Finding 3-Digit Even Numbers

You are given an integer array digits, where each element is a digit. The array may contain duplicates.

You need to find all the unique integers that follow the given requirements:

• The integer consists of the concatenation of three elements from digits in any arbitrary order.
• The integer does not have leading zeros.
• The integer is even.

For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements.

Return a sorted array of the unique integers.

Example 1:

Input: digits = [2,1,3,0]
Output: [102,120,130,132,210,230,302,310,312,320]
Explanation: All the possible integers that follow the requirements are in the output array.
Notice that there are no odd integers or integers with leading zeros.

Example 2:

Input: digits = [2,2,8,8,2]
Output: [222,228,282,288,822,828,882]
Explanation: The same digit can be used as many times as it appears in digits.
In this example, the digit 8 is used twice each time in 288, 828, and 882.

Example 3:

Input: digits = [3,7,5]
Output: []
Explanation: No even integers can be formed using the given digits.

Constraints:

• 3 <= digits.length <= 100
• 0 <= digits[i] <= 9

SOLUTION:

import bisect

class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        n = len(digits)
        f = [i for i in range(n) if digits[i] != 0]
        s = [i for i in range(n)]
        t = [i for i in range(n) if digits[i] in {0, 2, 4, 6, 8}]
        nums = []
        for i in f:
            for j in s:
                if i == j:
                    continue
                for k in t:
                    if i != k and j != k:
                        val = 100 * digits[i] + 10 * digits[j] + digits[k]
                        pos = bisect.bisect_left(nums, val)
                        if pos >= len(nums) or nums[pos] != val:
                            bisect.insort(nums, val)
        return nums