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Binary search technique:-
This technique is applied to an ordered list where elements are arranged either in ascending order or descending order. The array is separated into two parts and the item being searched for is compared with item at the centre of the array. If they are equal the search is successful. Or else the search is performed in either the first half or second half of the array. If the middle element is greater than the item
If a node in a binary tree is not containing left or right child or it is a leaf node then that absence of child node can be represented by the null pointers. The space engaged by
Merge sort: Merge sort is a sorting algorithm that uses the idea of split and conquers. This algorithm splits the array into two halves, sorts them separately and then merges t
1) What will call a graph that have no cycle? 2) Adjacency matrix of an undirected graph is------------- on main diagonal. 3) Represent the following graphs by adjacency matr
Sequential files are most frequently utilized in commercial batch oriented data processing where there is the concept of a master file on which details are inserted periodically. F
If a Dequeue is implemented via arrays, then this will suffer with the similar problems which a linear queue had suffered. Program 8 gives the array implementation of Dequeue.
Linked list representations contain great advantages of flexibility on the contiguous representation of data structures. However, they contain few disadvantages also. Data structur
Speed cameras read the time a vehicle passes a point (A) on road and then reads time it passes a second point (B) on the same road (points A and B are 100 metres apart). Speed of t
a. Explain the sum of subset problem. Apply backtracking to solve the following instance of sum of subset problem: w= (3, 4, 5, 6} and d = 13. Briefly define the method using a sta
In file access: what is the difference between serial, sequential and indexed sequential searching
Program segment for All pairs shortest paths algorithm AllPairsShortestPaths(int N, Matrix C, Matrix P, Matrix D) { int i, j, k if i = j then C[i][j] = 0 for ( i =
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