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A Stack has an ordered list of elements & an array is also utilized to store ordered list of elements. Therefore, it would be very simple to manage a stack by using an array. Though, the problem along with an array is that we are needed to declare the size of the array before using it in a program. Thus, the size of stack would be fixed. However, an array & a stack are completely different data structures, an array may be utilized to store the elements of a stack. We may declare the array along a maximum size large sufficient to manage a stack. Program1 implements a stack by using an array.
Abstract data type The thing which makes an abstract data type abstract is that its carrier set and its operations are mathematical entities, like geometric objects or numbers;
A graph with n vertices will absolutely have a parallel edge or self loop if the total number of edges is greater than n-1
algorithm for multiplication of two sparse matrices using linked lists..
Decision Tree - ID3 algorithm: Imagine you only ever do one of the following four things for any weekend: go shopping watch a movie play tennis just
The algorithm to delete any node having key from a binary search tree is not simple where as several cases has to be considered. If the node to be deleted contains no sons,
This is a unit of which targeted on the emerging data structures. Red- Black trees, Splay trees, AA-trees & Treaps are introduced. The learner must explore the possibilities of app
Q. Write down an algorithm to insert a node in between any two nodes in a linked list Ans. Insertion of a the node after the given element of the listis as follows
A telephone directory having n = 10 records and Name field as key. Let us assume that the names are stored in array 'm' i.e. m(0) to m(9) and the search has to be made for name "X"
(i) Consider a system using flooding with hop counter. Suppose that the hop counter is originally set to the "diameter" (number of hops in the longest path without traversing any
In a sorted list, Binary search is carried out by dividing the list into two parts depends on the comparison of the key. Since the search interval halves each time, the iteration o
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