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23
Heap Data Structure Challenges Written by Vincent Ngo

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Think you have a handle on heaps? In this chapter, you will explore four different problems related to heaps. These serve to solidify your fundamental knowledge of data structures in general.

Challenge 1: Find the nth smallest integer

Write a function to find the nth smallest integer in an unsorted array. For example:

let integers = [3, 10, 18, 5, 21, 100]

If n = 3, the result should be 10.

Challenge 2: Step-by-Step diagram

Given the following array, visually construct a min heap. Provide a step-by-step diagram of how the min heap is constructed.

[21, 10, 18, 5, 3, 100, 1]

Challenge 3: Combining two heaps

Write a method that combines two heaps.

Challenge 4: A Min Heap?

Write a function to check if a given array is a min heap.

Solutions

Solution to Challenge 1

There are many ways to solve for the nth smallest integer in an unsorted array. For example, you could choose a sorting algorithm you learned about in this chapter, sort the array, and just grab the element at the nth index.

func getNthSmallestElement(n: Int, elements: [Int]) -> Int? {
  var heap = Heap(sort: <, elements: elements) // 1
  var current = 1 // 2
  while !heap.isEmpty { // 3
    let element = heap.remove() // 4
    if current == n { // 5
      return element
    }
    current += 1 // 6
  }
  return nil // 7
}

Solution to Challenge 2

[21, 10, 18, 5, 3, 100, 1]

Solution to Challenge 3

Add this as an additional function of Heap.swift:

mutating public func merge(_ heap: Heap) {
  elements = elements + heap.elements
  buildHeap()
}

Solution to Challenge 4

To check if the given array is a min heap, you only need to go through all the parent nodes of the binary heap. To satisfy the min heap, every parent node must be less than or equal to its left and right child node.

func leftChildIndex(ofParentAt index: Int) -> Int {
  (2 * index) + 1
}

func rightChildIndex(ofParentAt index: Int) -> Int {
  (2 * index) + 2
}
func isMinHeap<Element: Comparable>(elements: [Element]) -> Bool {
  guard !elements.isEmpty else { // 1
    return true
  }
  // 2
  for i in stride(from: elements.count / 2 - 1, through: 0, by: -1) {
    let left = leftChildIndex(ofParentAt: i) // 3
    let right = rightChildIndex(ofParentAt: i)
    if elements[left] < elements[i] { // 4
      return false
    }
    if right < elements.count && elements[right] < elements[i]  { // 5
      return false
    }
  }
  return true // 6
}

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