# Word Break Leetcode Solution

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` and `wordDict[i]` consist of only lowercase English letters.
• All the strings of `wordDict` are unique.

Now, lets 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 Solutionin 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.

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