Introduction

In this blog post, we will explore how to solve the LeetCode problem "3176" 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 non-negative integer k. A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1]. Return the maximum possible length of a good subsequence of nums. Example 1: Input: nums = [1,2,1,1,3], k = 2 Output: 4 Explanation: The maximum length subsequence is [1,2,1,1,3]. Example 2: Input: nums = [1,2,3,4,5,1], k = 0 Output: 2 Explanation: The maximum length subsequence is [1,2,3,4,5,1]. Constraints: 1 <= nums.length <= 500 1 <= nums[i] <= 109 0 <= k <= min(nums.length, 25)

Explanation

Here's a solution to the problem, focusing on efficiency and clarity:

• Core Idea: We use dynamic programming to explore all possible subsequences. The DP state dp[i][j] represents the maximum length of a good subsequence ending at index i of nums, using at most j allowed differences.
• Transitions: When considering whether to include nums[i] in our subsequence, we either extend an existing subsequence or start a new one. If we extend, we check if it creates a difference and update our j (allowed differences) accordingly.

• Optimization: The constraint k <= min(nums.length, 25) is crucial. It limits the second dimension of the DP table, keeping the space and time complexity manageable.

• Complexity: Time: O(n²k), Space: O(nk), where n is the length of nums and k is the given integer.

Code

    def max_good_subsequence(nums, k):
    n = len(nums)
    dp = [[0] * (k + 1) for _ in range(n)]

    for i in range(n):
        for j in range(k + 1):
            dp[i][j] = 1  # Each element itself is a subsequence of length 1

            for l in range(i):
                cost = 0 if nums[i] == nums[l] else 1
                if j >= cost:
                    dp[i][j] = max(dp[i][j], dp[l][j - cost] + 1)

    return max(dp[i][k] for i in range(n))