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Oracle 1Z0-809 Exam - Topic 9 Question 27 Discussion

Given the code fragment:List listVal = Arrays.asList(''Joe'', ''Paul'', ''Alice'', ''Tom'');System.out.println (// line n1);Which code fragment, when inserted at line n1, enables the code to print the count of string elements whose length is greater than three?
D) listVal.stream().filter(x -> x.length()>3).mapToInt(x -> x).count()
A) listVal.stream().filter(x -> x.length()>3).count()
B) listVal.stream().map(x -> x.length()>3).count()
C) listVal.stream().peek(x -> x.length()>3).count().get()

Oracle 1Z0-809 Exam - Topic 9 Question 27 Discussion

Actual exam question for Oracle's 1Z0-809 exam
Question #: 27
Topic #: 9
[All 1Z0-809 Questions]

Given the code fragment:

List listVal = Arrays.asList(''Joe'', ''Paul'', ''Alice'', ''Tom'');

System.out.println (

// line n1

);

Which code fragment, when inserted at line n1, enables the code to print the count of string elements whose length is greater than three?

Show Suggested Answer Hide Answer
Suggested Answer: D

by Johnetta at May 27, 2022, 02:35 PM

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Justine
9 months ago
A is the best option, no doubt about it!
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Vicky
9 months ago
Wait, why would D use mapToInt? That seems weird.
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Mertie
9 months ago
C seems off, peek doesn't return what you expect.
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Tommy
10 months ago
I think B looks good too, but not quite right.
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Gennie
10 months ago
A is definitely the right choice!
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Reita
10 months ago
Option D is confusing to me; I feel like using mapToInt is unnecessary here since we just need a count of strings.
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Mozell
10 months ago
I remember practicing with streams, but option C seems off. I don't think peek is meant for counting like that.
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Luis
10 months ago
I'm not sure about option B; it seems like it might not count correctly since it maps to a boolean instead of filtering.
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Buck
10 months ago
I think option A looks familiar; it uses filter correctly to count elements based on their length.
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Gary
10 months ago
Hmm, I'm not too sure about this one. Let me think it through - I know Cucumber is a BDD framework, and Selenium API is for web automation, but I'm not sure if they have Appian-specific packages. I'll have to double-check the options.
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Vivan
10 months ago
Hmm, I'm a bit unsure about how to approach this. The question provides a lot of specifics, but I'm not sure I fully understand how to apply them to calculate the paid-up value. I'll need to re-read the question and think through the steps carefully.
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Carey
10 months ago
I'm a bit confused by the wording of this question. Let me re-read it carefully and think through the options.
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Kayleigh
10 months ago
I think the third-party DLP integration might be one of the features. We practiced that in the last session, but I'm not totally sure about the second one.
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Azzie
10 months ago
Remember, false billing is about fraudulently making the company spend money on unnecessary things. Option D captures that perfectly.
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Stephanie
1 year ago
I thought this was a question about cooking recipes, not Java programming! Guess I need to brush up on my programming skills before the next exam.
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Brent
1 year ago
B) listVal.stream().map(x -> x.length()>3).count()
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Clorinda
1 year ago
I know, I was surprised too! I need to study more for the next exam.
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Jonell
1 year ago
A) listVal.stream().filter(x -> x.length()>3).count()
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Keneth
1 year ago
Option D looks good, but it's unnecessary to use mapToInt() to convert the String elements to integers. The filter() method alone should do the trick.
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Lina
1 year ago
Option C is interesting, but the peek() method is used for side effects, not for filtering. It wouldn't give us the correct count.
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Jesusa
1 year ago
D) listVal.stream().filter(x -> x.length()>3).mapToInt(x -> x).count()
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Cortney
1 year ago
That's correct. The filter method will only keep elements with a length greater than 3.
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Carin
1 year ago
A) listVal.stream().filter(x -> x.length()>3).count()
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Shenika
1 year ago
Option B doesn't seem right because the map() method is used to transform the elements, not to filter them. The count() method would return the number of elements that are true, which is not what we want.
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Laticia
1 year ago
User1: So, we all agree that option A is the correct answer. It filters the elements based on the length condition and then counts them.
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Johnetta
1 year ago
User3: Yeah, option A is the right choice. It filters the elements with length greater than three and then counts them.
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Rocco
1 year ago
User2: I agree with you. Option B is incorrect because map() is for transformation, not filtering.
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Yasuko
1 year ago
User1: I think option A is correct. It filters the elements based on the length condition and then counts them.
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Jamal
1 year ago
Yes, using mapToInt() after filter() will give us the count of elements with length greater than three.
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Alpha
1 year ago
D) listVal.stream().filter(x -> x.length()>3).mapToInt(x -> x).count()
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Edwin
1 year ago
That makes sense, we want to filter out the elements with length greater than three.
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Nana
1 year ago
A) listVal.stream().filter(x -> x.length()>3).count()
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Janine
1 year ago
I think option A is the correct answer. The stream() method is used to create a stream of the list elements, and the filter() method is used to filter the elements based on the condition x.length() > 3, which gives us the count of the string elements whose length is greater than three.
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Chu
1 year ago
User2: Yes, you're right. The filter() method is used to filter elements based on the condition x.length() > 3.
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Paris
1 year ago
User1: I think option A is the correct answer.
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Venita
1 year ago
I'm not sure about the answer. Can someone explain why option A is correct?
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Elliott
1 year ago
I agree with Ressie. Option A makes sense as it filters the elements based on length and then counts them.
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Ressie
1 year ago
I think the answer is A) listVal.stream().filter(x -> x.length()>3).count(). Because it filters out the elements with length greater than three and then counts them.
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