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Data Structures and Algorithms

(CS210/ESO207/ESO211)

Lecture 25 Binary Heap : Operations: Insert(), DecreaseKey(), Merge()

Buildinga Binary heap on n elements in O(n) time.

Applications of Binary heap : sorting

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Recap from the last lecture

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Heap

Definition:a tree data structure satisfying the following property:

value stored in a node is smaller than the value stored in its children.

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Operations on a heap

Query Operations

Find-min: report the smallest key stored in the heap.

Update Operations CreateHeap(H): Create an empty heap H.

Insert(x,H): Insert a new key with value xinto the heap H.

Extract-min(H): delete the smallest key from H.

Decrease-key(p, , H): decrease the value of the key pby amount .

Merge(H1,H2): Merge two heaps H1and H2.

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A complete binary tree

Let vbe a node with label.

Label of left child(v) =

Label of right child(v) =

Label of parent(v) = 5

0 1

Level nodes

1 2

2 4

0

1 2

3 4 5 6

7 8 9 10 11

2 +1

2 +2

( 1)/2

# leaf nodes = /

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Binary heap

a complete binary tree satisfyingheap property at each node.

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H

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Implementation of a Binary heap

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If nis the maximum number of keys in the binary heap at any moment of

time, then we keep

H : an array of size nused for storing the binary heap. size: a variable for the total number of keys currently in the heap.

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Insert(x,H)

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Insert(x,H)

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Insert(x,H)

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Insert(x,H)

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Insert(x,H)

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Insert(x,H)

Insert(x,H)

{ size(H);

H()x;

size(H)size(H)+ 1;

While( ?? and ?? )

{

??

??

}}

Time complexity: O(log n)

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H() 0

( 1)/2 ;

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Other heaps

There are heaps which achieve much better time complexities than Binary

heap. One such heap is Fibonacci heap. Fibonacci heap is a link based data

structure.

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Binary heap Fibonacciheap

Find-min(H) O(1) O(1)

Insert(x,H) O(logn) O(1)

Extract-min(H) O(log n) O(log n)

Decrease-key(p, , H) O(log n) O(1)Merge(H1,H2) O(n) O(1)

Looking at the amazing bounds in the 3rdcolumn, dont

you feel excited to study Fibonaccci heap ?

We shall study it during CS345.

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Building a Binary heap

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Building a Binary heap

Problem:Given nelements {0, , }, build a binary heap Hstoring them.

Trivial solution: (Building the Binary heap incrementally)

CreateHeap(H);

For(= 0to )

Insert(,H);

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Building a Binary heap incrementally

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Time complexity of incrementally building a Binary heap

Theorem: Time complexity of building a binary heap incrementally is O(nlogn).

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Exploring the way to build a Binary heap in O(n) time

Question1:what is the cause of O( log)time complexity for incrementally building heap ?

Answer: # leaf nodes = / , the time complexity for inserting each of them takes O(log )

time.

Aim:an algorithm where leaf nodes take O(1) time.

Hint:Revisit the heap property: Every node stores value smaller than its children

Question2: In a complete binary tree (not necessary a heap), how many nodes satisfy heap

property ?Answer: all leaf nodes(since none of them has any child).

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What approach for building a heap comes to your mind based

on the above questions and their answers ? Ponder over it

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A new approach to build binary heap

1. Just copy the givenelements {0, , } into an array H.

2. The heap property holds for all the leaf nodes in the corresponding

complete binary tree.

3. Leaving all the leaf nodes, process the elements in the decreasing order

of their numbering and set the heap property for each of them.

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A new approach to build binary heap

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The first node to beprocessed.

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A new approach to build binary heap

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The second node tobe processed.

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A new approach to build binary heap

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The third node tobe processed.

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A new approach to build binary heap

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v

Let v be a node corresponding to index iin H.The process of restoring heap property at icalled Heapify(i,H).

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Heapify(,H)

Heapify(,H)

{ size(H) -1;

While( ?? )

{ ;

If(H[]>H[ + ]) + ;

If( + and H[]>H[ + ]) + ;

If( )

H()H();

elseFlag false;

;

}

}28

and Flag= true (1)/2

Flag true;

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Building Binary heap in O(n) time

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Time to heapify node v ?

v

Height(v)

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Building Binary heap in O(n) time

Time complexity of heapifying a node v= height(v)

Question:How many nodes of height are there in a Binary heap of size n ?

Answer: Let v be a node of height in the binary heap of size n.

The subtree(v) is surely a complete binary tree of height .

So subtree(v) must store > elements.

All subtrees of height are mutually disjoint. So the total number of elements

stored in all of them is bounded by .

Hence the number of subtrees of height is bounded by

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Time complexity to build the heap =

()log

=

=

log

=

= O()

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As an exercise (using knowledge from your JEE

preparation days), show that

log = is bounded by 2

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Sorting using a Binary heap

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Sorting using heap

Build heap H on the given nelements;

While(His not empty)

{ xExtract-min(H);print x;

}

This is HEAP SORT algorithm

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Heap sort

Advantages over merge sort and quick sort:

A non-recursive algorithm.

No extra space.

Disadvantage:

Average case number of comparisons performed during heap sort is more (by a

constant factor) than average number of comparisons performed during quick sort. Soquick sort outperformsheap sort in real applications.

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