Given an integer array digits, form all unique three digit positive even integers that can be built by selecting exactly three elements from digits without reusing an element more times than it appears in digits.

Return these integers in increasing order. A valid number must not have a leading zero, and its last digit must be even.

Note: Each occurrence of a value in digits can be used at most once per constructed number, so repeated values in digits allow repeated use up to their frequency.

Constraints:

• $3 \leq$ digits.length $\leq 100$

• $0 \leq$ digits[i] $\leq 9$

Input: digits = [2, 1, 3, 0]-List[int]
Output: [102, 120, 130, 132, 210, 230, 302, 310, 312, 320]-List[int]
Explanation: Form 3-digit numbers using three distinct indices, where the first digit cannot be 0 and the last digit must be even (0 or 2). Listing all valid unique results and sorting gives the output.

Input: digits = [2, 2, 8, 8, 2]-List[int]
Output: [222, 228, 282, 288, 822, 828, 882]-List[int]
Explanation: Digits repeat, so you may use 2 up to three times and 8 up to two times in a number (but not more than their counts). Valid even numbers (ending in 2 or 8) are generated, duplicates are removed, and the results are sorted.

Input: digits = [0, 1, 0, 4]-List[int]
Output: [100, 104, 140, 400, 410]-List[int]
Explanation: A leading zero is not allowed, so numbers like 010 are invalid. Using the available digits (two 0s, one 1, one 4), the valid even numbers (ending in 0 or 4) are collected uniquely and sorted.

Input: digits = [1, 3, 5]-List[int]
Output: []-List[int]
Explanation: No digit is even, so it’s impossible to form a 3-digit even number (the last digit must be even).

Input: digits = [0, 0, 0, 2]-List[int]
Output: [200]-List[int]
Explanation: The only way to make a 3-digit number without a leading zero is to start with 2, and the remaining two digits must be 0 and 0, producing 200 (even and valid).

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