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 derivative. For example, when the root"help" is followed by the word "ful", we can form a derivative "helpful".
Given a dictionary consisting of many roots and a sentence consisting of words separated by spaces, replace all the derivatives in the sentence with the root forming it. If a derivative can be replaced by more than one root, replace it with the root that has the shortest length.
Return the sentence after the replacement.
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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"
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Constraints:
1 <= dictionary.length <= 1000
1 <= dictionary[i].length <= 100
dictionary[i] consists of only lower-case letters.
1 <= sentence.length <= 106
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]
Every two consecutive words in sentence will be separated by exactly one space.
sentence does not have leading or trailing spaces.
Solutions
Solution 1: Trie
We can use a trie to store all the roots in the dictionary. Define the trie node class \(\text{Trie}\), which contains an array \(\text{children}\) of length \(26\) to store child nodes, and a boolean variable \(\text{is\_end}\) to mark whether it is a complete root.
For each root, we insert it into the trie. For each word in the sentence, we search for its shortest root in the trie. If found, we replace the word; otherwise, we keep it unchanged.
The time complexity is \(O(n \times |w| + L)\), and the space complexity is \(O(n \times |w|)\), where \(n\) and \(|w|\) are the number of roots in the dictionary and the average length, respectively, and \(L\) is the total length of words in the sentence.