package com.app.main.LeetCode.dynamic;

import java.util.HashSet;
import java.util.List;
import java.util.Set;

/**
 * 139
 *
 * medium
 *
 * https://leetcode.com/problems/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
 *
 *
 * Created with IDEA
 * author:Dingsheng Huang
 * Date:2019/12/17
 * Time:下午6:42
 */
public class WordBreak {
    public boolean wordBreak(String s, List<String> wordDict) {

        Set<String> set = new HashSet<>();
        set.addAll(wordDict);
        if (s.length() < 2) {
            return set.contains(s);
        }

        boolean[] dp = new boolean[s.length()];

        if (set.contains(s.substring(0, 1))) {
            dp[0] = true;
        } else {
            dp[0] = false;
        }
        for(int i = 1; i < s.length(); i++) {
            for (int j = i; j >= 0; j--) {
                if (j == 0) {
                    dp[i] = set.contains(s.substring(j, i + 1));
                } else {
                    dp[i] = dp[j - 1] && set.contains(s.substring(j, i + 1));
                }
                if (dp[i]) {
                    break;
                }
            }
        }
        return dp[s.length() - 1];
    }
}