L139. Word Break

Problem:

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.

letter can't be used again;

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

Solution:

• two pointers

• check hashset

public boolean wordBreak(String s, List<String> wordDict) {
    if(s == null || s.length() == 0) return true;
    if(wordDict == null) throw new IllegalArgumentException();
        
    Set<String> dict = new HashSet(wordDict);
    boolean[] dp = new boolean[s.length() + 1];
    dp[0] = true;
        
    for(int i = 1; i <= s.length(); i++){
        for(int j = 0; j < i; j++){
            if(dp[j] && dict.contains(s.substring(j,i))){
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.length()];
}