Arrays

Question 1

A
4
B
5
C
6
D
7
       Data-Structures       Arrays       GATE 2015 -(Set-2)
Question 1 Explanation: 
Question 2
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?
A
An array of 50 numbers
B
An array of 100 numbers
C
An array of 500 numbers
D
A dynamically allocated array of 550 numbers
       Data-Structures       Arrays       Gate-2005
Question 2 Explanation: 
→ Here we are storing values above 50 and we are ignoring the scores which is less than 50.
→ Then using array of 50 numbers is the best way to store the frequencies.
Question 3
 
A
I, II, and IV only
B
II, III, and IV only
C
II and IV only
D
IV only
       Data-Structures       Arrays       Gate-2003
Question 3 Explanation: 
i) A[2] can be consider as a pointer and this will not give any compile-time error.
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 4
 
A
4040
B
4050
C
5040
D
5050
       Data-Structures       Arrays       Gate-2002
Question 4 Explanation: 
Address for a[40][50] = BaseAddress + [40 * 100 * element size] + [50 * element size]
= 0 + [40 * 100 * 1] + [50 * 1]
= 4000 + 50
= 4050
Question 5
 
A
Rotates s left by k positions
B
Leaves s unchanged
C
Reverses all elements of s
D
None of the above
       Data-Structures       Arrays       Gate-2000
Question 5 Explanation: 
If we perform the three given open operations it will result left rotation by K positions. If we perform n time it will result the initial array.
Question 6
 
A
15i + j + 84
B
15j + i + 84
C
10i + j + 89
D
10j + i + 89
       Data-Structures       Arrays       Gate-1998
Question 6 Explanation: 
The address of element A[i][j] will be,
100 + 15 * (i-1) + (j-1)
= 100 + 15i - 15 + j - 1
= 15i + j + 84
There are 6 questions to complete.