Editorial
Let the completed string be .
If the prefix of length is a palindrome, its first and last characters must be equal. Therefore, for every (). Thus, every character of must be the same.
It remains to inspect the uppercase letters already fixed in .
- If there is no fixed uppercase letter, any of the 26 uppercase letters can be the common character, so the answer is .
- If all fixed uppercase letters are equal, the common character is uniquely determined, so the answer is .
- If two different fixed uppercase letters exist, making all characters equal is impossible, so the answer is .
Scanning each test case once is sufficient. The time complexity is , and the extra space complexity is .