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Finding 3-Digit Even Numbers

Difficulty: Easy

Problem 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.

Return a sorted array of the unique integers.

Key Insights

  • The integer must be formed by concatenating exactly three digits from the digits array.
  • The last digit of the integer must be even (0, 2, 4, 6, or 8) to ensure the entire number is even.
  • We need to avoid leading zeros unless the number itself is zero.
  • We can use a set to store unique integers to avoid duplicates.

Space and Time Complexity

Time Complexity: O(n^3) - where n is the number of digits, as we may need to check all combinations of three digits. Space Complexity: O(m) - where m is the number of unique 3-digit even numbers generated.

Solution

To solve the problem, we can use a combination of nested loops to generate all possible combinations of three digits while ensuring the last digit is even. We will check if the combination has a leading zero and if it does, we will skip that combination. Finally, we will store the valid combinations in a set to avoid duplicates, then convert it to a sorted list before returning.

Code Solutions

def findEvenNumbers(digits):
    result = set()
    n = len(digits)
    
    for i in range(n):
        for j in range(n):
            for k in range(n):
                # Check if the digits are distinct and the last digit is even
                if i != j and j != k and i != k and digits[k] % 2 == 0:
                    # Form the number and check for leading zero
                    if digits[i] != 0:  # Avoid leading zero
                        num = digits[i] * 100 + digits[j] * 10 + digits[k]
                        result.add(num)
    
    return sorted(result)
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