Sorted arrays and binary search trees can both be used as a dictionary data structure. Suppose that a dictionary is required to store n items that will be inserted in a randomized order.

Which statement is not correct?

Question

Sorted arrays and binary search trees can both be used as a dictionary data structure. Suppose that a dictionary is required to store n items that will be inserted in a randomized order.

Which statement is not correct?

BlackTom AI · Accepted Answer

Question restatement: A dictionary storing n items inserted in random order can be implemented with a sorted array or a binary search tree (BST). Which statement is not correct?

Option 1: Insertion into a sorted array takes O(n) time in the worst case, and insertion into a binary search tree also takes O(n) time in the worst case.
- Sorted array insertion requires shifting elements to make space for the new item, which is indeed O(n) in the worst case. 
- BST insertion in the worst case (when the tree is degenerate, e.g., a linked list due to insert order) also incurs O(n) time. 
- This option is internally consistent with known bounds, so it is a correct statement.

Option 2: Search in a sorted array takes O(log n) time in the average case, and search in a binary search tree also takes O(log n) time in the average case.
- For a sorted array, binary search yields O(log n) average and worst-case search time. The statement mentions average-case; that remains O(log n). 
- For a BST with random insertions, the expected height is O(log n), giving average-case search time O(log n). The statement is correct.

Option 3: Search in a sorted array takes O(log n) time in the worst case, whereas search in a binary search tree takes O(n) time in the worst case.
- Binary search on a sorted array consistently achieves O(log n) in both average and worst cases; there is no O(n) worst-case for a standard binary search on a sorted array. So the first part is correct (O(log n) worst case). 
- In a BST, worst-case search time is O(n) when the tree is highly unbalanced. This part is also correct. Overall, this option is consistent and thus correct.

Option 4: Insertion into a sorted array takes O(n) time in the average case, whereas insertion into a binary search tree takes O(log n) time in the average case.
- Insertion into a sorted array is dominated by shifting elements and remains O(n) in the average case as well as the worst case; stating O(n) for average is acceptable. 
- For a BST with random insertions, the average-case insertion time is O(log n). This aligns with typical expectations for a reasonably balanced BST. 
- This option is consistent and correct.

Option 5: Search in a sorted array takes O(n) time in the worst case, and search in a binary search tree also takes O(n) time in the worst case.
- The claim that a sorted array search can be O(n) in the worst case is inaccurate because binary search on a sorted array yields O(log n) time even in the worst case. This makes the whole statement false.
- While the BST worst-case search time can be O(n) in degenerate cases, the first part alone already contains the incorrect assertion about the sorted array, thereby rendering the entire option incorrect.

Summary: The only statement that is not correct is the one asserting that a sorted array search can take O(n) time in the worst case; in reality, it takes O(log n) worst-case (via binary search). The BST worst-case bound of O(n) is separately true, but it does not rescue the incorrect claim about the sorted array.

COMP10002_2025_SM2 Exam: Foundations of Algorithms (COMP10002_2025_SM2)- Requires Respondus LockDown Browser

Sorted arrays and binary search trees can both be used as a dictionary data structure. Suppose that a dictionary is required to store n items that will be inserted in a randomized order. Which statement is not correct?

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