Introduction

In this blog post, we will explore how to solve the LeetCode problem "3202" 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 integer array nums and a positive integer k. A subsequence sub of nums with length x is called valid if it satisfies: (sub[0] + sub[1]) % k == (sub[1] + sub[2]) % k == ... == (sub[x - 2] + sub[x - 1]) % k. Return the length of the longest valid subsequence of nums. Example 1: Input: nums = [1,2,3,4,5], k = 2 Output: 5 Explanation: The longest valid subsequence is [1, 2, 3, 4, 5]. Example 2: Input: nums = [1,4,2,3,1,4], k = 3 Output: 4 Explanation: The longest valid subsequence is [1, 4, 1, 4]. Constraints: 2 <= nums.length <= 103 1 <= nums[i] <= 107 1 <= k <= 103

Explanation

Here's the solution to find the longest valid subsequence, along with an explanation:

• Key Idea: The condition (sub[i] + sub[i+1]) % k == (sub[i+1] + sub[i+2]) % k simplifies to sub[i] % k == sub[i+2] % k. This means that elements at even positions must have the same remainder when divided by k, and elements at odd positions must also have the same remainder when divided by k. We can thus partition the array into even and odd subsequences based on remainders modulo k, and find the longest possible valid sequence.
• Optimization: Iterate through all possible remainders (0 to k-1). For each pair of remainders (remainder for even positions and remainder for odd positions), calculate the length of the valid subsequence.

• Maximization: Keep track of the maximum length found so far.

• Complexity: O(n + k^2), where n is the length of nums and k is the given integer. Space Complexity: O(k).

Code

    def longest_valid_subsequence(nums: list[int], k: int) -> int:
    """
    Finds the length of the longest valid subsequence of nums.

    Args:
        nums: An integer array.
        k: A positive integer.

    Returns:
        The length of the longest valid subsequence of nums.
    """

    counts = {}
    for num in nums:
        remainder = num % k
        counts[remainder] = counts.get(remainder, 0) + 1

    max_length = 0
    for rem1 in range(k):
        for rem2 in range(k):
            if rem1 in counts and rem2 in counts:
                if rem1 == rem2:
                    max_length = max(max_length, counts[rem1])
                else:
                    max_length = max(max_length, counts[rem1] + counts[rem2])
            elif rem1 in counts:
                max_length = max(max_length, counts[rem1])
            elif rem2 in counts:
                max_length = max(max_length, counts[rem2])

    return max_length