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heaps



  • heap is a data structure capable of giving you the smallest (or the largest) element in constant time, while adding or removing the smallest (or the largest) element on logarithmic time.

  • a heap is a binary tree with these properties:

    • it must have all of its nodes in a specific order, and
    • its shape must be complete (all the levels of the tree must be completely filled except maybe for the last one and the last level must have the left-most nodes filled, always).
    • a max heap's root node must have all its children either greater than or equal to its children. a min heap is the opposite. duplicate values are allowed.
  • heaps can be represented with linked lists or queues (arrays) or binary trees. in the case of a tree to array heap:

    • the parent node index is given by n / 2
    • the left children index is 2 * n
    • the right children index is 2 * n + 1
    • a node is a leaf when its index > n / 2
  • it's cheaper to heapify an array of data (O(N)) than creating an empty heap and inserting each element ((O(N log(N))). heapify means create a binary tree and then comparing each nodes with their child (and swapping when necessary). leaves are left out. parents node can simply exchange with their smallest child (so the max number of exchanges is N/2).

  • common applications of heap are: sort (heap sort), getting the top-k elements, and finding the kth element.




priority queues


  • a priority queue is an abstract data type with the following properties:

    1. every item has a priority (usually an integer).
    2. an item with a high priority is dequeued before an item with low priority. 3. two items with an equal priority are dequeued based on their order in the queue.
  • priority queues can be implemented with a stack, queue, or linked list data structures. however, heaps are the structure that guarantees both insertion and deletion to have time complexity O(log N) (while maintaining get_max/get_min at O(1)).



min heaps


  • a min heap is a complete binary tree where each node is smaller than its children (the root is the min element).

  • insert: - insert the element at the bottom, by finding the most rightmost node and checking its children: if left is empty, insert there, otherwise, insert on right. - then compare this node to each parent, exchanging them until the tree's properties are correct.

  • extract_min: - first, remove/return the top and then replace the tree's top with its latest element (the bottom most rightmost). - then bubble down, swapping it with one of its children until the min-heap is properly restored - there is no need for order between right and left, so this operation would only take O(log N) runtime.

  • the code below is an example of an ad-hoc heap class in python.


class MinHeap:

    def __init__(self, size):

        self.heapsize = size
        self.minheap = [0] * (size + 1)
        self.realsize = 0

    def add(self, element):

        if self.realsize + 1 > self.heapsize:
            print("Too many elements!")
            return False

        self.realsize += 1
        self.minheap[self.realsize] = element

        index = self.realsize
        parent = index // 2

        while self.minheap[index] < self.minheap[parent] and index > 1:

            self.minheap[parent], self.minheap[index] = self.minheap[index], self.minheap[parent]
            index = parent
            parent = index // 2
    
    def peek(self):

        return self.minheap[1]
    
    def pop(self):

        if self.realsize < 1:
            print("Heap is empty.")
            return False

        else:
            remove_element = self.minheap[1]
            self.minheap[1] = self.minheap[self.realsize]
            self.realsize -= 1
            index = 1

            while index <= self.realsize // 2:

                left_children = index * 2
                right_children = (index * 2) + 1

                if self.minheap[index] > self.minheap[left_children] or \
                   self.minheap[index] > self.minheap[right_children]:

                    if self.minheap[left_children] < self.minheap[right_children]:

                        self.minheap[left_children], self.minheap[index] = self.minheap[index], self.minheap[left_children]
                        index = left_children
                    
                    else:

                        self.minheap[right_children], self.minheap[index] = self.minheap[index], self.minheap[right_children]
                        index = right_children
                else:
                    break

            return remove_element
    
    def size(self):
        return self.realsize
    
    def __str__(self):
        return str(self.minheap[1 : self.realsize + 1])



max heaps


  • a max heap is a complete binary tree where each node is larger than its children (the root is the max element).

  • insert:

    • insert the element at the bottom, at the leftmost node.
    • then compare the node to each parent, exchanging them until the tree's properties are correct.
  • extreact_max:

    • remove/return the top and then replace the tree's top with its bottom rightmost element.
    • swap up until the max element is on the top.
  • the code below is an example of a max heap class built in python:


class MaxHeap:
    def __init__(self, heapsize):

        self.heapsize = heapsize
        self.maxheap = [0] * (heapsize + 1)
        self.realsize = 0

    def add(self, element):

        self.realsize += 1
        if self.realsize > self.heapsize:
            print("Too many elements!")
            self.realsize -= 1
            return False

        self.maxheap[self.realsize] = element
        index = self.realsize
        parent = index // 2

        while self.maxheap[index] > self.maxheap[parent] and index > 1:
            self.maxheap[parent], self.maxheap[index] = self.maxheap[index], self.maxheap[parent]
            index = parent
            parent = index // 2
            
    def peek(self):

        return self.maxheap[1]
    
    def pop(self):

        if self.realsize < 1:
            print("Heap is empty.")
            return False
        else:
            remove_element = self.maxheap[1]
            self.maxheap[1] = self.maxheap[self.realsize]
            self.realsize -= 1
            index = 1

            while (index <= self.realsize // 2):

                left_children = index * 2
                right_children = (index * 2) + 1

                if (self.maxheap[index] < self.maxheap[left_children] or self.maxheap[index] < self.maxheap[right_children]):
                    if self.maxheap[left_children] > self.maxheap[right_children]:
                        self.maxheap[left_children], self.maxheap[index] = self.maxheap[index], self.maxheap[left_children]
                        index = left_children
                    else:
                        self.maxheap[right_children], self.maxheap[index] = self.maxheap[index], self.maxheap[right_children]
                        index = right_children
                else:
                    break
            return remove_element
    
    def size(self):
        return self.realsize
    
    def __str__(self):
        return str(self.maxheap[1 : self.realsize + 1])


heap sort


  • the core concept of the heap sort involves constructing a heap from our input and then repeatedly removing the min/max element to sort the array.
  • the key idea for in-place heap sort involves a balance of these two ideas:
    • building a heap from an unsorted array through a "bottom-up heapification" process, and
    • using the heap to sort the input array
  • heapsort traditionally uses a max-heap to sort the array, although a min-heap also works.
  • this is not a stable sort.


top k elements problem (O(klog(N) + N))


  1. construct a max (min) heap and add all elements into it (O(N))
  2. traverse and delete the top element, storing the value into a resulting array
  3. repeat 2. until all k elements are removed (O(k * log(N)))


the kth-element problem


  1. construct a max (min) heap and add all elements into it (O(N))
  2. traverse and delete the top element
  3. repeat 2. until k-th largest (smallest) is found (O(k * log(N)))