D-30 submission.md

⚡ Question

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. Note: The same word in the dictionary may be reused multiple times in the segmentation. You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]
Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.
Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

🍑 Solution

class Solution {

    public List<String> wordBreak(String s, List<String> dict) {
    //create an array of ArrayList<String>
    List<String> dp[] = new ArrayList[s.length()+1];
    dp[0] = new ArrayList<String>();
 
    for(int i=0; i<s.length(); i++){
        if( dp[i] == null ) 
            continue; 
 
        for(String word:dict){
            int len = word.length();
            int end = i+len;
            if(end > s.length()) 
                continue;
 
            if(s.substring(i,end).equals(word)){
                if(dp[end] == null){
                    dp[end] = new ArrayList<String>();
                }
                dp[end].add(word);
            }
        }
    }
 
    List<String> result = new LinkedList<String>();
    if(dp[s.length()] == null) 
        return result; 
 
    ArrayList<String> temp = new ArrayList<String>();
    dfs(dp, s.length(), result, temp);
 
    return result;
}
 
public static void dfs(List<String> dp[],int end,List<String> result, ArrayList<String> tmp){
    if(end <= 0){
        String path = tmp.get(tmp.size()-1);
        for(int i=tmp.size()-2; i>=0; i--){
            path += " " + tmp.get(i) ;
        }
 
        result.add(path);
        return;
    }
 
    for(String str : dp[end]){
        tmp.add(str);
        dfs(dp, end-str.length(), result, tmp);
        tmp.remove(tmp.size()-1);
    }
}
}