Introduction

In this blog post, we will explore how to solve the LeetCode problem "2160" 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 a positive integer num consisting of exactly four digits. Split num into two new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in new1 and new2, and all the digits found in num must be used. For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3. Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329]. Return the minimum possible sum of new1 and new2. Example 1: Input: num = 2932 Output: 52 Explanation: Some possible pairs [new1, new2] are [29, 23], [223, 9], etc. The minimum sum can be obtained by the pair [29, 23]: 29 + 23 = 52. Example 2: Input: num = 4009 Output: 13 Explanation: Some possible pairs [new1, new2] are [0, 49], [490, 0], etc. The minimum sum can be obtained by the pair [4, 9]: 4 + 9 = 13. Constraints: 1000 <= num <= 9999

Explanation

Here's the solution to the problem:

• Core Idea: To minimize the sum, we want to make the two numbers new1 and new2 as small as possible. This can be achieved by distributing the smallest digits to the most significant places of new1 and new2.
• Algorithm: Sort the digits of the input number in ascending order. Construct new1 by combining the smallest two digits (first and second smallest). Construct new2 by combining the next two smallest digits (third and fourth smallest). Return the sum of new1 and new2.

• Leading Zero Handling: The problem states that leading zeros are allowed. Sorting ensures that if zeros exist, they will be placed in the most significant places.

• Complexity:

  • Runtime Complexity: O(1) since we're dealing with a fixed number of digits (4), making sorting also O(1).
  • Storage Complexity: O(1) as we use a fixed amount of extra space.

Code

    def minimum_sum(num: int) -> int:
    """
    Splits a 4-digit number into two new numbers and returns the minimum possible sum.
    """
    digits = sorted(str(num))
    new1 = int(digits[0] + digits[2])
    new2 = int(digits[1] + digits[3])
    return new1 + new2