Introduction

In this blog post, we will explore how to solve the LeetCode problem "3483" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Note: Each copy of a digit can only be used once per number, and there may not be leading zeros. Example 1: Input: digits = [1,2,3,4] Output: 12 Explanation: The 12 distinct 3-digit even numbers that can be formed are 124, 132, 134, 142, 214, 234, 312, 314, 324, 342, 412, and 432. Note that 222 cannot be formed because there is only 1 copy of the digit 2. Example 2: Input: digits = [0,2,2] Output: 2 Explanation: The only 3-digit even numbers that can be formed are 202 and 220. Note that the digit 2 can be used twice because it appears twice in the array. Example 3: Input: digits = [6,6,6] Output: 1 Explanation: Only 666 can be formed. Example 4: Input: digits = [1,3,5] Output: 0 Explanation: No even 3-digit numbers can be formed. Constraints: 3 <= digits.length <= 10 0 <= digits[i] <= 9

Explanation

Here's a breakdown of the approach and the Python code:

• High-Level Approach:

  • Iterate through the digits array to find each even digit that can be used as the last digit (units place).
  • For each even digit, iterate through the remaining digits to find suitable digits for the hundreds and tens places, ensuring the hundreds digit is not zero.
  • Use a frequency count to track the availability of each digit and avoid using the same digit multiple times if it's not present in the input array with sufficient frequency.

• Complexity:

  • Runtime Complexity: O(N^3), where N is the length of the digits array.
  • Storage Complexity: O(1) - Uses a fixed-size frequency count array.

Code

    def distinct_even_numbers(digits):
    """
    Calculates the number of distinct three-digit even numbers that can be formed using the given digits.

    Args:
        digits: A list of integers representing the available digits.

    Returns:
        The number of distinct three-digit even numbers.
    """

    count = 0
    distinct_numbers = set()  # Use a set to store distinct numbers

    for i in range(len(digits)):
        if digits[i] % 2 == 0:  # Check if the digit at index i is even

            for j in range(len(digits)):
                if i == j:
                    continue

                for k in range(len(digits)):
                    if i == k or j == k:
                        continue

                    if digits[j] != 0: # Hundreds digit cannot be zero
                        number = digits[j] * 100 + digits[k] * 10 + digits[i]
                        distinct_numbers.add(number)

    return len(distinct_numbers)