Introduction

In this blog post, we will explore how to solve the LeetCode problem "2605" 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

Given two arrays of unique digits nums1 and nums2, return the smallest number that contains at least one digit from each array. Example 1: Input: nums1 = [4,1,3], nums2 = [5,7] Output: 15 Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have. Example 2: Input: nums1 = [3,5,2,6], nums2 = [3,1,7] Output: 3 Explanation: The number 3 contains the digit 3 which exists in both arrays. Constraints: 1 <= nums1.length, nums2.length <= 9 1 <= nums1[i], nums2[i] <= 9 All digits in each array are unique.

Explanation

Here's a breakdown of the solution:

• Find Common Digits: First, check for any common digits between the two arrays. If a common digit exists, it's the smallest possible result, and we can return it immediately.
• Form Two-Digit Numbers: If no common digits are found, we need to form two-digit numbers using one digit from each array.

• Minimize: Compare the two possible two-digit numbers (smallest digit from nums1 followed by the smallest digit from nums2, and vice-versa) to find the smaller one.

• Runtime Complexity: O(m + n), where m and n are the lengths of nums1 and nums2 respectively. Storage Complexity: O(1).

Code

    def smallest_number_containing_digits(nums1, nums2):
    """
    Finds the smallest number containing at least one digit from each array.

    Args:
        nums1: A list of unique digits.
        nums2: A list of unique digits.

    Returns:
        The smallest number that contains at least one digit from each array.
    """

    common_digits = set(nums1) & set(nums2)
    if common_digits:
        return min(common_digits)

    min1 = min(nums1)
    min2 = min(nums2)

    return min(min1 * 10 + min2, min2 * 10 + min1)