LeetCode 2781. Length of the Longest Valid Substring

原题链接在这里：https://leetcode.com/problems/length-of-the-longest-valid-substring/description/

题目：

You are given a string word and an array of strings forbidden.

A string is called valid if none of its substrings are present in forbidden.

Return the length of the longest valid substring of the string word.

A substring is a contiguous sequence of characters in a string, possibly empty.

Example 1:

Input: word = "cbaaaabc", forbidden = ["aaa","cb"]
Output: 4
Explanation: There are 11 valid substrings in word: "c", "b", "a", "ba", "aa", "bc", "baa", "aab", "ab", "abc" and "aabc". The length of the longest valid substring is 4. 
It can be shown that all other substrings contain either "aaa" or "cb" as a substring. 

Example 2:

Input: word = "leetcode", forbidden = ["de","le","e"]
Output: 4
Explanation: There are 11 valid substrings in word: "l", "t", "c", "o", "d", "tc", "co", "od", "tco", "cod", and "tcod". The length of the longest valid substring is 4.
It can be shown that all other substrings contain either "de", "le", or "e" as a substring. 

Constraints:

• 1 <= word.length <= 105
• word consists only of lowercase English letters.
• 1 <= forbidden.length <= 105
• 1 <= forbidden[i].length <= 10
• forbidden[i] consists only of lowercase English letters.

题解：

The question is asking for the longest substring with no forbidden word.

Iterate from right to left using index i, and from to right index j, if word[i, j] is forbidden, mark the right position.

Update the result for each index i.

Time Complexity: O(m * maxLen + n * maxLen * maxLen). m = forbidden.size(). n = word.length(). 

Space: O(m * maxLen).

AC Java:

class Solution {
    public int longestValidSubstring(String word, List<String> forbidden) {
        Set<String> hs = new HashSet<>();
        int maxLen = 0;
        for(String s : forbidden){
            hs.add(s);
            maxLen = Math.max(maxLen, s.length());
        }

        int res = 0;
        int n = word.length();
        int right = n - 1;
        for(int i = n - 1; i >= 0; i--){
            for(int j = i; j <= Math.min(i + maxLen, right); j++){
                if(hs.contains(word.substring(i, j + 1))){
                    right = j - 1;
                    break;
                }
            }

            res = Math.max(res, right - i + 1);
        }

        return res;
    }
}