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):

Data Structures - Quiz(MCQ)

&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.

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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

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&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)

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