Introduction

In this blog post, we will explore how to solve the LeetCode problem "2829" 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 two integers, n and k. An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k. Return the minimum possible sum of a k-avoiding array of length n. Example 1: Input: n = 5, k = 4 Output: 18 Explanation: Consider the k-avoiding array [1,2,4,5,6], which has a sum of 18. It can be proven that there is no k-avoiding array with a sum less than 18. Example 2: Input: n = 2, k = 6 Output: 3 Explanation: We can construct the array [1,2], which has a sum of 3. It can be proven that there is no k-avoiding array with a sum less than 3. Constraints: 1 <= n, k <= 50

Explanation

Here's the solution to the problem:

• Key Idea: Construct the k-avoiding array greedily. Start by including small numbers. If including a number x would create a pair that sums to k (i.e., k - x is already in the array), skip x and move to the next number.
• Construction: Prioritize numbers greater than or equal to k. If n is small enough that all numbers greater than or equal to k aren't needed, then fill in remaining slots with numbers less than k/2.

• Optimization: Stop adding elements once the array reaches the specified length n.

• Complexity: O(n), O(n)

Code

    def k_avoiding_array(n: int, k: int) -> int:
    """
    Given two integers, n and k. An array of distinct positive integers is called a k-avoiding array if there
    does not exist any pair of distinct elements that sum to k. Return the minimum possible sum of a k-avoiding
    array of length n.

    For example:
    k_avoiding_array(5, 4) == 18
    k_avoiding_array(2, 6) == 3
    """
    result = 0
    count = 0
    used = set()
    i = 1

    while count < n:
        if k - i in used:
            i += 1
            continue

        result += i
        used.add(i)
        count += 1
        i += 1

    return result