Problem Informations

• Problem Index: 2900
• Problem Link: Longest Unequal Adjacent Groups Subsequence I
• Topics: Array, String, Dynamic Programming, Greedy

Problem Description

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 $[0, 1, \ldots, n - 1]$ denoted as $[i_0, i_1, \ldots, i_{k-1}]$, such that $groups[i_j] \neq groups[i_{j+1}]$ for each $0 \leq j < k - 1$ 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 $groups[0] \neq groups[2]$. Another subsequence that can be selected is ["a","b"] because $groups[1] \neq groups[2]$. 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 $groups[0] \neq groups[1]$ and $groups[1] \neq groups[2]$. Another subsequence that can be selected is ["a","b","d"] because $groups[0] \neq groups[1]$ and $groups[1] \neq groups[3]$. It can be shown that the length of the longest subsequence of indices that satisfies the condition is $3$.

Constraints:

• $1 \leq n == words.length == groups.length \leq 100$
• $1 \leq words[i].length \leq 10$
• $groups[i]$ is either $0$ or $1$.
• words consists of distinct strings.
• $words[i]$ consists of lowercase English letters.

Intuition

The task requires us to find the longest alternating subsequence from a list of words, which is dictated by an associated binary array indicating the “group” of each word. The primary objective is to select words such that the group of any consecutive selected words differs. This alternating pattern is crucial for the subsequence. Given this, our approach will capitalize on the constraints: ensuring each consecutive pair in the subsequence has differing group values to maximize the length of this subsequence.

Approach

The solution iteratively examines each word and its associated group value in the given arrays. It initiates the process by including the first word, naturally establishing the starting point of our subsequence. As it progresses through the list of words, it compares the group value of the current word to the previously selected word’s group value:

1. Initial Setup: We start by establishing a list for our result which initially contains the first word, as this sets the foundational element of our subsequence.

2. Iterative Evaluation:

   • Iterate through the words array starting from the second element.
   • For each word, check if its corresponding group (from the groups array) differs from the group of the previously included word in the result subsequence.
   • If the group is different, append the current word to our result list, thereby extending our subsequence.
   • Update the record of the last encountered group to reflect the group of the newly added word.

3. Result Compilation: Once the iteration over the words is complete, the resultant list will contain the longest possible subsequence that adheres to the alternating group condition.

The algorithm operates linearly with respect to the number of words, examining each one only once while making each decision based on immediate comparisons. This straightforward greedy approach efficiently constructs the longest alternating subsequence by capitalizing on each qualifying opportunity to extend the sequence.

Code

C++

class Solution {
public:
    vector<string> getLongestSubsequence(vector<string>& words, vector<int>& groups) {
        int len = words.size();  // Length of the words array
        int last_group = groups[0];  // Initialize the last group status with the first element's group
        vector<string> result = {words[0]};  // Start the result with the first word

        for (int i = 1; i < len; ++i) {
            // Check if the current group is different from the last group
            if (groups[i] != last_group) {
                result.emplace_back(words[i]);  // Add the word to the result if it alternates
                last_group = groups[i];  // Update the last group status
            }
        }
        return result;  // Return the longest alternating subsequence
    }
};

Python

class Solution:
    def getLongestSubsequence(self, words: list[str], groups: list[int]) -> list[str]:
        len_ = len(words)  # Length of the words array
        last_group = groups[0]  # Initialize the last group status with the first element's group
        result = [words[0]]  # Start the result with the first word

        for i in range(1, len_):
            # Check if the current group is different from the last group
            if groups[i] != last_group:
                result.append(words[i])  # Add the word to the result if it alternates
                last_group = groups[i]  # Update the last group status
        
        return result  # Return the longest alternating subsequence

Complexity

• Time complexity: $O(n)$
  The algorithm processes each element of the input arrays words and groups once in a single pass, leading to a time complexity of $O(n)$, where $n$ is the number of elements in the arrays. For each element, it checks a simple condition and potentially appends a word to the result array.
• Space complexity: $O(n)$
  The space complexity is $O(n)$ due to the space required to store the result array. In the worst-case scenario, the result array could contain up to $n$ elements, as each word could potentially be part of the alternating subsequence. The space used by auxiliary variables and input parameters is negligible in comparison.