Tejav
#3483

Unique 3-Digit Even Numbers

Easy
ArrayHash TableRecursionEnumerationHash MapArray
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Approaches

Brute ForceOptimal
Complexity Comparison
Brute ForceOptimal Solution★
Time
O(n³)
O(n)
Space
O(n)
O(n)
💡

Intuition

Time O(n)Space O(n)

Count valid even numbers directly by fixing the last digit as even and permuting the remaining digits. This reduces unnecessary checks.

⚙️

Algorithm

3 steps
  1. 1Step 1: Identify even digits from the input array.
  2. 2Step 2: For each even digit, fix it as the last digit and count permutations of the remaining digits ensuring no leading zeros.
  3. 3Step 3: Sum all valid counts for each even digit.
solution.py14 lines
1from itertools import permutations
2
3def count_even_numbers(digits):
4    from collections import Counter
5    count = 0
6    digit_count = Counter(digits)
7    for d in set(digits):
8        if d % 2 == 0:
9            digit_count[d] -= 1
10            for perm in permutations(digit_count.elements(), 2):
11                if perm[0] != 0:
12                    count += 1
13            digit_count[d] += 1
14    return count
ℹ

Complexity note: The complexity is linear as we only traverse the digits a few times.

  • 1Only even digits can be in the last position.
  • 2Leading zeros are not allowed.

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