My approach was a monotonic stack where I'd pop the top if it was smaller than the incoming digit and had an odd frequency, figuring that'd maximize the result.
Use a greedy strategy: process digits from left to right, maintaining a stack of selected digits. For each digit, decide whether to include it based on whether it can improve the final number while ensuring all counts are even, and handle the last occurrence of each digit carefully to avoid odd counts.
Pro tip: Clarify that the result should have no leading zeros unless the number is zero, and discuss how to handle that constraint within the greedy approach.
Restate the problem: delete a subset of digits to form the largest number where each digit appears an even number of times. Note that the order of remaining digits is preserved, and the result can be empty (which represents 0).
To maximize the number, we want the leftmost digits as large as possible. For each digit, we can decide to include it if it helps to form a larger number, but we must ensure that all digits end up with even counts.
Use a stack to build the result. Iterate through the digits, and for each digit, while the stack is not empty and the top of the stack is less than the current digit and the top digit can be removed (i.e., it appears again later to maintain even count), pop it. Then push the current digit if it can be part of an even count (e.g., if it's not the last occurrence or if we can pair it).
After processing all digits, ensure all counts are even by possibly removing the last occurrence of any digit with odd count. Remove leading zeros if the result is not empty. If the result is empty, return '0'.
The algorithm runs in O(n) time and O(n) space. Test with examples like '1234' (result '0'), '4444' (result '4444'), '123321' (result '123321'), and '1111' (result '1111').
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