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

Longest Unequal Adjacent Groups Subsequence II (LeetCode #2901) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps build solutions incrementally.

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

The brute force approach for Longest Unequal Adjacent Groups Subsequence II is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will generate all possible subsequences of indices and check each one for the given conditions. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n²).

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

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

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

Clarify the problem requirements before jumping to code. Think about edge cases, such as all words being the same length. Discuss your thought process while coding to show your understanding.

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

Not checking for unequal groups properly. Overlooking the need for equal lengths before Hamming distance check.

#2901

Longest Unequal Adjacent Groups Subsequence II

Medium

ArrayStringDynamic ProgrammingDynamic ProgrammingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

We can use dynamic programming to build up the solution by storing the lengths of the longest valid subsequences ending at each index. This reduces the number of checks needed compared to the brute force approach.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] represents the length of the longest valid subsequence ending at index i.
2Step 2: For each index i, check all previous indices j to see if they can form a valid pair with i based on the conditions.
3Step 3: Update dp[i] accordingly and keep track of the maximum length found.

solution.py27 lines

1# Full working Python code
2def longest_unequal_subsequence(words, groups):
3    n = len(words)
4    dp = [1] * n
5    prev = [-1] * n
6
7    for i in range(n):
8        for j in range(i):
9            if (groups[i] != groups[j] and \
10                len(words[i]) == len(words[j]) and \
11                hamming_distance(words[i], words[j]) == 1):
12                if dp[j] + 1 > dp[i]:
13                    dp[i] = dp[j] + 1
14                    prev[i] = j
15
16    max_length = max(dp)
17    index = dp.index(max_length)
18    result = []
19    while index != -1:
20        result.append(words[index])
21        index = prev[index]
22    return result[::-1]
23
24
25def hamming_distance(s1, s2):
26    return sum(c1 != c2 for c1, c2 in zip(s1, s2))
27

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops for checking pairs, but we use O(n) space for the dp and prev arrays to store lengths and previous indices.

1Dynamic programming helps build solutions incrementally.
2Hamming distance checks can be efficiently implemented.

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