Top MCQs on Divide and Conquer Algorithm with Answers

Top MCQs on Divide and Conquer Algorithm with Answers

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Question 1

Which of the following algorithms is NOT a divide & conquer algorithm by nature?
  • Euclidean algorithm to compute the greatest common divisor
  • Heap Sort
  • Cooley-Tukey fast Fourier transform
  • Quick Sort
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Question 2

Consider the polynomial p(x) = a0 + a1x + a2x^2 +a3x^3, where ai != 0, for all i. The minimum number of multiplications needed to evaluate p on an input x is:
  • 3
  • 4
  • 6
  • 9
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Question 3

Consider a situation where you don't have function to calculate power (pow() function in C) and you need to calculate x^n where x can be any number and n is a positive integer. What can be the best possible time complexity of your power function?
  • O(n)
  • O(nLogn)
  • O(LogLogn)
  • O(Logn)
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Question 4

Consider the problem of searching an element x in an array 'arr[]' of size n. The problem can be solved in O(Logn) time if. 1) Array is sorted 2) Array is sorted and rotated by k. k is given to you and k <= n 3) Array is sorted and rotated by k. k is NOT given to you and k <= n 4) Array is not sorted
  • 1 Only
  • 1 & 2 only
  • 1, 2 and 3 only
  • 1, 2, 3 and 4
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Question 5

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x

a

and x

b

for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(x

b

) is very small and then x

b

is the solution. The procedure is given below. Observe that there is an expression which is missing and is marked by? Which is the suitable expression that is to be put in place of? So that it follows all steps of the secant method?

Secant

Initialize: xa, xb, ε, N     // ε = convergence indicator
fb = f(xb) i = 0
while (i < N and |fb| > ε) do
   i = i + 1                 // update counter
   xt = ?                    // missing expression for
                             // intermediate value
   xa = xb                   // reset xa
   xb = xt                   // reset xb
   fb = f(xb)                // function value at new xb
end while
if |fb| > ε
  then // loop is terminated with i = N
  write “Non-convergence”
else
  write “return xb”
end if

  • xb – (fb– f(xa)) fb/ (xb – xa)

  • xa– (fa– f(xa)) fa/ (xb – xa)

  • xb – (fb – xa) fb/ (xb – fb(xa)

  • xa – (xb – xa) fa/ (fb – f(xa))

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Question 6

Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?
  • a1 < a2
  • a1 > a2
  • a1 = a2
  • Depends on the input
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Question 7

Consider the following array. Which algorithm out of the following options uses the least number of comparisons (among the array elements) to sort the above array in ascending order?

  • Selection sort
  • Mergesort
  • Insertion sort
  • Quicksort using the last element as pivot
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Question 8

A binary search tree T contains n distinct elements. What is the time complexity of picking an element in T that is smaller than the maximum element in T?

  • Θ(nlogn)

  • Θ(n)

  • Θ(logn)

  • Θ(1)

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Question 9

Suppose you are provided with the following function declaration in the C programming language. 
   int partition (int a[], int n);
The function treats the first element of a[] as a pivot, and rearranges the array so that all elements less than or equal to the pivot is in the left part of the array, and all elements greater than the pivot is in the right part. In addition, it moves the pivot so that the pivot is the last element of the left part. The return value is the number of elements in the left part. The following partially given function in the C programming language is used to find the kth smallest element in an array a[ ] of size n using the partition function. We assume k ≤ n C 
int kth_smallest (int a[], int n, int k)
{
   int left_end = partition (a, n);
   if (left_end+1==k)
   {
       return a [left_end];
   }
   if (left_end+1 > k)
   {
      return kth_smallest (____________________);
   }
   else
   {
      return kth_smallest (____________________);
    }
}
The missing argument lists are respectively
  • (a, left_end, k) and (a+left_end+1, n–left_end–1, k–left_end–1)
  • (a, left_end, k) and (a, n–left_end–1, k–left_end–1)
  • (a, left_end+1, N–left_end–1, K–left_end–1) and(a, left_end, k)
  • (a, n–left_end–1, k–left_end–1) and (a, left_end, k)
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Question 10

Let P be an array containing n integers. Let t be the lowest upper bound on the number of comparisons of the array elements, required to find the minimum and maximum values in an arbitrary array of n elements. Which one of the following choices is correct?

  • t>2n−2

  • t>3⌈n/2⌉ and t≤2n−2

  • t>n and t≤3⌈n/2⌉

  • t>⌈log2(n)⌉ and t≤n

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