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Solving Leetcode Interviews in Seconds with AI: Find the Maximum Length of Valid Subsequence I

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "3201" 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. A subsequence sub of nums with length x is called valid if it satisfies: (sub[0] + sub[1]) % 2 == (sub[1] + sub[2]) % 2 == ... == (sub[x - 2] + sub[x - 1]) % 2. Return the length of the longest valid subsequence of nums. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: nums = [1,2,3,4] Output: 4 Explanation: The longest valid subsequence is [1, 2, 3, 4]. Example 2: Input: nums = [1,2,1,1,2,1,2] Output: 6 Explanation: The longest valid subsequence is [1, 2, 1, 2, 1, 2]. Example 3: Input: nums = [1,3] Output: 2 Explanation: The longest valid subsequence is [1, 3]. Constraints: 2 <= nums.length <= 2 * 105 1 <= nums[i] <= 107

Explanation

Here's the breakdown of the solution:

  • Key Insight: The condition (sub[i] + sub[i+1]) % 2 being the same for all i implies that the parities of adjacent elements in the subsequence must either alternate (even, odd, even, odd...) or be the same (even, even, even... or odd, odd, odd...).
  • Counting: We can iterate through the nums array and count the number of even and odd numbers. The longest valid subsequence will either be all even numbers, all odd numbers, or an alternating sequence of even and odd numbers. The alternating sequence can be constructed by simply taking all even and odd numbers available.
  • Optimization: We don't need to actually construct the subsequences; we only need to count the maximum possible length.

  • Complexity: Runtime: O(n), Storage: O(1)

Code

    def longestValidSubsequence(nums):
    """
    Finds the length of the longest valid subsequence of nums.

    Args:
        nums: An integer array.

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

    even_count = 0
    odd_count = 0

    for num in nums:
        if num % 2 == 0:
            even_count += 1
        else:
            odd_count += 1

    if even_count == 0 or odd_count == 0:
        return even_count + odd_count
    else:
        return even_count + odd_count

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