648. Replace Words

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:




Solution 1 : hash map

We can check the prefixes directly. For each word in the sentence, we'll look at successive prefixes and see if we saw them before.

def replaceWords(self, roots, sentence):
    rootset = set(roots)

    def replace(word):
        for i in xrange(1, len(word)):
            if word[:i] in rootset:
                return word[:i]
        return word

    return " ".join(map(replace, sentence.split()))





class Solution {
    public String replaceWords(List<String> dict, String sentence) {
        // convert the list of words to set of words 
        // traversing the string sentence, check every possible prefix of each word,
        // see if exists in the dictionary, if it is , replace ith 
        // if not exists,  keep the original word
        
        if(dict == null || dict.size() == 0) return sentence;
        Set<String> set = new HashSet<>(dict);
        StringBuilder sb = new StringBuilder();
        for(String word : sentence.split(" ")){
            String prefix = "";
            for(int i = 1; i <= word.length(); i++){
                // i = 1, i <= word.length()
                prefix = word.substring(0, i);
                if(set.contains(prefix)) break;
            }
            // if found a root, or prefix, append, if not, we will have prefix has all the chars at the end 
            sb.append(prefix).append(" ");
        }
        sb.setLength(sb.length() - 1); // have to setLength first before convert it to string 
       
        return sb.toString();
    }
}





Solution 2 : trie 

build trie with words from the dictionary 
Traverse the sentence word by word, check If each word has a prefix in the dictionary, if so, replace the word in the sentence with the prefix 





// not tested , others code 

public String replaceWords(List<String> dict, String sentence){
    String[] tokens = sentence.split(" ");
    TrieNode trie = buildTrie(dict);
    return replaceWords(tokens, trie);
    
}

    private String replaceWords(String[] tokens, TrieNode root){
        StringBuilder sb = new StringBuilder();
        for(String token : tokens){
            sb.append(getShortestReplacement(token, root)).append(" ");
        }
        return sb.substring(0, sb.length() - 1); // sb.susbtirng() returns a string ??? 
        // String substring() : This method has two variants and returns a new string that is a substring of this string. 
    }



    private String getShortestReplacement(String token, TrieNode root){
        TrieNode cur = root;
        StringBuilder sb = new StringBuilder();
        for(char c : token.toCharArray()){
            sb.append(c);
            if(cur.children[c - 'a'] != null){
                if(cur.children[c - 'a'].isWord){
                    return sb.toString();
                }
                cur = cur.children[c - 'a'];
            }else{
                return token; // ? 
            }
        }
        return token; // ? 
    }


    
    private TrieNode buildTrie(List<String> dict){
        TrieNode root = new TrieNode(' '); // not the normal way of building trie 
        for(String word : dict){
            TrieNode cur = root;
            for(char c : word.toCharArray()){
                if(cur.children[c - 'a'] == null){
                    cur.children[c - 'a'] = new TrieNode(c);
                }
                cur = cur.children[c - 'a'];
            }
            cur.isWord = true;
        }
        return root;
    }
    public class TrieNode{
        char val;
        TrieNode[] children;
        boolean isWord;
        
        public TrieNode(char val){
            this.val = val;
            this.children = new TrieNode[26];
            this.isWord = false;
        }
    }
}