Question 474 of 519

Working with Streams and Lambda Expressions →hardMultiple ChoiceObjective-mapped

1Z0-829 Working with Streams and Lambda Expressions Practice Question

This 1Z0-829 practice question tests your understanding of working with streams and lambda expressions. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.

A stream pipeline filters strings, sorts them, and returns the first match: .filter(s -> s.length() > 3).sorted().findFirst(). This is inefficient because sorted() processes all elements. Which alternative achieves the same result with better performance?

Clue words in this question

Noticing these words before you look at the options changes how you read each choice.

Clue: "first"
Why it matters: Order matters here. You are being tested on which action comes before the others — not which action is generally useful.

A
.filter(s -> s.length() > 3).min(Comparator.naturalOrder())
min() uses a reduction that processes each element once, maintaining the minimum without sorting.
B
.filter(s -> s.length() > 3).sorted().limit(1).findFirst()
Why wrong: limit(1) after sorted() does not prevent sorting all elements; sorted() still requires the full sorted order.
C
.filter(s -> s.length() > 3).parallel().sorted().findFirst()
Why wrong: parallel does not fix the sorting overhead; sorted() still requires a full sort in parallel, which may not be beneficial.
D
.filter(s -> s.length() > 3).sorted().collect(Collectors.toList()).get(0)
Why wrong: Still sorts all elements, and collect triggers full processing.

Full breakdown with real-world context →

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

✓

.filter(s -> s.length() > 3).min(Comparator.naturalOrder())

Option A is correct because `min(Comparator.naturalOrder())` is a terminal operation that finds the smallest element according to natural ordering without sorting the entire stream. It uses a single pass reduction, which is O(n) in time complexity, whereas `sorted().findFirst()` must sort all elements (O(n log n)) before picking the first. This makes `min()` significantly more efficient for this use case.

Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

✓
.filter(s -> s.length() > 3).min(Comparator.naturalOrder())
Why this is correct
min() uses a reduction that processes each element once, maintaining the minimum without sorting.
Clue confirmation
The clue word "first" in the question point toward this answer.
Related concept
Read the scenario before looking for a memorised answer.
✗
.filter(s -> s.length() > 3).sorted().limit(1).findFirst()
Why it's wrong here
limit(1) after sorted() does not prevent sorting all elements; sorted() still requires the full sorted order.
✗
.filter(s -> s.length() > 3).parallel().sorted().findFirst()
Why it's wrong here
parallel does not fix the sorting overhead; sorted() still requires a full sort in parallel, which may not be beneficial.
✗
.filter(s -> s.length() > 3).sorted().collect(Collectors.toList()).get(0)
Why it's wrong here
Still sorts all elements, and collect triggers full processing.

Common exam traps

Common exam trap: answer the scenario, not the keyword

The trap here is that candidates often assume `sorted().findFirst()` is optimized to stop early, but in Java streams, `sorted()` is a stateful intermediate operation that must process all elements before any downstream operation can begin, making it inherently inefficient for finding a single minimum or maximum.

Detailed technical explanation

How to think about this question

Under the hood, `min()` uses a `BinaryOperator` reduction that compares elements pairwise, maintaining only the current minimum in memory. This contrasts with `sorted()`, which internally uses a `TimSort` algorithm (O(n log n)) and requires storing all elements in an intermediate buffer. In real-world scenarios with large datasets, such as processing millions of log entries, using `min()` instead of `sorted().findFirst()` can reduce memory footprint from O(n) to O(1) and cut processing time dramatically.

KKey Concepts to Remember

Read the scenario before looking for a memorised answer.
Find the constraint that changes the correct option.
Eliminate answers that are true in general but not in this case.

TExam Day Tips

Watch for words such as best, first, most likely and least administrative effort.
Review why wrong options are wrong, not only why the correct option is correct.

Key takeaway

Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.

Real-world example

How this comes up in practice

A practitioner preparing for the 1Z0-829 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.

What to study next

Got this wrong? Here's your next step.

Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.

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FAQ

Questions learners often ask

What does this 1Z0-829 question test?

Working with Streams and Lambda Expressions — This question tests Working with Streams and Lambda Expressions — Read the scenario before looking for a memorised answer..

What is the correct answer to this question?

The correct answer is: .filter(s -> s.length() > 3).min(Comparator.naturalOrder()) — Option A is correct because `min(Comparator.naturalOrder())` is a terminal operation that finds the smallest element according to natural ordering without sorting the entire stream. It uses a single pass reduction, which is O(n) in time complexity, whereas `sorted().findFirst()` must sort all elements (O(n log n)) before picking the first. This makes `min()` significantly more efficient for this use case.

What should I do if I get this 1Z0-829 question wrong?

Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.

Are there clue words in this question I should notice?

Yes — watch for: "first". Order matters here. You are being tested on which action comes before the others — not which action is generally useful.

What is the key concept behind this question?

Read the scenario before looking for a memorised answer.

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