Introduction

In this blog post, we will explore how to solve the LeetCode problem "3020" 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 positive integers nums. You need to select a subset of nums which satisfies the following condition: You can place the selected elements in a 0-indexed array such that it follows the pattern: [x, x2, x4, ..., xk/2, xk, xk/2, ..., x4, x2, x] (Note that k can be be any non-negative power of 2). For example, [2, 4, 16, 4, 2] and [3, 9, 3] follow the pattern while [2, 4, 8, 4, 2] does not. Return the maximum number of elements in a subset that satisfies these conditions. Example 1: Input: nums = [5,4,1,2,2] Output: 3 Explanation: We can select the subset {4,2,2}, which can be placed in the array as [2,4,2] which follows the pattern and 22 == 4. Hence the answer is 3. Example 2: Input: nums = [1,3,2,4] Output: 1 Explanation: We can select the subset {1}, which can be placed in the array as [1] which follows the pattern. Hence the answer is 1. Note that we could have also selected the subsets {2}, {3}, or {4}, there may be multiple subsets which provide the same answer. Constraints: 2 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

Here's the breakdown of the solution:

• Greedy Approach: Prioritize extending existing sequences that adhere to the pattern. Start by finding pairs (x, x^2), and if found, try to extend the sequence in both directions (x^4, x^8, ...).
• Frequency Counting: Use a dictionary (or Counter) to efficiently track the number of occurrences of each number in the input array. This allows for quick checks if a required element (x^2) exists.

• Iterative Extension: Build the sequences iteratively. For each element, check if its square exists in the remaining available elements. If it does, construct a sequence, remove the elements from the count, and keep extending until no more squares can be found.

• Time Complexity: O(N * log(max(nums))), where N is the length of nums. Space Complexity: O(N)

Code

    from collections import Counter

def solve():
    n = int(input())
    nums = list(map(int, input().split()))
    counts = Counter(nums)
    ans = 0

    for num in sorted(list(counts.keys())):
        if counts[num] > 0:
            counts[num] -= 1
            curr = [num]
            curr_num = num

            while True:
                next_num = curr_num * curr_num
                if next_num in counts and counts[next_num] > 0:
                    curr.append(next_num)
                    counts[next_num] -= 1
                    curr_num = next_num
                else:
                    break

            ans = max(ans, 2 * (len(curr) - 1) + 1 if len(curr) > 1 else 1 if len(curr) == 1 else 0)

    print(ans)

def max_subset_size(nums):
    counts = Counter(nums)
    ans = 0

    for num in sorted(list(counts.keys())):
        if counts[num] > 0:
            counts[num] -= 1
            curr = [num]
            curr_num = num

            while True:
                next_num = curr_num * curr_num
                if next_num in counts and counts[next_num] > 0:
                    curr.append(next_num)
                    counts[next_num] -= 1
                    curr_num = next_num
                else:
                    break

            ans = max(ans, 2 * (len(curr) - 1) + 1 if len(curr) > 1 else 1 if len(curr) == 1 else 0)

    return ans