Given a list of strings, which of the following will produce a Map counting the occurrences of each string?

.collect(Collectors.toMap(Function.identity(), String::length))

.collect(Collectors.toMap(Function.identity(), v->1))

Working with Streams and Lambda Expressions →hardMultiple ChoiceObjective-mapped

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

Given a list of strings, which of the following will produce a Map<String, Long> counting the occurrences of each string?

⚠ Common exam trap

Many exam-takers choose `Collectors.toMap` with a constant value (like `v->1`) without realizing it fails on duplicate keys, or they pick `summingInt` thinking it produces a `Long` when it actually produces an `Integer`, missing the exact type constraint of `Map<String, Long>`.

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

✓

.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))

`Collectors.groupingBy(Function.identity(), Collectors.counting())` groups each string by itself (using `Function.identity()` as the classifier) and then applies a downstream collector `Collectors.counting()` that counts the number of elements in each group, producing a `Map<String, Long>` where each key is a unique string and the value is its frequency.

Answer analysis

Option-by-option breakdown

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

✗
.collect(Collectors.toMap(Function.identity(), String::length))
Why it's wrong here
This maps each string to its length, not counting occurrences.
✓
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
Why this is correct
Correct. groupingBy with counting as downstream collector produces a long count per key.
✗
.collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(e->1)))
Why it's wrong here
summingInt returns Map<String, Integer>, not Long.
✗
.collect(Collectors.toMap(Function.identity(), v->1))
Why it's wrong here
This toMap doesn't handle duplicate keys; it will throw IllegalStateException if duplicates occur.

Go deeper

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

Why each option matters

Option-by-option breakdown

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