Word Break II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Word Break II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemWord Break II– LeetCode Problem

Word Break II– LeetCode Problem


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 Solutions
Word Break II Solution in C++:
class Solution {
  vector<string> wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> wordSet{begin(wordDict), end(wordDict)};
    unordered_map<string, vector<string>> memo;
    return wordBreak(s, wordSet, memo);

  vector<string> wordBreak(const string& s,
                           const unordered_set<string>& wordSet,
                           unordered_map<string, vector<string>>& memo) {
    if (memo.count(s))
      return memo[s];

    vector<string> ans;

    // 1 <= prefix.length() < s.length()
    for (int i = 1; i < s.length(); ++i) {
      const string& prefix = s.substr(0, i);
      const string& suffix = s.substr(i);
      if (wordSet.count(prefix))
        for (const string& word : wordBreak(suffix, wordSet, memo))
          ans.push_back(prefix + " " + word);

    // contains whole string, so don't add any space
    if (wordSet.count(s))

    return memo[s] = ans;
Word Break II Solution in Java:
class Solution {
  public List<String> wordBreak(String s, List<String> wordDict) {
    Set<String> wordSet = new HashSet<>(wordDict);
    Map<String, List<String>> memo = new HashMap<>();
    return wordBreak(s, wordSet, memo);

  private List<String> wordBreak(final String s, Set<String> wordSet,
                                 Map<String, List<String>> memo) {
    if (memo.containsKey(s))
      return memo.get(s);

    List<String> ans = new ArrayList<>();

    // 1 <= prefix.length() < s.length()
    for (int i = 1; i < s.length(); ++i) {
      final String prefix = s.substring(0, i);
      final String suffix = s.substring(i);
      if (wordSet.contains(prefix))
        for (final String word : wordBreak(suffix, wordSet, memo))
          ans.add(prefix + " " + word);

    // contains whole string, so don't add any space
    if (wordSet.contains(s))

    memo.put(s, ans);
    return ans;
Word Break II Solution in Python:
class Solution:
  def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
    wordSet = set(wordDict)

    def wordBreak(s: str) -> List[str]:
      ans = []

      # 1 <= len(prefix) < len(s)
      for i in range(1, len(s)):
        prefix = s[0:i]
        suffix = s[i:]
        if prefix in wordSet:
          for word in wordBreak(suffix):
            ans.append(prefix + ' ' + word)

      # contains whole string, so don't add any space
      if s in wordSet:

      return ans

    return wordBreak(s)

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