In English, we have a concept called root
, which can be followed by some other words to form another longer word - let's call this word successor
. For example, the root an
, followed by other
, which can form another word another
.
Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor
in the sentence with the root
forming it. If a successor
has many roots
can form it, replace it with the root with the shortest length.
You need to output the sentence after the replacement.
Example 1:
Input: dict = ["cat", "bat", "rat"] sentence = "the cattle was rattled by the battery" Output: "the cat was rat by the bat"
Note:
- The input will only have lower-case letters.
- 1 <= dict words number <= 1000
- 1 <= sentence words number <= 1000
- 1 <= root length <= 100
- 1 <= sentence words length <= 1000
Related Topics:
Hash Table, Trie
// OJ: https://leetcode.com/problems/replace-words/
// Author: github.com/lzl124631x
// Time: O(D + S) where D is size of all contents in dictionary, S is size of all contents in sentence
// Space: O(D)
class TrieNode {
public:
TrieNode *next[26] = {};
bool isWord = false;
};
class Trie {
private:
TrieNode root;
public:
void insert(string &s) {
auto node = &root;
for (char c : s) {
if (!node->next[c - 'a']) node->next[c - 'a'] = new TrieNode();
node = node->next[c - 'a'];
}
node->isWord = true;
}
string getWord(string &s) {
auto node = &root;
for (int i = 0; i < s.size(); ++i) {
if (!node->next[s[i] - 'a']) return s;
node = node->next[s[i] - 'a'];
if (node->isWord) return s.substr(0, i + 1);
}
return s;
}
};
class Solution {
public:
string replaceWords(vector<string>& dict, string sentence) {
istringstream ss(sentence);
string word, ans;
Trie trie;
for (auto s : dict) trie.insert(s);
while (ss >> word) ans += trie.getWord(word)+ " ";
ans.pop_back();
return ans;
}
};