Longest Unequal Adjacent Groups Subsequence I

Longest Unequal Adjacent Groups Subsequence I

What is this problem about?

In this problem, you are given an array of words and a corresponding binary array called groups (containing 0s and 1s). Your goal is to select the longest possible subsequence of words such that no two adjacent words in your selected subsequence belong to the same group. Essentially, the groups must alternate 0, 1, 0, 1 or 1, 0, 1, 0.

Why is this asked in interviews?

This is an entry-level Array and Greedy algorithm problem. It tests a candidate's ability to map elements from two parallel arrays based on a shifting constraint. Interviewers use it to see if candidates can avoid overcomplicating things with Dynamic Programming when a simple linear Greedy scan works perfectly. It assesses basic sequence building logic.

Algorithmic pattern used

This problem perfectly fits the Greedy pattern. Because you just need the longest alternating sequence and the specific words chosen don't have lengths or weights that conflict, you can simply iterate through the arrays and always pick the first available word that belongs to a different group than your previously chosen word.

Example explanation

Words: ["apple", "banana", "cherry", "date"] Groups: [0, 0, 1, 0]

We start building our subsequence:

1. We always take the first element. Subsequence: ["apple"]. Last group used: 0.
2. Look at "banana". Group is 0. We need a 1. Skip it.
3. Look at "cherry". Group is 1. Perfect! Subsequence: ["apple", "cherry"]. Last group used: 1.
4. Look at "date". Group is 0. Perfect! Subsequence: ["apple", "cherry", "date"]. Last group used: 0.

The longest unequal adjacent groups subsequence is ["apple", "cherry", "date"].

Common mistakes candidates make

The biggest mistake candidates make is assuming this requires Dynamic Programming because the word "subsequence" is in the title. Applying DP creates unnecessary $O(N^2)$ time and $O(N)$ space overhead. Because the constraints (groups) only have two states (0 and 1) and no other variables (like string length) matter, the greedy approach of picking the earliest valid transition is mathematically optimal.

Interview preparation tip

When evaluating sequence-building problems, always ask yourself: "Do the elements have varying weights or costs?" If the problem only asks for the longest alternating sequence based on a binary state (0 or 1), a Greedy $O(N)$ algorithm is almost always the solution. Save the heavy DP templates for when the elements carry different numerical values that must be maximized.