Word_Break.java

/**
 * 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.
 */
class Solution {
    public boolean wordBreak(String s, List<String> words) {
        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[0] = true;

        for(int i = 0; i < n; i++){
            for(String word : words){
                int j = i, k = word.length() - 1;
                while(j >= 0 && k >= 0){
                    if(s.charAt(j) == word.charAt(k)){
                        j--;
                        k--;
                    }
                    else break;
                }
                if(k == -1) dp[i + 1] = dp[j + 1];
                if(dp[i + 1]) break;
            }
        }
        return dp[n];
    }
}