648. Replace Words

1. Description

In English, we have a concept called root, which can be followed by some other word to form another longer word - let’s call this word successor. For example, when the root “an” is followed by the successor word “other”, we can form a new word “another”.
Given a dictionary consisting of many roots and a sentence consisting of words separated by spaces, replace all the successors in the sentence with the root forming it. If a successor can be replaced by more than one root, replace it with the root that has the shortest length.
Return the sentence after the replacement.

2. Example

Example 1

Input: dictionary = [“cat”,“bat”,“rat”], sentence = “the cattle was rattled by the battery”
Output: “the cat was rat by the bat”

Example 2

Input: dictionary = [“a”,“b”,“c”], sentence = “aadsfasf absbs bbab cadsfafs”
Output: “a a b c”

Example 3

Input: dictionary = [“a”, “aa”, “aaa”, “aaaa”], sentence = “a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa”
Output: “a a a a a a a a bbb baba a”

Example 4

Input: dictionary = [“catt”,“cat”,“bat”,“rat”], sentence = “the cattle was rattled by the battery”
Output: “the cat was rat by the bat”

Example 5

Input: dictionary = [“ac”,“ab”], sentence = “it is abnormal that this solution is accepted”
Output: “it is ab that this solution is ac”

3. Constraints

1 <= dictionary.length <= 1000
1 <= dictionary[i].length <= 100
dictionary[i] consists of only lower-case letters.
1 <= sentence.length <= 10$^{6}$
sentence consists of only lower-case letters and spaces.
The number of words in sentence is in the range [1, 1000]
The length of each word in sentence is in the range [1, 1000]
Each two consecutive words in sentence will be separated by exactly one space.
sentence does not have leading or trailing spaces.

4. Solutions

Bit Manipulation

m is the number of letters in i_dictionary, n is the number of letters in i_sentence
Time complexity: O(m)
Space complexity: O(n)

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class TrieNode {
public:
    array<TrieNode*, 26> childs;  // 26 lower case letters
    bool isEnd;

    TrieNode() : isEnd(false) {
        for (int i = 0; i < childs.size(); ++i) {
            childs[i] = nullptr;
        }
    }
};

class Solution {
public:
    // string replaceWords(vector<string>& dictionary, string sentence)
    string replaceWords(const vector<string>& i_dictionary, const string& i_sentence) {
        TrieNode* trieRoot = new TrieNode();
        constructTrieTree(trieRoot, i_dictionary);

        string replacedSentence;
        stringstream sentenceStream(i_sentence);
        for (string word; sentenceStream >> word;) {
            replacedSentence += getReplacedWord(trieRoot, word) + " ";
        }

        if (!replacedSentence.empty() && replacedSentence.back() == ' ') {
            replacedSentence.pop_back();
        }

        return replacedSentence;
    }

private:
    int getLetterIndex(char letter) {
        return letter - 'a';
    }

    void constructTrieTree(TrieNode* o_trieRoot, const vector<string>& i_dictionary) {
        for (auto word : i_dictionary) {
            auto iter = o_trieRoot;
            for (auto letter : word) {
                if (iter->childs[getLetterIndex(letter)] == nullptr) {
                    iter->childs[getLetterIndex(letter)] = new TrieNode();
                }

                iter = iter->childs[getLetterIndex(letter)];
            }

            iter->isEnd = true;
        }
    }

    string getReplacedWord(TrieNode* i_trieRoot, const string& i_word) {
        auto iter = i_trieRoot;
        string replacedWord = i_word;
        for (int i = 0; i < i_word.size(); ++i) {
            int letter = i_word[i];
            if (iter->isEnd) {
                replacedWord = i_word.substr(0, i);
                break;
            }

            if (iter->childs[getLetterIndex(letter)] != nullptr) {
                iter = iter->childs[getLetterIndex(letter)];
            } else {
                break;
            }
        }

        return replacedWord;
    }
};

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