Question 269 of 481
1Z0-811 Arrays and Methods Practice Question
In a login system, the authentication method receives an array of user roles as a String[] and checks if a specific role is present. The array may be large (thousands of roles), and the method is called frequently for each user request. Performance is critical. The array is static and does not change after initialization. Which approach is most efficient for repeated checks?
⚠ Common exam trap
Oracle often tests the misconception that sorting and binary search is the most efficient approach for repeated lookups, but they overlook the upfront sorting cost and the fact that HashSet provides O(1) average-time complexity, which is superior for static, frequently queried data.
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
✓
Convert the array to a HashSet once and reuse it for lookups.
Converting the array to a HashSet once provides O(1) average-time complexity for subsequent contains() checks, which is far more efficient than O(n) linear search or O(log n) binary search when the method is called frequently on a static array. The HashSet leverages hash codes for direct bucket lookup, making it ideal for repeated membership tests on large, unchanging data sets.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Use a for loop to iterate and compare each string.
Why it's wrong here
O(n) per check, slower for frequent checks.
- ✗
Sort the array and use Arrays.binarySearch().
Why it's wrong here
Sorting is O(n log n) once, but binary search is O(log n). Still, HashSet is faster for large n.
- ✓
Convert the array to a HashSet once and reuse it for lookups.
Why this is correct
HashSet provides O(1) average lookup time.
- ✗
Use a List and the contains() method.
Why it's wrong here
List.contains() is O(n) for ArrayList.
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Last reviewed: Jun 30, 2026
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