Priority Queue (Heap)

Nattee Niparnan

Flash Back

• Still recall BST? and AVL?• What about Hash?

• What they can do?

The need of another ADT

• Queue with priority– Hospital– Bank

• Can’t be realized by a normal queue• Previous ADTs are not efficient enough

Priority Queue (Heap)

Insert Find Delete Find Min

Delete Min

List O(1) O(n) O(n) O(n) O(n)BST,AVL O(lg n) O(lg n) O(lg n) O(lg n) O(lg n)

Ordered List

O( n ) O( n ) O( n ) O( 1 ) O( 1 )

Hash O(1) O(1) O(1) - -Heap O(lg n) - - O(1) O(lg n)

Excel in “find min”, “delete min”

What is Heap?

• Another kind of trees• Which is

– A binary tree (not BST)– (Almost) Complete

Still recall a Binary Tree?

• A Binary Tree is either– An empty node– A node with at most two children– The children are also a Binary Tree

Specialized Binary Tree

• BST– Add

• “all value in nodes of left tree are less than the value of the node”

• “all value in nodes of right tree are more than the value of the node”

• AVL– Add

• The same as BST• The height of the left subtree and the right subtree must not

be more than 1

Specialized Binary Tree

• Heap– Add

A. “all value in nodes of left tree and right tree are more than the value of the node”

B. Tree is “complete binary tree”

What is Complete Binary Tree

• A binary tree which is– Completely filled

• Except the bottom level which must be filled from left to right

What is Complete Binary Tree

• A binary tree which is– Completely filled

• Except the bottom level which must be filled from left to right

Example of a Binary Heap

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Why we need such properties?

• A. because we need to find minimal– Helps us in the process

• B. Also helps us in the process– But also allow us to store a tree on an array

• Easy to identify children and parent

Binary Heap on Array26 40 31 48 50 85 36 107 48 55 57 88

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Identifying child

• Parent at heap[i]– Left child at heap[(i * 2)]– Right child at heap[(i * 2) + 1]

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3

7 8

0

1 2

4 5 6

9 10 11

Identifying parent

• child at heap[i]– Parent at heap[(i – 1) / 2]

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3

7 8

0

1 2

4 5 6

9 10 11

Operation on Heap

• FindMin– Simple, it is at the top

• Insert• DeleteMin

How to insert?

• Where to put?– Have to maintain the property of the heap– Do we know the exact position?

– Solution• Add to the bottom

– Preserve B. but A. might fails• Fix it!!!

Percolate Up (Bubble Up)

• To bring smaller item up• How?

While I am less than my parentswap myself with my parent

Adding 30 (Bubble Up)

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Adding 30 (Bubble Up)

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SWAP!!

Adding 30 (Bubble Up)

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99 51 55 57 88 85

DONE!

SWAP!!

What is the Big O of bubble Up?

• O (lg n)

• But… assume that the value to be add more than the median of all value– The “bigger” half must be the leave of the tree– So, we usually bubble up only one time.

How to Delete

• We know where to delete– At the root– But…. What happen after that?

• Fix it!!

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Headless tree

How to Delete

• Bring someone else as the head– Who?

• The easiest to moved one• The tail

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Headless tree88

How to Delete

• Now, we fix B.– But A. fail

• Fix it

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Headless tree88

Percolate Down (sieve down)

• To bring bigger item down• How?

While I am more than any of my childrenswap myself with “smaller” child

Percolate Up (Bubble Up)

• To bring smaller item up• How?

While I am less than my parentswap myself with my parent

See it in action

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See it in action

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More than?

Which is smaller?

See it in action

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SWAP!!

See it in action

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More than?

Which is smaller?

See it in action

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99 51 55 57Done!!

SWAP!!

What is the Big O of sieve down?

• O (lg n)

• Unlike the bubble up case– The value we put at the top is usually the bigger

one• Hence, most of the time, we will have to actually do the

O(lg n)

That’s it

• Now, we look at the code

Java Code of Heap

public class BinaryHeap<AnyType extends Comparable<? super AnyType>>{ public BinaryHeap( ) public BinaryHeap( int capacity ) public BinaryHeap( AnyType [ ] items )

public void insert( AnyType x )

public AnyType findMin( )

public AnyType deleteMin( )

private static final int DEFAULT_CAPACITY = 10; private int currentSize; private AnyType [ ] array;

private void percolateDown( int hole )}

/** * Construct the binary heap. * @param capacity the capacity of the binary heap. */ public BinaryHeap( int capacity ) { currentSize = 0; array = (AnyType[]) new Comparable[ capacity + 1 ]; }

Constructor

public AnyType findMin( ){ if( isEmpty( ) ) throw new UnderflowException( ); return array[ 1 ];}

FindMin

public void insert( AnyType x ){ if( currentSize == array.length - 1 ) enlargeArray( array.length * 2 + 1 );

// Percolate up int hole = ++currentSize; for( ; hole > 1 && x.compareTo( array[ hole / 2 ] ) < 0; hole /= 2 ) array[ hole ] = array[ hole / 2 ];

array[ hole ] = x;}

Insert & Percolate Up

Loop until root

Father is larger?

Delete Min

public AnyType deleteMin( ){ if( isEmpty( ) ) throw new UnderflowException( );

AnyType minItem = findMin( ); array[ 1 ] = array[ currentSize-- ]; percolateDown( 1 );

return minItem;}

Percolate Downprivate void percolateDown( int hole ){ int child; AnyType tmp = array[ hole ];

For( ; hole * 2 <= currentSize; hole = child ) { child = hole * 2; if( child != currentSize && array[ child + 1 ].compareTo( array[ child ] ) < 0 ) child++; if( array[ child ].compareTo( tmp ) < 0 ) array[ hole ] = array[ child ]; else break; } array[ hole ] = tmp;}

We have two children?

The right one is smaller?

The smaller one is less than me?

Loop until the bottom

Main (testing)

public static void main( String [ ] args ) { int numItems = 10000; BinaryHeap<Integer> h = new BinaryHeap<Integer>( ); int i = 37;

for( i = 37; i != 0; i = ( i + 37 ) % numItems ) h.insert( i ); for( i = 1; i < numItems; i++ ) if( h.deleteMin( ) != i ) System.out.println( "Oops! " + i ); }

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