LeetCode #2900 — EASY

Longest Unequal Adjacent Groups Subsequence I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a string array words and a binary array groups both of length n.

A subsequence of words is alternating if for any two consecutive strings in the sequence, their corresponding elements at the same indices in groups are different (that is, there cannot be consecutive 0 or 1).

Your task is to select the longest alternating subsequence from words.

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] != 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"]

Explanation: 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.

Step 01

Brute Force Baseline

Problem summary: You are given a string array words and a binary array groups both of length n. A subsequence of words is alternating if for any two consecutive strings in the sequence, their corresponding elements at the same indices in groups are different (that is, there cannot be consecutive 0 or 1). Your task is to select the longest alternating subsequence from words. Return the selected subsequence. If there are multiple answers, return any of them. Note: The elements in words are distinct.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

["c"]
[0]

Example 2

["d"]
[1]

Step 02

Core Insight

What unlocks the optimal approach

This problem can be solved greedily.
Begin by constructing the answer starting with the first number in <code>groups</code>.
For each index <code>i</code> in the range <code>[1, n - 1]</code>, add <code>i</code> to the answer if <code>groups[i] != groups[i - 1]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2900: Longest Unequal Adjacent Groups Subsequence I
class Solution {
    public List<String> getLongestSubsequence(String[] words, int[] groups) {
        int n = groups.length;
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (i == 0 || groups[i] != groups[i - 1]) {
                ans.add(words[i]);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2900: Longest Unequal Adjacent Groups Subsequence I
func getLongestSubsequence(words []string, groups []int) (ans []string) {
	for i, x := range groups {
		if i == 0 || x != groups[i-1] {
			ans = append(ans, words[i])
		}
	}
	return
}

# Accepted solution for LeetCode #2900: Longest Unequal Adjacent Groups Subsequence I
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]]

// Accepted solution for LeetCode #2900: Longest Unequal Adjacent Groups Subsequence I
impl Solution {
    pub fn get_longest_subsequence(words: Vec<String>, groups: Vec<i32>) -> Vec<String> {
        let mut ans = Vec::new();
        for (i, &g) in groups.iter().enumerate() {
            if i == 0 || g != groups[i - 1] {
                ans.push(words[i].clone());
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2900: Longest Unequal Adjacent Groups Subsequence I
function getLongestSubsequence(words: string[], groups: number[]): string[] {
    const ans: string[] = [];
    for (let i = 0; i < groups.length; ++i) {
        if (i === 0 || groups[i] !== groups[i - 1]) {
            ans.push(words[i]);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.