In this post, we are going to solve the Word Break II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
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: []
Constraints:
1 <= s.length <= 201 <= wordDict.length <= 10001 <= wordDict[i].length <= 10sandwordDict[i]consist of only lowercase English letters.- All the strings of
wordDictare unique.
Now, let’s see the leetcode solution of Word Break II Leetcode Solution.
Word Break II Leetcode Solution in Python
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
wordSet = set(wordDict)
@functools.lru_cache(None)
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:
ans.append(s)
return ans
return wordBreak(s)Word Break II Leetcode Solution in CPP
class Solution {
public:
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);
}
private:
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))
ans.push_back(s);
return memo[s] = ans;
}
};Word Break II Leetcode 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))
ans.add(s);
memo.put(s, ans);
return ans;
}
}Note: This problem Word Break II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purpose.
