Arrays/Stacks/Queues

Single correct

• A284
• B213
• C142
• D71

#include<stdio.h>
int main( )
{
    static int a[ ] = {10, 20, 30, 40, 50};
    static int *p[ ] = {a, a+3, a+4, a+1, a+2};
    int **ptr = p;
    ptr++;
    printf("%d%d", ptr-p, **ptr);
}

Numerical answer

Code segment 1:
    //initialize elements of Y to 0
    //initialize elements X[i][j] of X to i+j
    for(i = 0; i < n; i++)
        Y[i] += X[0][i];

Code Segment 2:
    //initialize elements of Y to 0
    //initialize elements X[i][j] of X to i+j
    for(i = 0; i < n; i++)
        Y[i] += X[i][0];

Single correct

• AOnly S2 is correct
• BOnly S3 is correct
• COnly S1 and S2 are correct
• DOnly S1 and S3 are correct

Single correct

• AA queue cannot be implemented using this stack.
• BA queue can be implemented where ENQUEUE takes a single instruction and DEQUEUE takes a sequence of two instructions.
• CA queue can be implemented where ENQUEUE takes a sequence of three instructions and DEQUEUE takes a single instruction.
• DA queue can be implemented where both ENQUEUE and DEQUEUE take a single instruction each.

C = 100;
for i = 1 to n do
    for j = 1 to n do
    {
        Temp = A[i][j] + C;
        A[i][j] = A[j][i];
        A[j][i] = Temp - C;
    }
for i = 1 to n do
    for j = 1 to n do
        output (A[i][j]);

Single correct

• AThe matrix A itself
• BTranspose of the matrix A
• CAdding 100 to the upper diagonal elements and subtracting 100 from lower diagonal elements of A
• DNone of the above

MultiDequeue(Q){
    m = k
    while (Q is not empty) and (m > 0) {
        Dequeue(Q)
        m = m - 1
    }
}

Single correct

• A$\Theta(n)$
• B$\Theta(n + k)$
• C$\Theta(nk)$
• D$\Theta(n^2)$

Answer key coming soon

Single correct

• Afull: (REAR+1) mod $n$ == FRONT; empty: REAR == FRONT
• Bfull: (REAR+1) mod $n$ == FRONT; empty: (FRONT+1) mod $n$ == REAR
• Cfull: REAR == FRONT; empty: (REAR+1) mod $n$ == FRONT
• Dfull: (FRONT+1) mod $n$ == REAR; empty: REAR == FRONT

Answer key coming soon