Day2

 Today, I solved an interesting problem: finding the length of the longest valid parentheses substring in a string. Using a two-pass approach, I iterated from left to right to count matching parentheses and calculate valid lengths while resetting counts for invalid cases.

To handle unbalanced scenarios, I reversed the process, iterating from right to left and applying the same logic. This ensured I captured all valid combinations efficiently. It was a great exercise in understanding edge cases and optimizing string traversal with an O(n) solution.

Excited to keep pushing my limits and tackling more challenges! 🚀

𝗧𝗼𝗱𝗮𝘆'𝘀 𝗽𝗿𝗼𝗯𝗹𝗲𝗺:
32. Longest Valid Parentheses

𝗘𝘅𝗽𝗹𝗮𝗻𝗮𝘁𝗶𝗼𝗻:
This Java code solves the problem of finding the length of the longest valid substring of parentheses in a given string s. It employs two passes to ensure all valid cases are considered. In the first pass, the code scans the string from left to right, using counters left and right to count the number of opening and closing parentheses, respectively.

Whenever left equals right, it calculates the maximum length of valid parentheses (left * 2). If right exceeds left, the counters are reset. The second pass scans from right to left to handle unbalanced cases where there are more opening parentheses.

Here, the same logic applies, resetting counters if left exceeds right. By combining both passes, the code accounts for valid substrings starting or ending in different positions. The result, stored in max, is returned as the length of the longest valid parentheses substring. This approach is efficient with O(n) time complexity.

𝗘𝘅𝗮𝗺𝗽𝗹𝗲 𝟭:
Input: s = "(()"
Output: 2

𝗘𝘅𝗮𝗺𝗽𝗹𝗲 𝟮:
Input: s = ")()())"
Output: 4