Introduction

In this blog post, we will explore how to solve the LeetCode problem "2900" 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 a string array words and a binary array groups both of length n, where words[i] is associated with groups[i]. Your task is to select the longest alternating subsequence from words. A subsequence of words is alternating if for any two consecutive strings in the sequence, their corresponding elements in the binary array groups differ. Essentially, you are to choose strings such that adjacent elements have non-matching corresponding bits in the groups array. Formally, you need to find the longest subsequence of an array of indices [0, 1, ..., n - 1] denoted as [i0, i1, ..., ik-1], such that groups[ij] != groups[ij+1] for each 0 <= j < k - 1 and then find the words corresponding to these indices. Return the selected subsequence. If there are multiple answers, return any of them. Note: The elements in words are distinct. Example 1: Input: words = ["e","a","b"], groups = [0,0,1] Output: ["e","b"] Explanation: A subsequence that can be selected is ["e","b"] because groups[0] != groups[2]. Another subsequence that can be selected is ["a","b"] because groups[1] != groups[2]. It can be demonstrated that the length of the longest subsequence of indices that satisfies the condition is 2. Example 2: Input: words = ["a","b","c","d"], groups = [1,0,1,1] Output: ["a","b","c"] Explanation: A subsequence that can be selected is ["a","b","c"] because groups[0] != groups[1] and groups[1] != groups[2]. Another subsequence that can be selected is ["a","b","d"] because groups[0] != groups[1] and groups[1] != groups[3]. It can be shown that the length of the longest subsequence of indices that satisfies the condition is 3. Constraints: 1 <= n == words.length == groups.length <= 100 1 <= words[i].length <= 10 groups[i] is either 0 or 1. words consists of distinct strings. words[i] consists of lowercase English letters.

Explanation

Here's the solution to the problem:

• Iterate through the groups array.

• Maintain the last selected group value and add the corresponding word to the result if the current group value is different from the last selected group value.

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

Code

    def longest_alternating_subsequence(words, groups):
    """
    Finds the longest alternating subsequence from words based on groups.

    Args:
        words: A list of strings.
        groups: A list of integers (0 or 1).

    Returns:
        A list of strings representing the longest alternating subsequence.
    """

    result = []
    last_group = -1  # Initialize with a value that is not 0 or 1

    for i in range(len(words)):
        if groups[i] != last_group:
            result.append(words[i])
            last_group = groups[i]

    return result