Problem Description

You have two arrays of the same length n:

• words: an array of distinct strings
• groups: a binary array (containing only 0s and 1s)

Your task is to find the longest possible subsequence from words where the subsequence is "alternating". A subsequence is alternating when the corresponding groups values for any two consecutive words in the subsequence are different.

In other words, if you pick words at positions i and j to be consecutive in your subsequence, then groups[i] must be different from groups[j] (one must be 0 and the other must be 1).

For example, if words = ["a", "b", "c", "d"] and groups = [0, 1, 1, 0]:

• A valid alternating subsequence could be ["a", "b", "d"] with groups [0, 1, 0]
• But ["b", "c"] would not be valid since both have group value 1

The solution uses a greedy approach: starting from the first element, include each word whose group value differs from the previously included word's group value. This works because we want the longest subsequence, so we should include a word whenever we can maintain the alternating property.

The implementation checks if the current index is 0 (first element) or if the current group value differs from the previous one (groups[i] != groups[i-1]). If either condition is true, the word at that position is included in the result.

Intuition

The key insight is that we want to maximize the length of our alternating subsequence. Since we're looking for the longest possible subsequence, we should be greedy - whenever we have an opportunity to add a word while maintaining the alternating property, we should take it.

Think about it this way: as we scan through the arrays from left to right, we face a decision at each position - should we include this word or skip it?

If we just included a word with group value 0, the next word we can include must have group value 1. But here's the crucial observation: if we encounter multiple consecutive words with group value 1, which one should we pick? The answer is the first one we encounter. Why? Because picking the first valid option gives us the most flexibility for future selections and doesn't cause us to miss any opportunities.

For example, if we have groups = [0, 1, 1, 0, 0, 1]:

• Starting with index 0 (value 0)
• At index 1, we see 1 - different from previous, so include it
• At index 2, we see 1 - same as previous, so skip
• At index 3, we see 0 - different from our last included (1), so include it
• At index 4, we see 0 - same as previous, so skip
• At index 5, we see 1 - different from our last included (0), so include it

This greedy strategy works because there's no benefit to skipping a valid word and waiting for a later one with the same group value. We're essentially looking for the points where the group values change, and including the word at each transition point. This naturally gives us the longest alternating subsequence.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation uses a simple one-pass greedy algorithm with list comprehension in Python.

The solution iterates through the groups array using enumerate() to get both the index i and the value x at each position. For each position, it checks two conditions:

1. i == 0: Is this the first element? If yes, we always include it since there's no previous element to compare against.

2. x != groups[i - 1]: Is the current group value different from the previous group value? If yes, we have a transition point and should include this word.

If either condition is true, we add words[i] to our result.

The entire solution is elegantly expressed as a list comprehension:

[words[i] for i, x in enumerate(groups) if i == 0 or x != groups[i - 1]]

Let's trace through an example:

• words = ["a", "b", "c", "d", "e"]
• groups = [1, 0, 0, 1, 1]

Iteration process:

• i=0, x=1: Condition i == 0 is true → Include "a"
• i=1, x=0: Condition x != groups[0] (0 ≠ 1) is true → Include "b"
• i=2, x=0: Condition x != groups[1] (0 ≠ 0) is false → Skip "c"
• i=3, x=1: Condition x != groups[2] (1 ≠ 0) is true → Include "d"
• i=4, x=1: Condition x != groups[3] (1 ≠ 1) is false → Skip "e"

Result: ["a", "b", "d"] with corresponding groups [1, 0, 1]

The time complexity is O(n) where n is the length of the arrays, as we make a single pass through the data. The space complexity is O(k) where k is the size of the output array (in the worst case, when groups alternate perfectly, k = n).

Example Walkthrough

Let's walk through a concrete example to understand how the greedy approach works:

Input:

• words = ["cat", "dog", "bird", "fish", "ant", "bee"]
• groups = [0, 1, 1, 0, 0, 1]

Goal: Find the longest alternating subsequence where consecutive words have different group values.

Step-by-step process:

1. Index 0: word = "cat", group = 0

   • This is the first element (i == 0), so we include it
   • Result so far: ["cat"], groups: [0]

2. Index 1: word = "dog", group = 1

   • Previous group was 0, current is 1 (different!)
   • Include "dog"
   • Result so far: ["cat", "dog"], groups: [0, 1]

3. Index 2: word = "bird", group = 1

   • Previous group was 1, current is 1 (same)
   • Skip "bird"
   • Result remains: ["cat", "dog"], groups: [0, 1]

4. Index 3: word = "fish", group = 0

   • Previous group was 1 (from "dog"), current is 0 (different!)
   • Include "fish"
   • Result so far: ["cat", "dog", "fish"], groups: [0, 1, 0]

5. Index 4: word = "ant", group = 0

   • Previous group was 0, current is 0 (same)
   • Skip "ant"
   • Result remains: ["cat", "dog", "fish"], groups: [0, 1, 0]

6. Index 5: word = "bee", group = 1

   • Previous group was 0 (from "fish"), current is 1 (different!)
   • Include "bee"
   • Final result: ["cat", "dog", "fish", "bee"], groups: [0, 1, 0, 1]

Key Insight: Notice how we only include words at "transition points" where the group value changes from the previous one. By greedily taking the first word at each transition, we maximize the length of our alternating subsequence. The final subsequence ["cat", "dog", "fish", "bee"] has length 4, which is the longest possible alternating subsequence for this input.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array groups (which is also the length of the array words). The algorithm iterates through the groups array exactly once using enumerate(), and for each element, it performs constant-time operations: checking if i == 0, comparing x != groups[i - 1], and appending to the result list when the condition is met.

Space Complexity: O(n) in the worst case. The space is used for storing the output list. In the worst-case scenario, when all adjacent elements in groups are different (alternating between 0 and 1), the result list will contain all n words from the input. The list comprehension creates a new list that could potentially hold all elements from the words array.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Problem as Finding Any Alternating Pattern

A common mistake is thinking we need to find subsequences that alternate between 0 and 1 in a specific pattern (like always starting with 0 or always following 0→1→0→1). The problem actually asks for the longest subsequence where consecutive elements have different group values, regardless of the starting value or specific pattern.

Incorrect approach:

# Wrong: Forcing a specific alternating pattern
result = []
expected = 0  # or 1
for i, group in enumerate(groups):
    if group == expected:
        result.append(words[i])
        expected = 1 - expected  # Toggle between 0 and 1

Correct approach: The greedy solution correctly handles any alternating pattern by simply checking if the current group differs from the previous one.

2. Attempting to Skip Elements to Find a "Better" Subsequence

Some might think skipping certain elements could lead to a longer subsequence later. For example, if we have groups [0, 1, 1, 0, 0, 1], one might consider skipping the second 1 to potentially include more elements later.

Why this doesn't work: The greedy approach is optimal because including an element whenever possible (when it differs from the previous) always maximizes the subsequence length. Skipping an includable element never helps since it doesn't change what elements can be included afterward.

3. Off-by-One Errors in Index Checking

When implementing manually without list comprehension, it's easy to make indexing errors:

Incorrect:

result = []
for i in range(1, len(groups)):  # Missing first element!
    if groups[i] != groups[i-1]:
        result.append(words[i])

Correct:

result = [words[0]]  # Don't forget the first element
for i in range(1, len(groups)):
    if groups[i] != groups[i-1]:
        result.append(words[i])

4. Confusing Subsequence with Subarray

Remember that a subsequence doesn't need to be contiguous. The solution correctly handles this by examining all elements in order but only including those that maintain the alternating property.