Question

How long would it take to remove the \(\lceil\log n\rceil\) smallest elements from a heap that contains \(n\) entries using the removeMin() operation?

Short answer

It takes approximately \(O(\log^2 n)\) time to remove the \(\lceil\log n\rceil\) smallest elements from the heap.

Step-by-step solution

- Understanding the Problem

The problem involves removing the \(\lceil\log n\rceil\) smallest elements from a heap containing \(n\) entries using the removeMin() operation. The removeMin() operation in a heap removes the smallest element (usually at the root) and performs a re-heapify operation to maintain the heap property.

- Determine the Number of Operations

Removing \(\lceil\log n\rceil\) elements means we need to call removeMin() \(\lceil\log n\rceil\) times.

- Cost of removeMin() Operation

In a heap, removing the smallest element (removeMin()) requires \(O(\log n)\) time since after removing the root element, we need to re-heapify the heap which takes logarithmic time.

- Total Time to Remove Elements

Since we need to perform removeMin() \(\lceil\log n\rceil\) times and each removeMin() operation takes \(O(\log n)\) time, the total time complexity is \((\lceil\log n\rceil \cdot O(\log n))\).

- Simplify the Expression

While \(\lceil\log n\rceil\) can be approximated to \(\log n\) (since the ceiling function increases the value by at most 1), the total time complexity is approximately \(O(\log^2 n)\).

Key concepts

removeMin operation

The removeMin operation is a fundamental function in the context of heaps. A heap is a specific type of complete binary tree where each parent node is less than or equal to its child nodes (min-heap). The smallest element in a min-heap is always at the root. The removeMin operation takes this smallest element and removes it from the heap.
Once the smallest element is removed, the heap's structure must be maintained. To do this, the last element in the heap is moved to the root position. However, this move disrupts the heap property. As a result, a process known as re-heapification or reheapify must be performed.
During reheapify, the new root element is compared with its children. If the new root is larger than either child, it is swapped with the smaller child. This process repeats until the heap property is restored.
When you remove the smallest element \(x\) times from the heap, you are effectively performing the removeMin operation \(x\) times. Each time, the heap will require reheapification to maintain its properties.

logarithmic time complexity

Logarithmic time complexity is a way to describe an algorithm's efficiency. In the context of heaps, operations like insertion, deletion, and access typically have logarithmic complexity.
This means that the time needed to perform the operation grows logarithmically with the number of elements \(n\) in the heap. Specifically, if the heap contains \(n\) elements, operations like removeMin will take \(O(\text{log} n)\) time.
This efficiency comes from the balanced nature of the complete binary tree structure of heaps. Each level in the tree has approximately half as many nodes as the level above it. As a result, the depth of the tree is proportional to the logarithm of the total number of nodes \(n\), making the operations efficient.

reheapify

Reheapify is an essential process that ensures the heap retains its properties after an element is removed. When you remove the smallest element from a min-heap, the last element in the heap is moved to the root.
This move often disrupts the heap property, and the heap must be rebalanced. Reheapify addresses this by comparing the new root with its children.
Here's a step-by-step of how reheapify works:

• Compare the new root with its children.
• Swap the root with the smaller of the two children if it is larger.
• Repeat the process at the new position of the swapped element until the heap property is restored.

Reheapify is efficient with a complexity of \(O(\text{log} n)\). This makes operations like removeMin feasible even for large heaps.