Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

There are only 450 three-digit even numbers, so testing candidate numbers is simpler than permuting digit positions and removing duplicates. A candidate is possible exactly when its digit frequencies do not exceed the available frequencies.

Approach

Count the supplied digits. Enumerate three-digit values in ascending order, retaining only even candidates, and count each candidate's three digits. Append it when every required count is available. Python filters evenness while scanning all three-digit values; JavaScript enumerates only even values. Starting at 100 excludes leading-zero results, and enumerating values rather than index triples automatically produces sorted, distinct output while still respecting repeated-digit availability.

class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        map = defaultdict(int)
        for x in digits: map[x] += 1
        res = []

        for i in range(100, 1000):
            a = i % 10
            b = (i // 10) % 10
            c = (i // 100) % 10

            temp = defaultdict(int)
            temp[a] += 1
            temp[b] += 1
            temp[c] += 1
            
            if temp[a] <= map[a] and temp[b] <= map[b] and temp[c] <= map[c] and i % 2 == 0:
                res.append(i)
        
        return res

Complexity analysis

Let d = digits.length. Python tests the 900 three-digit values and JavaScript directly tests the 450 even values, so both take O(d + 900) = O(d) time under the fixed decimal domain. Auxiliary space is O(1) apart from the bounded result array.