#2900

Easy

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Longest Unequal Adjacent Groups Subsequence I

Find the longest alternating subsequence in a string array based on a binary group array.

Problem Statement

You are given a string array words and a binary array groups, both of length n. A subsequence of words is considered alternating if, for any two consecutive strings in the sequence, their corresponding elements in groups are different. In other words, two consecutive elements cannot have the same value in groups.

Your task is to select the longest alternating subsequence from words based on the values in groups that meet the condition described above.

Examples

Example 1

Input: words = ["e","a","b"], groups = [0,0,1]

Output: ["e","b"]

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"]

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.

Solution Approach

State Transition Dynamic Programming

The solution can be modeled using dynamic programming, where we transition between states depending on whether the current element in groups differs from the previous one. We maintain a state for the length of the current alternating subsequence and update it based on the values in groups.

Greedy Approach for Subsequence Selection

A greedy approach can be used by selecting elements that alternate between 0 and 1 in the groups array, ensuring the longest possible subsequence by following this alternating pattern as we traverse the list of words.

Efficient Traversal

To solve this problem efficiently, we can iterate through the list of words in linear time, adjusting our alternating subsequence as we progress, ensuring a solution with a time complexity of O(n).

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity of this solution is O(n), where n is the length of the words array. Since we only traverse the arrays once, and the space complexity is O(1) as we use a constant amount of extra space for maintaining the subsequence length and current state.

What Interviewers Usually Probe

Look for understanding of dynamic programming with state transitions.
Evaluate how well the candidate applies the greedy approach in practice.
Assess whether the candidate can identify and avoid unnecessary computations to maintain efficiency.

Common Pitfalls or Variants

Common pitfalls

Failing to properly track the alternating sequence when encountering elements with the same group value.
Overcomplicating the solution with unnecessary nested loops or data structures.
Not taking advantage of the greedy nature of the problem, leading to suboptimal subsequences.

Follow-up variants

Consider variations where the input contains multiple possible subsequences with different lengths.
Explore alternative solutions where the goal is to find subsequences that maximize both length and distinctness of elements.
Consider a version of the problem with larger input sizes, testing the efficiency of the solution.

FAQ

What is the approach for solving Longest Unequal Adjacent Groups Subsequence I?

The approach involves using dynamic programming with state transitions while applying a greedy strategy to select alternating elements from the groups array.

What is the time complexity of the solution?

The time complexity is O(n), where n is the length of the words array, because the solution only requires a single pass through the list.

Can this problem be solved using a brute-force method?

While a brute-force method can be used, it would result in an inefficient solution. A dynamic programming approach ensures optimal performance and avoids unnecessary computations.

What is a state transition in dynamic programming for this problem?

A state transition refers to updating the subsequence length based on whether the current word can alternate with the previous word according to the groups array.

How does GhostInterview assist with this problem?

GhostInterview helps by providing clear explanations of dynamic programming and greedy strategies, along with practice examples to build confidence in solving the problem.

terminal

Solution

Solution 1: Greedy

We can traverse the array $groups$, and for the current index $i$, if $i=0$ or $groups[i] \neq groups[i - 1]$, we add $words[i]$ to the answer array.

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class Solution:
    def getLongestSubsequence(self, words: List[str], groups: List[int]) -> List[str]:
        return [words[i] for i, x in enumerate(groups) if i == 0 or x != groups[i - 1]]

Continue Practicing

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