10. Word Break

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.

Approach

• Create array with first of out of bounds index set as true
• Loop through the string length and then loop through the dictionary and check if substring matches
• If everything works then true will be set as true by the end of the loop

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] d = new boolean[s.length() + 1];
        d[s.length()] = true;
        for (int i = s.length() - 1; i >= 0; i--) {
            for (String w: wordDict) {
                if (i + w.length() <= s.length() && 
                    s.substring(i, i + w.length()).equals(w)) {
                        d[i] = d[i + w.length()];
                }
                if (d[i]) break;
 
            }
        }
        return d[0];
    }
}