new TreeNodes();
            }
            temp = temp->next[ch - 'a'];
        }
        temp->isWord = true;
    }

    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    bool search(string word) {
        TreeNodes *temp = root;
        return search(word, root);
    }
};

Time Complexity: O(n)

Improvement

However, it doesn’t have to use Trie tree. The solution can be very easy just use map. Just the code shown, the data structure map can store the length of word as key, and the set of word with the corresponding length of word as value, Then it will be efficient to search. It will be better to use unorder_map because we don’t need to keep the ordered map.

class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary() {

    }

    /** Adds a word into the data structure. */
    void addWord(string word) {
        map[(int)word.length()].insert(word);
    }

    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    bool search(string word) {
        int len = word.length();

        for (auto& s: map[len]) {
            bool flag = true;
            for (int i=0; i<s.length(); i++) {
                if (word[i] == '.')
                    continue;
                if (word[i] != s[i]) {
                    flag = false;
                    break;
                }
            }
            if (flag)
                return true;
        }

        return false;
    }

private:
    unordered_map<int, unordered_set<string>> map;
};

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