- A
The for loop uses an index that becomes invalid after removal.
After removal, elements shift left, so incrementing i causes skipping of the next element.
- B
The list is unmodifiable, so remove throws UnsupportedOperationException.
Why wrong: The list is modifiable as elements are removed.
- C
The remove method is called with the wrong index.
Why wrong: The index is correct at the time of call, but subsequent indices shift.
- D
The for loop uses an iterator that throws ConcurrentModificationException.
Why wrong: This is an index-based loop, not using an iterator.
Question 116 of 519
1Z0-829 Index shifting during removal Practice Question
This 1Z0-829 practice question tests your understanding of controlling program flow. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. A key principle to apply: index shifting during removal. 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 developer writes a method that uses a for loop to iterate over a list of strings and remove elements that match a specific pattern using the list's remove(int index) method. The developer uses an index variable that increments normally. However, after running the method, some elements that should have been removed are still present, and some elements are skipped. The list initially contains [A, B, C, D, E] and the developer expects to remove B and D. After the loop, the list is [A, C, E] as expected? Actually, the developer observes that after the loop, the list contains [A, C, D, E] (D was not removed). What is the most likely cause?
Clue words in this question
Noticing these words before you look at the options changes how you read each choice.
Clue:
"most likely"Why it matters: Probability qualifier — the question wants the most probable cause or outcome, not a guaranteed one. Eliminate low-probability options.
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
✓
The for loop uses an index that becomes invalid after removal.
Option A is correct. When using a for loop with an index variable that increments normally, removing an element via `remove(int index)` causes all subsequent elements to shift left. This means the loop effectively skips the element that moves into the current index after a removal. In this scenario, after removing B at index 1, element C moves to index 1, but the loop increments i to 2, so C is never checked. Then at i=2, the loop sees D (originally at index 3). If the removal condition for D is checked at this point and matches, D would be removed, yielding [A, C, E]. However, the observed result is [A, C, D, E], indicating D was not removed. This can happen if the loop's condition uses a fixed bound (e.g., original size) causing an `IndexOutOfBoundsException` (not shown) or if the removal pattern is based on original indices that become invalid after the first removal. The most likely cause is that the index variable becomes stale after removal, leading to skipped elements or missed removals.
Key principle: Index shifting during removal
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
The for loop uses an index that becomes invalid after removal.
Why this is correct
After removal, elements shift left, so incrementing i causes skipping of the next element.
Clue confirmation
The clue word "most likely" in the question point toward this answer.
Related concept
Index shifting during removal
- ✗
The list is unmodifiable, so remove throws UnsupportedOperationException.
Why it's wrong here
The list is modifiable as elements are removed.
- ✗
The remove method is called with the wrong index.
Why it's wrong here
The index is correct at the time of call, but subsequent indices shift.
- ✗
The for loop uses an iterator that throws ConcurrentModificationException.
Why it's wrong here
This is an index-based loop, not using an iterator.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Oracle often tests the subtlety that removing elements from a list while iterating forward with an index variable causes elements to be skipped due to index shifting, leading candidates to mistakenly think the issue is with the remove method's index or an iterator exception.
Detailed technical explanation
How to think about this question
Under the hood, ArrayList's remove(int index) method calls System.arraycopy() to shift all elements after the removed index left by one position. When using a forward-iterating for loop, the increment of the index variable after removal causes the next element to be skipped. A common fix is to iterate backward (from last index to 0) or decrement the index after removal. This behavior is specific to index-based removal and does not occur when using an Iterator's remove() method, which handles index adjustment internally.
KKey Concepts to Remember
- Index shifting during removal
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
Index shifting during removal
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. Index shifting during removal 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.
Review index shifting during removal, then practise related 1Z0-829 questions on the same topic to reinforce the concept.
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FAQ
Questions learners often ask
What does this 1Z0-829 question test?
Controlling Program Flow — This question tests Controlling Program Flow — Index shifting during removal.
What is the correct answer to this question?
The correct answer is: The for loop uses an index that becomes invalid after removal. — Option A is correct. When using a for loop with an index variable that increments normally, removing an element via `remove(int index)` causes all subsequent elements to shift left. This means the loop effectively skips the element that moves into the current index after a removal. In this scenario, after removing B at index 1, element C moves to index 1, but the loop increments i to 2, so C is never checked. Then at i=2, the loop sees D (originally at index 3). If the removal condition for D is checked at this point and matches, D would be removed, yielding [A, C, E]. However, the observed result is [A, C, D, E], indicating D was not removed. This can happen if the loop's condition uses a fixed bound (e.g., original size) causing an `IndexOutOfBoundsException` (not shown) or if the removal pattern is based on original indices that become invalid after the first removal. The most likely cause is that the index variable becomes stale after removal, leading to skipped elements or missed removals.
What should I do if I get this 1Z0-829 question wrong?
Review index shifting during removal, then practise related 1Z0-829 questions on the same topic to reinforce the concept.
Are there clue words in this question I should notice?
Yes — watch for: "most likely". Probability qualifier — the question wants the most probable cause or outcome, not a guaranteed one. Eliminate low-probability options.
What is the key concept behind this question?
Index shifting during removal
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Last reviewed: Jun 30, 2026
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