Data Structures

Pop operation in Stack

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POP Operation

In this  operation topmost element is deleted from the stack.Before deleting check if TOP = NULL.If yes,it means that stack is empty and no deletion can be done.If an attempt to delete element is made in this case the UNDERFLOW message will be printed on screen.If no,then element is deleted then value of TOP is decremented by 1.

For example

  • We want to delete/pop 100 from this stack(Fig 9).
  • TOP = 4
  • TOP is not equal to null.
  • 100 is popped out/deleted.
  • TOP variable gets decremented to 3.

 

Screen Shot 2014-04-08 at 5.17.45 PM

Fig 9 : Pop Operation in Stack

Program 2

#include<stdio.h>

int main()
{
	int max;
	int stack[10];
	int num,val,i,temp;
	int top = -1;
	//printf("Enter the maximum size of stack");
	scanf("%d",&max);
	//printf("Enter the no of elements in stack");
	scanf("%d",&num);
	//printf("Enter the elements of stack");
	for(i=0;i<num;i++)
	{
		scanf("%d\n",&stack[i]);
		top++;
	}
	temp =top;
	if(top == -1)
	{
		printf("Underflow");
	}
	else
	{
		top =top -1;
		printf("The stack after pop operation is \n");
		for(i=0;i<temp;i++)
		{
		 printf("%d\n",stack[i]);
		top++;
	}
	}
	return 0;
}

Run
 Illustration of Program 2

  • Refer Program 1 for working of for loop and consider the same values.
  • top = 4
  • temp = top i.e temp = 4
  • if( top ==-1) i.e if no element is present in the stack then Underflow message will be printed on screen as no element is present to pop out.
  • top = top -1 =4-1=3.
  • for(i=0;i<temp;i++) - i would iterate till it is less than temp(4).
  • Elements of stack after pop operation is printed.

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Push operation in Stack

Push Operation

In this operation,the element is inserted at the top of stack(Fig 4).In order to insert element we need to check whether TOP = MAX-1.If  yes,then insertion  is not possible.If  element is still  tried for insertion , OVERFLOW message will be printed on screen as the stack do not have extra space to handle new element.If space is present and element is inserted,then the value of top will be incremented by 1 .

For example

  • Say we want to insert/push 800 in the following stack(Fig 4).
  • TOP = 4(index starts from 0)
  • MAX= 13
  • The value of TOP is 4 and MAX-1 is 13-1=12.TOP and MAX are not equal,thus element 800 can be pushed into the stack.
  • After insertion value of TOP gets incremented t0 5.
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Fig 5 : Push Operation in stack

Program 1

#include<stdio.h>
int main()
{
	int stack[10];
	int max;
	int num;
	int top = -1;
	int i;
	int item;
	//printf("Enter the maximum size of stack");
	scanf("%d",&max);
	//printf("Enter the no of elements in stack");
	scanf("%d",&num);
        //printf("Enter the elements of stack");
	for(i=0;i<num;i++)//Fig 7
	{
		scanf("%d\n",&stack[i]);
		top++;
	}

	if(top == max -1)//Fig 8
	{
		printf("Overflow");
	}
	else
	{
		//printf("Enter the item you want to insert\n");
		scanf("%d",&item);
		top = top + 1;//Fig 6
		stack[top] = item;
		printf("The stack after push operation is \n");
		for(i=0;i<=num;i++)
	{
	 printf("%d\n",stack[i]);
		top++;
	}
	}

	return 0;
}

Run

 Illustration of Program 1

  • Integer array of stack with size 10 is defined.
  • max is the maximum capacity of stack(Say 10).
  • num is the actual number of elements in stack(Say 5).
  • Variable top is initialized to -1.
  • i is used in for loop and item is the element we want to push in stack.
  • for loop(Fig 7) and scanf are used to read the values of stack.
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Fig 6 : Push Operation

 

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Fig 7 : Working of for loop

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Fig 8 : Working of if statement

 

 


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Operations on Binomial Heap – Extract-min

In Extract-Min operation node with minimum key is deleted from the binomial heap h.The running time to extract minimum value is  O(log n).The steps followed  are :

  • Find the root (say x) with minimum key.
  • Delete the root.
  • Break the binomial heap into h and h’.
  • Perform the union operation to h and h’.

Given the binomial heap h in Fig 1.The first step is to find the root with minimum key.Node with value 1 is smallest so it is deleted and h is broken down into h and h’(Fig 2).

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Fig 1 : Binomial heap h

Fig 2 : Binomial heaps h and h'

Fig 2 : Binomial heaps h and h’

 

Perform the union operation on h and h’.

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Fig 3 : Union on h and h’

 

After the union operation on h and h’ the structure seems like as given in Fig 4

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Fig 4 : Structure after union of h and h’

 

Different cases are performed according to the situation.x points to 11,next-x points to 2,sibling[next-x] points to 3 as shown in Fig 5

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Fig 5 : Structure after union of h and h’

 

Case 1 is applied in Fig 6.x now points to 2,next-x points to 3 and sibling[next-x] points to 7.

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Fig 6 : Case 1 applied

 

Case 3 is applied to Fig 7.Trees with root 2 and 3 are of same order i.e. 1 and 2 is less than 3 so 3 becomes the child of 2.next-x will now point to 7 and sibling[next-x] will point to 4.

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Fig 7 : Case 3 applied

 

 

Case 3 is applied again in Fig 8.2 is less than 7 so 7 becomes the child of 2.next-x will now point to 4.

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Fig 8 : Case 3

 

Case 3 is applied in Fig 9 as 2<4 .So 4 becomes the child of 2.x now points to 2 and next-x points to null.Thus the process stops here.

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Fig 9 : Case 3 applied


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By Cusp2207 on June 11, 2014 | Computer Science, Data Structures | 2 comments
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Operations On Binomial Heap – Union

The union of two heaps is the merging root lists by root degree.But if we simply merge two heaps then a problem can arise.There may be a chance that we get two or more trees of same root degree.This violates the property of binomial heap.To deal with this problem we have four cases and solution to these cases respectively. Referring to all figures we choose arbitrary names like :

  • x is the second node of merged heaps.
  • prev-x points to previous node of x.
  • next-x points to next node of x.
  • sibling[next-x] points to next node of next-x.
  • Bk and  B1 are Binomial Heaps of order k and 1 respectively.
  • We apply different cases according to the situation until next-x becomes nil.

Cases in union of binomial heaps

Case 1

If we merge  heaps and have a case like  x points to binomial heap of order k and next-x points to binomial heap of order 1 then ,then we need to shift our pointers towards right.x will now point to binomial heap with order 1.next-k will point to heap with root node d and prev-x will point to heap with root node b and order k.

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Fig 1 : Case 1

 

Case 2

If in union operation the heaps with root b,c,d are of same order k and x points to heap of order k and root node b,next-x points to next heap of x with same order and root node c and sibling[next-x] points to next heap of next-x again with same order and root node d then pointers shift by one. x now points to heap with root node c,next-x points heap with root d and prev-x points to heap with root b.

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Fig 2 : Case 2

 

Case 3

After the union operation,if  b,c are of same order and key[x] is less than key[next[x]] then c becomes the left child of b and x remain on pointing to b but next-z now points to points to d which was earlier pointed by sibling[next-x].

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Fig 3 : Case 3

 

Case 4

The pointer points to same binomial heaps as in case 3 except the difference that key of x is greater than key[next[x]].In this case,b becomes the child of c.c is pointed by x and d is pointed by next-x.

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Fig 4 : Case 4

 

Example

Consider two heaps h1 and h2 as given in Fig 5.The union operation has to be performed on these two heaps.h1 has binomial trees of order 0,1 and 2.h2 has binomial trees of order 0,1 and 3.At the point of union operation both the heaps are combined together but the resultant structure does not follows heap property i.e the resultant structure has binomial trees of same order.In Fig 5,two trees with root 2 and 5 are of order 0 and two trees with roots 1 and 3 are of order 1 violating the property of heaps.In order to make a structure satisfying heap properties we follow case according to the situation.The union of h1 and h2 is shown in Fig 6.

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Fig 5 : Heaps h1 and h2

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Fig 6 : Union of h1 and h2

 

 

In Fig 7,2 is less than 5 and both the trees are of same order i.e 0 so case 3 is followed in which root pointed by next-x(5) becomes the child of x(2) and pointer next-x shifts to right i.e it shifts to 1 and sibling[next-x] also shifts pointer to right i.e to 3.

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Fig 7 : Case 3 applied

 

In Fig 8,all three binomial trees pointed by x,next-x and sibling[next-x] are of same order i.e 1 so case 2 is followed in which all three pointers shift to right by one i.e x will point to 1,next-x will point to  3 and sibling[next-x] will point to 7.

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Fig 8 : Case 2 applied

 

 

In Fig 9,Case 3 is followed as root 1 is less than root 3 and their trees are of same order.Tree with root 3 becomes child of tree with root 1.

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Fig 9 : Case 3 applied

 

In Fig 10,Case 3 is followed again as root 1 is less than root 7 and their trees are of same order i.e. 2.Tree with root 1 becomes the child of tree with root 7.

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Fig 10 : Case 3 applied

 

In Fig 11,Case 3 is followed as 1 is less than 4 and their trees are of same order i.e.3.Tree with root 1 becomes the child of tree with root 4. The next-x becomes nil  and so the process of applying cases stop here.We get two binomial trees of different orders.They are of order 1 and 4 respectively.Union of heaps is done satisfying all the properties of binomial heap.

Screen Shot 2014-06-10 at 6.15.30 PM

Fig 11 : Case 3 applied

 


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Binomial Heaps

Binomial Heaps are similar to binary heaps with additional feature of implementing binomial series as sequence of trees.The heap(Fig 1) is represented using left child right sibling pointers.It has three links per node (parent,left,right) and the roots of tree are connected using single linked list.The degrees of tree decrease from left to right.

Properties of Binomial Heap

  • Each Binomial tree in the heap obey min-heap property.
  • There can be only 0 or 1 binomial trees for each order .

The root  value of every binomial tree is smaller than its children and there will be at most log n +1 binomial trees of binomial heap with n nodes.

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Fig 1 : Binomial Heap

 

Each binomial tree corresponds to one digit in binary representation of a number n.For e.g.the binary representation of 13(Fig 2) is 1101(23+22+20=8+4+1).The binomial heap will have 13 nodes with binomial trees of order 3,2 and 0 respectively.The orders came from the powers of 2. Either there will be no tree or there will one tree of each order i.e we cannot have any two trees of same order in the binomial heap.Similarly for a number like 15,the binary representation is 1111(23+22 +21 +20 =8+4+2+1).The binomial heap will have 15 nodes(Fig 4)with binomial trees of order 3,2,1,0 respectively.

Binomial Heap of 13

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Fig 2 : Binomial Heap of 13

 

Details of Binomial Heap in Fig 2

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Fig 3 : Details of Binomial Heap in Fig 2

 

Binomial Heap of 15

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Fig 3 : Binomial Heap of 15

 

Details of Binomial heap in Fig 3

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Fig 4 : Details of Binomial Heap in Fig 3

 

Representing Binomial Heaps

The nodes in the binomial heap can be represented by a structure with parent,key,degree,child and sibling attributes ( Fig 5).

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Fig 5 : Structure to represent node of Binomial Heap

 

Example

Consider the binomial heap as given in Fig 6.Say we want to represent 9(Say) then the structure would be like as given in Fig 7 or if we want represent 31 the structure would be like as given in Fig 8

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Fig 6 : Binomial Heap

 

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Fig 7 : Structure to represent node 9

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Fig 8 : Structure to represent node 31

 

Operations on Binomial Heaps

The various operations that can be performed on Binomial heaps are:

  1. Union
  2. Delete Min
  3. Decrease Key
  4. Delete
  5. Insert

 

 


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By Cusp2207 on June 4, 2014 | Computer Science, Data Structures | A comment?
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Binomial Trees

Binomial Trees are one of  the type of trees that are defined  recursively.A Binomial tree of order 0 is a single node and a binomial tree of order n has a root node whose children are roots of binomial trees of order n-1,n-2,n-3,n-4,……3,2,1,0.

Properties of Binomial Tree

  • There are 2n nodes in a binomial tree of order where n is the order and degree of tree(Fig 1).
  • Deleting roots yield binomial trees Bn-1,Bn-2,….0.
  • Bhas \tbinom n d nodes at depth d.
Slide1

Fig 1 : Binomial Tree

Binomial Tree of height/order(n)=0

When height =0 then single node will be present in Binomial tree(Fig 2).

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Fig 2 : Binomial Tree of Order 0

 

 

Binomial Tree at height (n)=1

When n=1 then 2=nodes will be present in the tree(Fig 3).

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Fig 3 : Binomial Tree of order 1

 

 

Binomial Tree of order=2

When n=2 then 2= 4 nodes will be present in the tree.The subtree is binomially  attached to the root node(Fig 4).

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Fig 4 : Binomial Tree of order 2

 

 

Binomial Tree of order 3

If order of tree is 3,then 23  nodes are present in the Binomial tree.The root is connected to subtrees of order 0(green color),1(red) and 2(black)  (Fig 5).

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Fig 5 : Binomial Tree of order 3

 

 

Bn  has  \tbinom n d nodes at depth d.

If we have a binomial tree of order n,then we can trace the number of nodes at any depth by  \tbinom n d.For e.g the no of nodes of binomial tree of order 4 at depth 2 will be 6.(Fig 6).In fig 7,number of nodes at level 2 is 6 and is shown by red highlighted area.

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Fig 6 : Binomial Tree of order 4


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By Cusp2207 on May 30, 2014 | Computer Science, Data Structures | A comment?
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Deletion In B-Trees

In order to delete elements from B-Tree we need to ensure that their properties(Refer : http://letslearncs.com/b-trees/ does not get violated.It should remain a binary search tree and number of pointers must be according to the order and keys should be one less than the order in each and every node.

Example

Consider a B-tree of Fig 1.Say we want to delete 2,21,10,3 and 4 step by step from the tree.

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Fig 1 : B-Tree

 

Delete 2

  • 2 is present in node b.Deletion of 2 do not violate any property of B-tree,so we delete 2 directly from node b.(Fig 2)

 

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Fig 2 : Delete 2

 

Delete 21

  • Deletion of 21 cannot be done directly.It causes node e to underflow so elements are redistributed among c,g and e(Fig 3).
    • 10 remains in c.
    • 12 is stored in g and 13 is stored in e.

 

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Fig 3 : Delete 21

 

Delete 10

  • Deletion of 10 again cannot be done directly.It causes node c to underflow which further causes parent node g to combine with f and a and tree shrinks by one level.3,7 and 9 are stored in a,1 in b,4 and 5 in d,8 in h ,12 and 13 in e (Fig 4).
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Fig 4 : Delete 10

 

Delete 3

  • Deletion of 3 causes 4 to move in a (Fig 5).
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Fig 5 : Delete 3

 

Delete 4

  • Deletion of needs redistribution of the keys in the subtrees of 4; however, nodes b and d do not have enough keys to redistribute without causing an underflow. Thus, nodes b and d are combined together(Fig 6).
Screen Shot 2014-05-29 at 5.47.33 PM

Fig 6 : Delete 4


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B-Trees

The B-tree is a generalization of a binary search tree  in which a node can have more than two children .The B-tree is optimized for systems that read and write large blocks of data.It has following properties

  • The data items are stored at leaves.
  • Every node has between M/2 and M children where M is a positive integer greater than 1.
  • M is the order as well as height of tree.
  • The non leaf nodes store up to M-1 keys.
  • The root is either a leaf or has between two and M children.
  • All leaves are at the same depth and have between [L/2] and L children.

Insertion in B-Tree

The insertion in b-tree is done such that they satisfy all the properties listed above.

Example

Consider a B-Tree of order 4 and we need to insert 5,3,21,9,1,13,2,7,10,12,4,8.Each node can have at most 4 pointers and 3 keys (M-1) or  at least 2 pointers and 1 key.Since it is a binary search tree the keys of left subtree will be less than the root value and keys of right subtree will be greater than root value.

  • Create a node a and insert 5.(Fig 1)
  • Insert 3 to the left of 5 (3<5).
  • Insert 21 to the right of 5(21 >5).
  • In order to insert 9 split node a into b and c as 9 cannot be accommodated in node a(Max keys =3).The left pointer  of 9 points to  3 and 5  ((3 ,5) <9)and the right pointer points to 21(21 >9),
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Fig 1 : Insertion in B-Tree

 

  • Insert 1 in node b (1<3)(Fig 2).
  • Insert 13 in node c(13>9 and 13 <21).
  • 2 cannot be directly inserted in node b
    • In order to insert 2 split b into b and d.
    • 3 gets stored in  node a.
    • Store 1 and 2 in node b.
    • Store 5 in node d.

 

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Fig 2 : Insert 1,13,2

 

  • Insert 7 in node d .(7 >5 and 7<9)(Fig 3)
  • Insert 10 in node c(10 >9 and 10 <13).
  • Now 12 cannot be directly inserted in node c.
    • Split c into c and e.
    • 12 is then inserted in node  c.
    • 21 is stored in e.

 

Screen Shot 2014-05-29 at 3.56.29 PM

Fig 3 : Insert 7,10,12

 

  • Insert 4 in node d(4 >3 and 4 <5).(Fig 4)
  • 8 cannot be inserted in d directly.
    • Split d into f and g
    • Split f into f and h.
    • 8 is inserted in h.
Screen Shot 2014-05-29 at 4.26.04 PM

Fig 4 : Insert 4,8

 


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By Cusp2207 on May 29, 2014 | Computer Science, Data Structures | 1 comment
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Heapsort

The Maxheap sorts the elements in increasing order whereas Minheap sorts the element in decreasing order.To sort the heap three steps have to be followed

  • Heapify -The process picks the largest child key and compare it to the parent key. If parent key is larger than key then heapify quits, otherwise it swaps the parent key with the largest child key. So that the parent  now becomes larger than its children.
  • Build Heap – Use the procedure ‘Heapify’ in a bottom-up fashion to convert an array  into a heap. The process starts from last nodes and not the leaves.
  • Heap Sort – Sorting is done by removing the max element and storing in array .The storage starts from last position and goes till first index of array.

Time Analysis

  • Build Heap Algorithm will run in O(n) time
  • There are n-1 calls to Heapify each call requires O(log n)time
  • Heap sort program combine Build Heap program and Heapify, therefore it has the running time of  O(n log n) time
  • Total time complexity: O(n log n)

Example

Given an array in Fig 1 .We need to convert it to Maxheap and sort it.

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Fig 1 : Array

 

Array is converted to heap in Fig 2

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Fig 2 : Heap

 

The further steps are illustrated through following figures

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Step 1

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Step 2

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Step 3

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Step 4

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Step 5

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Step 6

 

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Step 7

 

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Step 8

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Step 9

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Step 10

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Step 11

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Step 12

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Step 13

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Step 14

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Step 15

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Step 16

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Step 17

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Step 18

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Step 19

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Step 20

 

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Step 21


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By Cusp2207 on May 22, 2014 | Computer Science, Data Structures | A comment?
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Implementation and Merging of Heaps

Implementation

Heaps of n keys can be represented by array of length n+1(Fig 2).For a node at rank i ,the left child is at rank 2i +1 and the right child is at rank  2i +2.The links between nodes are not stored explicitly.

  • In Fig 1,heap has 7 keys i.e n =7.
  • Array will be of length  8(n+1=7+1).

 

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Fig 1 : Heap

 

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Fig 2 : Implementation of Heap

 

Merging two heaps

If two heaps are to be merged and a key(k) has to be inserted then we follow these steps :

  • A new heap is created with root node storing k and two given heaps as subtrees.
  • Perform downheap to restore  relational property of heap.

Example

Given are the two heaps h1 and h2 in Fig 3.We need to merge them and add 5 to it.

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Fig 3 : h1 and h2

 

A new heap h3  is created with root node storing 5 and h1,h2 as subtrees(Fig 4).

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Fig 4 : Merging of h1 and h2 with insertion of key 5.

 

Perform downheap operation to restore heap property(Fig 5).

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Fig 5 : Downheap operation


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