1Z0-829 Working with Streams and Lambda Expressions Practice Question
A developer writes: list.stream().collect(Collectors.groupingBy(Function.identity(), Collectors.counting())); What is the result type and content?
⚠ Common exam trap
The trap is that candidates may focus on individual correct statements and overlook that option A combines both B and C. Recognizing that the result is both a Map<T, Long> and a frequency map is key.
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
✓
Both B and C.
Option A, 'Both B and C,' is the correct answer because both statements accurately describe the result. Collectors.groupingBy with Function.identity() and Collectors.counting() produces a Map<T, Long> (where T is the element type) containing the frequency of each element, which can also be described as a Map<Object, Long> with occurrence counts. Options B and C individually are true but incomplete; the question expects selection of the combination. Option D is incorrect because groupingBy without a map supplier returns a HashMap, not a ConcurrentMap.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
Both B and C.
Why this is correct
This option is self-referential and meaningless; it does not provide a valid description of the result.
- ✗
Map<T, Long> where T is element type.
Why it's wrong here
Correct. The stream elements are of type T, and groupingBy with identity produces a Map<T, Long>.
- ✗
Map<Object, Long> with number of occurrences of each element.
Why it's wrong here
Correct. The map stores each distinct element as a key and its count as a Long value.
- ✗
A ConcurrentMap.
Why it's wrong here
Incorrect. The default map type is HashMap, not ConcurrentMap.
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