Ace Test Series: Algorithms - Time Complexity

Question

Which of the following statement is true?

• A. Binary insertion sorting (insertion sort that uses binary search to find each insertion point) requires $O (n \log n)$ total operations.
• B. In the merge-sort execution tree, roughly the same amount of work is done at each level of the tree.
• C. In a min-heap, the next largest element of any element can be found in $O(\log n)$ time.
• D. In a BST, we can find the next smallest element to a given element in $O(1)$ time.

Answer 1

Insertion Sort with Binary search : yes no comparison surely reduced bt no of swaps still be there.. Time complexity remain O(n²).
Time complexity of  Merge Sort T(n) = T(n/2) + O(n) here O(n) is level cost.
In min heap for next max it will take O(n) time .
Let element is X .. then Find X will take O(n) in worst  case then constant time to find next min. Total O(n)

only B is correct .