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2094. Finding 3-Digit Even Numbers

Difficulty Topics

Description

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

finding-3-digit-even-numbers.py
class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        res = set()
        n = len(digits)

        for i in range(n):
            if digits[i] == 0: continue
            for j in range(n):
                if i == j: continue
                for k in range(n):
                    if i == k or j == k or digits[k] & 1: continue
                    res.add(digits[i] * 100 + digits[j] * 10 + digits[k])

        return sorted(list(res))