How do you solve Longest Unequal Adjacent Groups Subsequence I on LeetCode?

Longest Unequal Adjacent Groups Subsequence I (LeetCode #2900) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Greedy approaches often yield optimal solutions for problems involving sequences.

What is the time complexity of Longest Unequal Adjacent Groups Subsequence I?

The optimal solution for Longest Unequal Adjacent Groups Subsequence I runs in O(n) time and uses O(n) space. The time complexity is O(n) because we make a single pass through the words array, and the space complexity is O(n) for storing the result.

What is the brute force solution for Longest Unequal Adjacent Groups Subsequence I?

The brute force approach for Longest Unequal Adjacent Groups Subsequence I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences of the given words and checking which ones meet the alternating condition based on the groups array. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Unequal Adjacent Groups Subsequence I?

Longest Unequal Adjacent Groups Subsequence I uses the following concepts: Array, String, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming, Array.

What are the interview tips for Longest Unequal Adjacent Groups Subsequence I?

Clearly explain your thought process while solving the problem. Start with a brute-force solution to demonstrate understanding before optimizing. Practice identifying patterns in problems to help recognize when to apply specific techniques.

What are common mistakes when solving Longest Unequal Adjacent Groups Subsequence I?

Students often try to optimize too early without understanding the brute-force approach. Some may overlook edge cases, such as all elements being in the same group.

#2900

Longest Unequal Adjacent Groups Subsequence I

Easy

ArrayStringDynamic ProgrammingGreedyGreedyDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy strategy to build the longest alternating subsequence by iterating through the words and selecting them based on the alternating condition directly, which is efficient and straightforward.

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Algorithm

3 steps

1Step 1: Initialize an empty result list and add the first word to it.
2Step 2: Iterate through the words starting from the second word.
3Step 3: For each word, check if its corresponding group is different from the last added word's group in the result. If yes, add it to the result.

solution.py6 lines

1def longestAlternatingSubsequence(words, groups):
2    result = [words[0]]
3    for i in range(1, len(words)):
4        if groups[i] != groups[result[-1]]:
5            result.append(words[i])
6    return result

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the words array, and the space complexity is O(n) for storing the result.

1Greedy approaches often yield optimal solutions for problems involving sequences.
2Understanding the conditions for subsequences is crucial for constructing valid solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.