# Solving Leetcode Interviews in Seconds with AI: Find the Maximum Length of a Good Subsequence I


	# 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](https://www.chatmagic.app), 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(n<sup>2</sup>k), Space: O(nk), where n is the length of `nums` and k is the given integer.

	
	# Code
	```python
	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))
	```