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Unique 3-Digit Even Numbers

Difficulty: Easy

Problem Description

You are given an array of digits called digits. Your task is to determine the number of distinct three-digit even numbers that can be formed using these digits. Each copy of a digit can only be used once per number, and there may not be leading zeros.

Key Insights

  • A valid three-digit number must have a non-zero digit as the first digit.
  • The last digit must be an even digit (0, 2, 4, 6, or 8) to ensure the number is even.
  • We need to avoid counting duplicate numbers due to repeated digits in the input array.
  • The approach involves iterating through combinations of the digits while accounting for the constraints on leading and trailing digits.

Space and Time Complexity

Time Complexity: O(n^3) where n is the number of digits (for combinations and checking). Space Complexity: O(n) for storing distinct numbers.

Solution

To solve the problem, we can use a combination of enumeration and set data structures to generate all possible three-digit combinations from the digits array. We will:

  1. Identify valid even digits to be used as the last digit.
  2. For each even digit, determine the remaining digits to form the first and second digits of the number.
  3. Use a set to keep track of unique three-digit numbers formed to avoid duplicates.

Code Solutions

from itertools import permutations

def uniqueEvenDigitsCount(digits):
    unique_numbers = set()
    even_digits = [d for d in digits if d % 2 == 0]
    
    for last_digit in even_digits:
        remaining_digits = [d for d in digits if d != last_digit] + ([last_digit] if digits.count(last_digit) > 1 else [])
        for first_digit in remaining_digits:
            if first_digit == 0:
                continue
            for second_digit in remaining_digits:
                if second_digit != first_digit:
                    number = 100 * first_digit + 10 * second_digit + last_digit
                    unique_numbers.add(number)
    
    return len(unique_numbers)

# Example usage
print(uniqueEvenDigitsCount([1, 2, 3, 4]))  # Output: 12
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