> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/dynamic-programming/word-break.md).

# Word Break

Given a **non-empty** string *s* and a dictionary *wordDict* containing a list of **non-empty** words, determine if *s* can be segmented into a space-separated sequence of one or more dictionary words.

**Note:**

* The same word in the dictionary may be reused multiple times in the segmentation.
* You may assume the dictionary does not contain duplicate words.

**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
```

```java
class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean dp[] = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (String x : wordDict) {
                if (i - x.length() >= 0 && s.substring(i - x.length(), i).compareTo(x) == 0) {
                    dp[i] = dp[i] || dp[i - x.length()];
                }
            }
        }
        return dp[s.length()];
    }
}
```