In this post, we are going to solve the Word Break 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
, return true
if s
can be segmented into a space-separated sequence of one or more dictionary words.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
Example 1:
Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".
Example 2:
Input: s = "applepenapple", wordDict = ["apple","pen"] Output: true Explanation: Return true because "applepenapple" can be segmented as "apple pen apple". Note that you are allowed to reuse a dictionary word.
Example 3:
Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] Output: false
Constraints:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s
andwordDict[i]
consist of only lowercase English letters.- All the strings of
wordDict
are unique.
Now, let’s see the leetcode solution of Word Break Leetcode Solution.
Word Break Leetcode Solution in Python
class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> bool: wordSet = set(wordDict) @functools.lru_cache(None) def wordBreak(s: str) -> bool: if s in wordSet: return True return any(s[:i] in wordSet and wordBreak(s[i:]) for i in range(len(s))) return wordBreak(s)
Word Break Leetcode Solution in CPP
class Solution { public: bool wordBreak(string s, vector<string>& wordDict) { return wordBreak(s, {begin(wordDict), end(wordDict)}, {}); } private: bool wordBreak(const string& s, const unordered_set<string>&& wordSet, unordered_map<string, bool>&& memo) { if (wordSet.count(s)) return true; if (memo.count(s)) return memo[s]; // 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) && wordBreak(suffix, move(wordSet), move(memo))) return memo[s] = true; } return memo[s] = false; } };
Word Break Leetcode Solution in Java
class Solution { public boolean wordBreak(String s, List<String> wordDict) { return wordBreak(s, new HashSet<>(wordDict), new HashMap<>()); } private boolean wordBreak(final String s, Set<String> wordSet, Map<String, Boolean> memo) { if (memo.containsKey(s)) return memo.get(s); if (wordSet.contains(s)) { memo.put(s, true); return true; } // 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) && wordBreak(suffix, wordSet, memo)) { memo.put(s, true); return true; } } memo.put(s, false); return false; } }
Note: This problem Word Break is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purpose.