LeetCode 2781: Length of the Longest Valid Substring Solution

Master LeetCode problem 2781 (Length of the Longest Valid Substring), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2781. Length of the Longest Valid Substring

Hard

Array Hash Table String Sliding Window

Problem Explanation

Explanation

To find the length of the longest valid substring, we can iterate through the string word and check if each substring of word is present in the forbidden list. We can use a sliding window approach to efficiently check for valid substrings. If a substring is valid, we update the length of the longest valid substring found so far.

Algorithm

Initialize a variable maxLen to store the length of the longest valid substring found so far.
Iterate through each character in the string word.
For each character, check all substrings starting from that character to the end of the string.
If a substring is not present in the forbidden list, update maxLen if the length of the current substring is greater.
Return maxLen as the length of the longest valid substring.

Time Complexity

The time complexity of this approach is O(n^3), where n is the length of the string word.

Space Complexity

The space complexity is O(1) as we are not using any additional data structures.

Solution Code

class Solution {
    public int longestValidSubstring(String word, String[] forbidden) {
        int n = word.length();
        int maxLen = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j <= n; j++) {
                String sub = word.substring(i, j);
                boolean isValid = true;
                for (String f : forbidden) {
                    if (sub.contains(f)) {
                        isValid = false;
                        break;
                    }
                }
                if (isValid) {
                    maxLen = Math.max(maxLen, j - i);
                }
            }
        }
        return maxLen;
    }
}

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