Word Break II LeetCode Solution – Queslers

Problem – Word Break II

Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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: []


  • 1 <= s.length <= 20
  • 1 <= wordDict.length <= 1000
  • 1 <= wordDict[i].length <= 10
  • s and wordDict[i] consist of only lowercase English letters.
  • All the strings of wordDict are unique.

Word Break II LeetCode Solution in Java

public List<String> wordBreak(String s, Set<String> wordDict) {
    return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());

// DFS function returns an array including all substrings derived from s.
List<String> DFS(String s, Set<String> wordDict, HashMap<String, LinkedList<String>>map) {
    if (map.containsKey(s)) 
        return map.get(s);
    LinkedList<String>res = new LinkedList<String>();     
    if (s.length() == 0) {
        return res;
    for (String word : wordDict) {
        if (s.startsWith(word)) {
            List<String>sublist = DFS(s.substring(word.length()), wordDict, map);
            for (String sub : sublist) 
                res.add(word + (sub.isEmpty() ? "" : " ") + sub);               
    map.put(s, res);
    return res;

Word Break II LeetCode Solution in Python

class Solution(object):
def wordBreak(self, s, wordDict):
    :type s: str
    :type wordDict: Set[str]
    :rtype: List[str]
    return self.helper(s, wordDict, {})
def helper(self, s, wordDict, memo):
    if s in memo: return memo[s]
    if not s: return []
    res = []
    for word in wordDict:
        if not s.startswith(word):
        if len(word) == len(s):
            resultOfTheRest = self.helper(s[len(word):], wordDict, memo)
            for item in resultOfTheRest:
                item = word + ' ' + item
    memo[s] = res
    return res

Word Break II LeetCode Solution in C++

class Solution {
    unordered_map<string, vector<string>> m;

    vector<string> combine(string word, vector<string> prev){
        for(int i=0;i<prev.size();++i){
            prev[i]+=" "+word;
        return prev;

    vector<string> wordBreak(string s, unordered_set<string>& dict) {
        if(m.count(s)) return m[s]; //take from memory
        vector<string> result;
        if(dict.count(s)){ //a whole string is a word
        for(int i=1;i<s.size();++i){
            string word=s.substr(i);
                string rem=s.substr(0,i);
                vector<string> prev=combine(word,wordBreak(rem,dict));
                result.insert(result.end(),prev.begin(), prev.end());
        m[s]=result; //memorize
        return result;
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