Which of the following that determines the need for the Circular Queue?

Data Structures - Quiz(MCQ)

Follows the FIFO principle

Avoid wastage of memory

Access the Queue using priority

None of the above

Correct Answer : Avoid wastage of memory

Explanation : Avoid wastage of memory. In a linear queue, there are chances of wastage of memory because if the rear is pointing to the last element whereas the front is pointing to the element other than the first element; it means that spaces allocated before the front are free, but it cannot be reused as rear cannot be incremented. In contrast, the last element is connected to the first element in a circular queue; if initial spaces are vacant, then rear can be incremented by using the statement (rear+1) mod max where max is the size of the array. Therefore, we conclude that the circular queue avoids wastage of memory.

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The five items: A, B, C, D, and E are pushed in a stack, one after other starting from A. The stack is popped four items and each element is inserted in a queue. The two elements are deleted from the queue and pushed back on the stack. Now one item is popped from the stack. The popped item is

Correct Answer : D

A three dimensional array in â€˜Câ€™ is declared as int A[x][y][z]. Consider that array elements are stored in row major order and indexing begins from 0. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer):

&A[0][0][0] + w(x * y * p + z * q+ r)

&A[0][0][0] + w(y * z * q + z * p + r)

&A[0][0][0] + w(y * z * p + z*q + r)

&A[0][0][0] + w(x * y * q + z * p + r)

Correct Answer : &A[0][0][0] + w(y * z * p + z*q + r)

Explanation : According to above question we have to find the address of A[p][q][r] To reach pth row we must have to cross 0 to p-1 row i.e. p rows and each rows contains y∗z elements Hence , = y∗z∗p Now to reach qth element in pth row we have to cross q rows and each row contains z(total columns) elements =z∗q to reach rth elements we have to cross r elements in (p+1)th row Total elements to cross =(y∗z∗p+z∗q+r) Now each element occupies m amount of space in memory Therefore total space occupied by these elements = m(y∗z∗p+z∗q+r) Hence , address of A[p][q][r]=base address+ Space Occupied by the Elements Before it. =&A[0][0][0]+m(y*z*p+z*q+r) Hence Option (C) is correct.

A B-Tree used as an index for a large database table has four levels including the root node. If a new key is inserted in this index, then the maximum number of nodes that could be newly created in the process are