Which of the following option is true if implementation of Queue is from the linked list?

Data Structures - Quiz(MCQ)

In enqueue operation, new nodes are inserted from the beginning and in dequeue operation, nodes are removed from the end.

In enqueue operation, new nodes are inserted from the end and in dequeue operation, nodes are deleted from the beginning.

Both (A) and (B)

In enqueue operation, new nodes are inserted from the end and in dequeue operation, nodes are deleted from the end.

Correct Answer : Both (A) and (B)

Explanation : As we know that Queue has two ends, i.e., one for the insertion and another one for the deletion. If Queue is implemented using Linked list then it can be done in either of the ways.

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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