2094. Finding 3-Digit Even Numbers

Problem Link

Description

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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

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Python3

class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        N = len(digits)
        visited = [False] * N
        res = []
        
        def backtrack(index, s):
            if index == 3:
                nonlocal res
                res.append(s)
                return
            
            for j in range(N):
                if visited[j] or (digits[j] == 0 and index == 0) or (index == 2 and digits[j] % 2 != 0): continue
                visited[j] = True
                backtrack(index + 1, s * 10 + digits[j])
                visited[j] = False
 
        backtrack(0, 0)
        return sorted(set(res))