Top MCQs on Array Data Structure with Answers

An array is a collection of items stored at contiguous memory locations. The idea is to store multiple items of the same type together. This makes it easier to calculate the position of each element by simply adding an offset to a base value, i.e., the memory location of the first element of the array (generally denoted by the name of the array). More On Array Data Structure
Array Quiz

Array Quiz

Question 1

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?
 

Tick

An array of 50 numbers
 

Cross

An array of 100 numbers
 

Cross

An array of 500 numbers
 

Cross

A dynamically allocated array of 550 numbers
 



Question 1-Explanation: 

An array of 50 numbers is correct.

Question 2

What will the output of the below code?

C++

#include <iostream>
using namespace std;

int main()
{

    int arr[2] = { 1, 2 };
    cout << 0 [arr] << ", " << 1 [arr] << endl;
    return 0;
}

Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 2};
        System.out.println(arr[0] + ", " + arr[1]);
    }
}
Tick

1, 2

Cross

Syntax error

Cross

Run time error

Cross

None



Question 2-Explanation: 

0[arr]] is a different way to represent array element, which represents the first element of the array.
similarly, 1[arr] is a different way to represent array element, which represents the second element of the array.

Hence the correct output is (A)

Question 3

The minimum number of comparisons required to determine if an integer appears more than n/2 times in a sorted array of n integers is

Cross

Θ(n)

Tick

Θ(logn)

Cross

Θ(n*logn)

Cross

Θ(1)



Question 3-Explanation: 

If you answered Theta(1), then think of examples {1, 2, 2, 2, 4, 4}, {1, 1, 2, 2, 2, 2, 3, 3} The Best way to find out whether an integer appears more than n/2 times in a sorted array(Ascending Order) of n integers, would be binary search approach.

  1. The First occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is i.
  2. The Last occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is j.
  3. Now number of occurrence of that element(count) is (j-i+1). Overall time complexity = log n +log n +1 = O(logn)

See Check for Majority Element in a sorted array This solution is contributed by Nirmal Bharadwaj

Question 4

Let A be a square matrix of size n x n. Consider the following program. What is the expected output? 

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]);
Tick

The matrix A itself

Cross

Transpose of matrix A

Cross

Adding 100 to the upper diagonal elements and subtracting 100 from diagonal elements of A

Cross

None of the above



Question 4-Explanation: 

If we take look at the inner statements of first loops, we can notice that the statements swap A[i][j] and A[j][i] for all i and j. Since the loop runs for all elements, every element A[l][m] would be swapped twice, once for i = l and j = m and then for i = m and j = l. Swapping twice means the matrix doesn’t change.

Question 5
An algorithm performs (logN)1/2 find operations, N insert operations, (logN)1/2 delete operations, and (logN)1/2 decrease-key operations on a set of data items with keys drawn from a linearly ordered set. For a delete operation, a pointer is provided to the record that must be deleted. For the decrease-key operation, a pointer is provided to the record that has its key decreased. Which one of the following data structures is the most suited for the algorithm to use, if the goal is to achieve the best total asymptotic complexity considering all the operations?
Tick
Unsorted array
Cross
Min-heap
Cross
Sorted array
Cross
Sorted doubly linked list


Question 5-Explanation: 
The time complexity of insert in unsorted array is O(1), O(Logn) in Min-Heap, O(n) in sorted array and sorted DLL.
  1. For unsorted array, we can always insert an element at end and do insert in O(1) time
  2. For Min Heap, insertion takes O(Log n) time. Refer Binary Heap operations for details.
  3. For sorted array, insert takes O(n) time as we may have to move all elements worst case.
  4. For sorted doubly linked list, insert takes O(n) time to find position of element to be inserted.
Since number of insertion operations is asymptotically higher, unsorted array is preferred.
Question 6

Consider an array consisting of –ve and +ve numbers. What would be the worst case time complexity of an algorithm to segregate the numbers having same sign altogether i.e all +ve on one side and then all -ve on the other ?

Cross

O(N)

Cross

O(N Log N)

Tick

O(N * N)

Cross

O(N Log Log N)



Question 6-Explanation: 

Here we can use the partition algorithm of quick sort for segregation and answer will be O(N*N). Choose the first element as pivot whatever may be its sign we don’t care and keep an extra index at pivot position .

Question 7
Let A[1...n] be an array of n distinct numbers. If i < j and A[i] > A[j], then the pair (i, j) is called an inversion of A. What is the expected number of inversions in any permutation on n elements ?
Cross
n(n-1)/2
Tick
n(n-1)/4
Cross
n(n+1)/4
Cross
2n[logn]


Question 7-Explanation: 
There are n(n-1)/2 pairs such that i < j. For a pair (ai, aj), probability of being inversion is 1/2. Therefore expected value of inversions = 1/2 * (n(n-1)/2) = n(n-1)/4.
Question 8

A three dimensional array in ‘C++’ 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):

Cross

&A[0][0][0] + w(y * z * q + z * p + r)

Tick

&A[0][0][0] + w(y * z * p + z*q + r)

Cross

&A[0][0][0] + w(x * y * p + z * q+ r)

Cross

&A[0][0][0] + w(x * y * q + z * p + r)



Question 8-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 (B) is correct.

Question 9

Which of the following correctly declares an array?

Tick

int geeks[20];

Cross

int geeks;

Cross

geeks{20};

Cross

array geeks[20];



Question 9-Explanation: 

Option A is correct. Int is the data type used, geeks is the name of the array and 20 is the size of the array.

Question 10
Consider a two dimensional array A[20][10]. Assume 4 words per memory cell, the base address of array A is 100, elements are stored in row-major order and first element is A[0][0]. What is the address of A[11][5] ?
Tick
560
Cross
460
Cross
570
Cross
575


Question 10-Explanation: 
Element A[11][0] is stored at \"Base Address + 11 * 10 * 4\" which is \"Base Address + 440\" = 540. So A[11][5] is stored at 540 + 5*4 = 560.
There are 20 questions to complete.

  • Last Updated : 27 Sep, 2023

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