Arrays/Stacks/Queues

int first=0, second=0, len1=0, len2=0, maxlen=0;
for (int i=0; i < n; i++) {
  if (X[i] == first) {
    len2++; len1++;
  } else if (X[i] == second) {
    len2++;
    len1 = (P);
    second = first;
  } else {
    len2 = (Q);
    len1 = 1; second = first;
  }
  if (len2 > maxlen) {
    maxlen = len2;
  }
  first = X[i];
}

Single correct

• A(P) len1+1 (Q) len2+1
• B(P) 1 (Q) len1+1
• C(P) 1 (Q) len2+1
• D(P) len2+1 (Q) len1+1

#include<stdio.h>
int main(int argc, char *argv[])
{
  int a[3][3][3] =
  {{1, 2, 3, 4, 5, 6, 7, 8, 9},
   {10, 11, 12, 13, 14, 15, 16, 17, 18},
   {19, 20, 21, 22, 23, 24, 25, 26, 27}};
  int i = 0, j = 0, k = 0;
  for( i = 0; i < 3; i++ ){
    for(k = 0; k < 3; k++ )
      printf("%d ", a[i][j][k]);
    printf("\n");
  }
  return 0;
}

Single correct

• A1 2 3 / 10 11 12 / 19 20 21 (three lines)
• B1 4 7 / 10 13 16 / 19 22 25 (three lines)
• C1 2 3 / 4 5 6 / 7 8 9 (three lines)
• D1 2 3 / 13 14 15 / 25 26 27 (three lines)

Answer key coming soon

#include <stdio.h>
int main(){
  int arr[4][5];
  int i, j;
  for (i=0; i<4; i++){
    for (j=0; j<5; j++){
      arr[i][j] = 10*i + j;
    }
  }
  printf("%d", *(arr[1] + 9));
  return 0;
}

Single correct

• A14
• B20
• C24
• D30

Answer key coming soon

Single correct

• A$\Theta(1), \Theta(1)$
• B$\Theta(1), \Theta(n)$
• C$\Theta(n), \Theta(1)$
• D$\Theta(n), \Theta(n)$

Single correct

• AI only
• BII only
• CBoth I and II
• DNeither I nor II

Answer key coming soon

Single correct

• ABoth operations can be performed in $O(1)$ time
• BAt most one operation can be performed in $O(1)$ time but the worst case time for the other operation will be $\Omega(n)$
• CThe worst case time complexity for both operations will be $\Omega(n)$
• DWorst case time complexity for both operations will be $\Omega(\log n)$

Answer key coming soon

Single correct

• AUnsorted array
• BMin-heap
• CSorted array
• DSorted doubly linked list

int main () {
    unsigned int x[4][3] =
        {{1,2,3},{4,5,6},{7,8,9},{10,11,12}};
    printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
}

Single correct

• A2036, 2036, 2036
• B2012, 4, 2204
• C2036, 10, 10
• D2012, 4, 6

Single correct

• A284
• B213
• C142
• D71

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