Arrays
Question 1 |
Consider the following C program:
#include <stdio.h> int main () { int arr [] = {1,2,3,4,5,6,7,8,9,0,1,2,5}, *ip = arr+4; printf ("%d\n", ip[1]); return 0; }
The number that will be displayed on execution of the program is _____.
5 | |
6 | |
7 | |
8 |

We know that arr is a pointer to arr[ ] & hence arr+4 is pointer to 4th index of array (starting from 0 to 4).
Now *ip is a pointer of int type pointing to memory location 108, which is part of arr.
Hence, when we will print ip[1] it will be equivalent to *(ip+1).
Address of ip will be incremented by 1 & value inside 110 will be printed.
Question 2 |
A Young tableau is a 2D array of integers increasing from left to right and from top to bottom. Any unfilled entries are marked with ∞, and hence there cannot be any entry to the right of, or below a ∞. The following Young tableau consists of unique entries.
1 2 5 14 3 4 6 23 10 12 18 25 31 ∞ ∞ ∞
When an element is removed from a Young tableau, other elements should be moved into its place so that the resulting table is still a Young tableau (unfilled entries may be filled in with a ∞). The minimum number of entries (other than 1) to be shifted, to remove 1 from the given Young tableau is ____________.
4 | |
5 | |
6 | |
7 |

Question 3 |
A program P reads in 500 integers in the range [0, 100] representing the scores of 500 students. It then prints the frequency of each score above 50. What would be the best way for P to store the frequencies?
An array of 50 numbers | |
An array of 100 numbers | |
An array of 500 numbers | |
A dynamically allocated array of 550 numbers |
→ Then using array of 50 numbers is the best way to store the frequencies.
Question 4 |
Assume the following C variable declaration
int *A [10], B[10][10];
Of the following expressions
I. A[2] II. A[2][3] III. B[1] IV. B[2][3]
which will not give compile-time errors if used as left hand sides of assignment statements in a C program?
I, II, and IV only | |
II, III, and IV only | |
II and IV only | |
IV only |
ii) A[2][3] This results an integer, no error will come.
iii) B[1] is a base address of an array. This will not be changed it will result a compile time error.
iv) B[2][3] This also results an integer. No error will come.
Question 5 |
Consider the following declaration of a two dimensional array in C:
char a[100][100];
Assuming that the main memory is byte-addressable and that the array is stored starting from memory address 0, the address of a [40][50] is:
4040 | |
4050 | |
5040 | |
5050 |
= 0 + [40 * 100 * 1] + [50 * 1]
= 4000 + 50
= 4050
Question 6 |
Suppose you are given an array s[1...n] and a procedure reverse (s,i,j) which reverses the order of elements in a between positions i and j (both inclusive). What does the following sequence do, where 1 ≤ k ≤ n:
reverse(s, 1, k) ; reverse(s, k + 1, n); reverse(s, l, n);
Rotates s left by k positions | |
Leaves s unchanged | |
Reverses all elements of s | |
None of the above |
Question 7 |
Let A be a two dimensional array declared as follows:
A: array [1 ... 10] [1 ... 15] of integer;
Assuming that each integer takes one memory location, the array is stored in row-major order and the first element of the array is stored at location 100, what is the address of the element a[i][j]?
15i + j + 84 | |
15j + i + 84 | |
10i + j + 89 | |
10j + i + 89 |
100 + 15 * (i-1) + (j-1)
= 100 + 15i - 15 + j - 1
= 15i + j + 84
Question 8 |
Consider the following C program.
#includeint main () { int a [4] [5] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}}; printf (“%d\n”, *(*(a+**a+2) +3)); return (0); }
The output of the program is _______.
19 |
#include
int main()
{
int a[4][5] = { {1,2,3,4,5},
{6,7,8,9,10},
{11,12,13,14,15},
{16,17,18,19,20}
};
printf("%d\n",a); //880 (consider base address = 880)
printf("%d\n",*a); //880
printf("%d\n",**a); //1
printf("%d\n",**a+2); //3
printf("%d\n",a+**a+2); //940
printf("%d\n",*(a+**a+2));//940
printf("%d\n",*(a+**a+2)+3);//952
printf("%d\n",*(*(a+**a+2)+3));//19
return 0;
}

Question 9 |
#include
{
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;
}
1 2 3 10 11 12 19 20 21 | |
1 4 7 10 13 16 19 22 25 | |
1 2 3 4 5 6 7 8 9 | |
1 2 3 13 14 15 25 26 27 |
Hence, 1, 2, 3 will be 1^st row
10 , 11, 12 will be 2^nd row
19, 20, 21 will be 3^rd row
Question 10 |

i > min; j!= (n+i)mod n; A[j + k]; temp; i + 1 ; | |
i < min; j!= (n+i)mod n; A[j + k]; temp; i + 1; | |
i > min; j!= (n+i+k)mod n; A[(j + k)]; temp; i + 1; | |
i < min; j!= (n+i-k)mod n; A[(j + k)mod n]; temp; i + 1; |
Step-1: Observe all options, 4 options they are given 4th and 5th blank is temp and i+1.
So, they given clue that, 4th and 5th options must be temp and i+1.
Step-2: Based on the 4th and 5th options, we can guess that 1st blank is i
Step-4: Assume 2nd blank is correct, we are considering j!= (n+i) mod n this condition in while loop.
→ The meaning of the condition is “j becomes equal to (n+i)mod n then control goes out of the while loop.”
Step-5: The condition (n+i)mod n=i and j is always equal to i because we are assigning the value of i to j in the code segment line 3.
Step-6: based on these constraint we can say that 4th option is correct.
Reason: The control never enters the 2nd while loop. Sometimes it will enter the 2nd while loop, when we shift the numbers. It means total K places left.
Question 11 |
Type COLOGNE : (LIME, PINE, MUSK, MENTHOL); var a : array [COLOGNE] of REAL; | |
var a : array [REAL] of REAL; | |
var a : array [‘A’…’Z’] of REAL; | |
var a : array [BOOLEAN] of REAL; |
Expect the option B, All remaining indexes are not real numbers.
Option A , takes enum value as index which is integer number.
Option C, takes character which is having equivalent decimal value.
Option D, has boolean value as index whose value may be 0 or 1
Question 12 |
1267 | |
1164 | |
1264 | |
1169 |
Base or starting address of the array is 1120.
The address of the 49th element = base address of array + number of elements before current element * size of element
= 1120 + 48 * 3 = 1264
Question 13 |
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]);
The matrix A itself | |
Transpose of matrix A | |
Adding 100 to the upper diagonal elements and subtracting 100 from diagonal elements of A | |
None of the option |

Question 14 |
i+j | |
j+i(i-1)/2
| |
i+j-1 | |
i+j(j-1)/2 |
Now at ith row, the jth element will be at j position.
So the index of (i, j)th element of lower triangular matrix in this new representation is
j = i(i-1)/2
Question 15 |
1 2 3 3 5 5 7 8 | |
1 2 3 4 5 6 7 8 | |
8 7 6 5 4 3 2 1 | |
1 2 3 5 4 6 7 8 |

For loop will execute for the i value 2,3,4,5
For i = 2, x[x[2]] = x[2]
= x[3] = 3 // since x[2] = 3
The array elements representation are

