Map Sum Pairs LeetCode Solution using Trie and DFS

A nice problem can be solved by using Trie, DFS, and a hash table.

677. Map Sum Pairs

You need to Design a map that allows you to do the following:

• Maps a string key to a given value.
• Returns the sum of the values that have a key with a prefix equal to a given string.

Implement the MapSum class:

• MapSum() Initializes the MapSum object.
• void insert(String key, int val) Inserts the key-val pair into the map. If the key already existed, the original key-value pair will be overridden to the new one.
• int sum(string prefix) Returns the sum of all the pairs' value whose key starts with the prefix.

Example 1:

Input
["MapSum", "insert", "sum", "insert", "sum"]
[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]
Output
[null, null, 3, null, 5]
Explanation
MapSum mapSum = new MapSum();
mapSum.insert("apple", 3);  
mapSum.sum("ap");           // return 3 (apple = 3)
mapSum.insert("app", 2);    
mapSum.sum("ap");           // return 5 (apple + app = 3 + 2 = 5)

Constraints:

• 1 <= key.length, prefix.length <= 50
• key and prefix consist of only lowercase English letters.
• 1 <= val <= 1000
• At most 50 calls will be made to insert and sum.

Reference Code

struct TrieNode{
    public:
    bool is_word = false;
    vector<TrieNode*> children;
    TrieNode(){
        children.assign(26, nullptr);
    }
};
class MapSum {
public:
    unordered_map<string, int> m;
    TrieNode* root;
    /** Initialize your data structure here. */
    MapSum() {
        root = new TrieNode();
    }
    
    void insert(string key, int val) {
        if (m.find(key) != m.end()) {
            m[key] = val;
            return;
        }
        m[key] = val;
        // insert this word to TrieNode
        auto node = root;
        const int n = key.size();
            for (int i = 0; i < n; ++i){
                int ch_idx = key[i] - 'a';
                if (node -> children[ch_idx] == nullptr){
                    node -> children[ch_idx] = new TrieNode();
                }
                node = node -> children[ch_idx];
                if (i == n - 1) node -> is_word = true;
            }
        return;
    }
    
    int dfs(TrieNode* node, string prefix){
        int ret = 0;
        if (node -> is_word) ret += m[prefix];
        for (char ch = 'a'; ch <= 'z'; ch++){
            if (node -> children[ch - 'a'] != nullptr){
                ret += dfs(node -> children[ch - 'a'], prefix + ch);
            }
        }
        return ret; 
    }
    
    int sum(string prefix) {
        auto node = root;
        for (int i = 0; i < prefix.size(); ++i){
            int ch_idx = prefix[i] - 'a';
            if (node -> children[ch_idx] == nullptr) return 0;
            node = node -> children[ch_idx];
        }
        return dfs(node, prefix);
    }
};
/**
 * Your MapSum object will be instantiated and called as such:
 * MapSum* obj = new MapSum();
 * obj->insert(key,val);
 * int param_2 = obj->sum(prefix);
 */

Problem

472. Concatenated Words

Solution

struct TrieNode{
    public:
    bool is_word = false;
    vector<TrieNode*> children;
    TrieNode(){
        children.assign(26, nullptr);
    }
};
class Trie{
    public:
    TrieNode* root;
    Trie(){
        root = new TrieNode();
    }
    void insert(string &word){
        if (word.empty())return;
        auto curr = root;
        for (auto c: word){
            int idx = c - 'a';
            if (curr->children[idx] == nullptr){
                curr->children[idx] = new TrieNode();
            }
            curr = curr->children[idx];
        }
        curr -> is_word = true;
    }
};

class Solution {
public:

    unordered_map<string, int> memo;
    int dfs(int i, string& word, TrieNode* node){
        string sub_str = word.substr(i);
        if (memo.find(sub_str) !=memo.end())return memo[sub_str];
        if (i == word.size()){
            return memo[sub_str] = 0;
        }
        auto curr = node;
        int max_words = -1;
        for (int j = i; j < (int) word.size(); ++j){
            curr = curr->children[word[j] - 'a'];
            if (curr == nullptr)break;
            if (curr -> is_word){
                int this_ret = dfs(j + 1, word, node);
                if (this_ret == -1) continue;
                else {
                    max_words = max(max_words, 1 + this_ret); 
                }
            }
        }
        return memo[sub_str] = max_words;
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Trie trie;
        for (auto &word: words){
            trie.insert(word);
        }
        vector<string> ret;
        for (auto &word: words){
            int num_of_concat_words = dfs(0, word, trie.root);
            if (num_of_concat_words >= 2){
                ret.push_back(word);
            }
        }
        return ret;
    }
};

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