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 <= 3001 <= wordDict.length <= 10001 <= wordDict[i].length <= 20sandwordDict[i]consist of only lowercase English letters.- All the strings of
wordDictare 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.
