Introduction

In this blog post, we will explore how to solve the LeetCode problem "2901" 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 an array groups, both arrays having length n. The hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. You need to select the longest subsequence from an array of indices [0, 1, ..., n - 1], such that for the subsequence denoted as [i0, i1, ..., ik-1] having length k, the following holds: For adjacent indices in the subsequence, their corresponding groups are unequal, i.e., groups[ij] != groups[ij+1], for each j where 0 < j + 1 < k. words[ij] and words[ij+1] are equal in length, and the hamming distance between them is 1, where 0 < j + 1 < k, for all indices in the subsequence. Return a string array containing the words corresponding to the indices (in order) in the selected subsequence. If there are multiple answers, return any of them. Note: strings in words may be unequal in length. Example 1: Input: words = ["bab","dab","cab"], groups = [1,2,2] Output: ["bab","cab"] Explanation: A subsequence that can be selected is [0,2]. groups[0] != groups[2] words[0].length == words[2].length, and the hamming distance between them is 1. So, a valid answer is [words[0],words[2]] = ["bab","cab"]. Another subsequence that can be selected is [0,1]. groups[0] != groups[1] words[0].length == words[1].length, and the hamming distance between them is 1. So, another valid answer is [words[0],words[1]] = ["bab","dab"]. It can be shown that the length of the longest subsequence of indices that satisfies the conditions is 2. Example 2: Input: words = ["a","b","c","d"], groups = [1,2,3,4] Output: ["a","b","c","d"] Explanation: We can select the subsequence [0,1,2,3]. It satisfies both conditions. Hence, the answer is [words[0],words[1],words[2],words[3]] = ["a","b","c","d"]. It has the longest length among all subsequences of indices that satisfy the conditions. Hence, it is the only answer. Constraints: 1 <= n == words.length == groups.length <= 1000 1 <= words[i].length <= 10 1 <= groups[i] <= n words consists of distinct strings. words[i] consists of lowercase English letters.

Explanation

Here's a breakdown of the solution:

• Dynamic Programming: We use dynamic programming to find the longest subsequence. dp[i] stores the length of the longest subsequence ending at index i.
• Hamming Distance Check: For each pair of words, we calculate the hamming distance to see if they can be adjacent in the subsequence.

• Group Check: We ensure that adjacent elements in the subsequence have different group assignments.

• Runtime Complexity: O(n² * m), where n is the number of words and m is the maximum length of a word.

• Storage Complexity: O(n), where n is the number of words.

Code

    def longest_subsequence(words, groups):
    n = len(words)
    dp = [1] * n
    predecessor = [-1] * n
    max_len = 1
    end_index = 0

    def hamming_distance(str1, str2):
        if len(str1) != len(str2):
            return float('inf')
        distance = 0
        for i in range(len(str1)):
            if str1[i] != str2[i]:
                distance += 1
        return distance

    for i in range(1, n):
        for j in range(i):
            if groups[i] != groups[j] and len(words[i]) == len(words[j]) and hamming_distance(words[i], words[j]) == 1:
                if dp[j] + 1 > dp[i]:
                    dp[i] = dp[j] + 1
                    predecessor[i] = j
        if dp[i] > max_len:
            max_len = dp[i]
            end_index = i

    result = []
    curr = end_index
    while curr != -1:
        result.append(words[curr])
        curr = predecessor[curr]

    return result[::-1]