Question 1
Which of the following algorithms is NOT a divide & conquer algorithm by nature?
A
Euclidean algorithm to compute the greatest common divisor
B
Heap Sort
C
Cooley-Tukey fast Fourier transform
D
Quick Sort
Divide and Conquer    
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Question 1 Explanation: 
Question 2
Consider the following C program
int main() 
{ 
   int x, y, m, n; 
   scanf ("%d %d", &x, &y); 
   /* x > 0 and y > 0 */
   m = x; n = y; 
   while (m != n) 
   { 
      if(m>n) 
         m = m - n; 
      else
         n = n - m; 
   } 
   printf("%d", n); 
}
What does the program compute? (GATE CS 2004)
A
x + y using repeated subtraction
B
x mod y using repeated subtraction
C
the greatest common divisor of x and y
D
the least common multiple of x and y
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Question 2 Explanation: 
This is an implementation of Euclid’s algorithm to find GCD
Question 3
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:
A
3
B
4
C
6
D
9
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Question 3 Explanation: 
Multiplications can be minimized using following order for evaluation of the given expression. p(x) = a0 + x(a1 + x(a2 + a3x))
Question 4
Maximum Subarray Sum problem is to find the subarray with maximum sum. For example, given an array {12, -13, -5, 25, -20, 30, 10}, the maximum subarray sum is 45. The naive solution for this problem is to calculate sum of all subarrays starting with every element and return the maximum of all. We can solve this using Divide and Conquer, what will be the worst case time complexity using Divide and Conquer.
A
O(n)
B
O(nLogn)
C
O(Logn)
D
O(n^2)
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Question 5
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?
A
O(n)
B
O(nLogn)
C
O(LogLogn)
D
O(Logn)
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Question 5 Explanation: 
We can calculate power using divide and conquer in O(Logn) time. See http://www.geeksforgeeks.org/write-a-c-program-to-calculate-powxn/.
Question 6
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
A
1 Only
B
1 & 2 only
C
1, 2 and 3 only
D
1, 2, 3 and 4
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Question 7
The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates xa and xb for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(xb) is very small and then xb 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
A
xb – (fb– f(xa)) fb/ (xb – xa)
B
xa– (fa– f(xa)) fa/ (xb – xa)
C
xb – (fb – xa) fb/ (xb – fb(xa)
D
xa – (xb – xa) fa/ (fb – f(xa))
Divide and Conquer    GATE-CS-2015 (Set 2)    
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Question 8
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
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
(a, left_end, k) and (a+left_end+1, n–left_end–1, k–left_end–1)
B
(a, left_end, k) and (a, n–left_end–1, k–left_end–1)
C
(a, left_end+1, N–left_end–1, K–left_end–1) and(a, left_end, k)
D
(a, n–left_end–1, k–left_end–1) and (a, left_end, k)
Divide and Conquer    GATE-CS-2015 (Set 2)    
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Question 9
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?
A
a1 < a2
B
a1 > a2
C
a1 = a2
D
Depends on the input
Divide and Conquer    GATE 2017 Mock    
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Question 9 Explanation: 
When Divide and Conquer is used to find the minimum-maximum element in an array, Recurrence relation for the number of comparisons is
T(n) = 2T(n/2) + 2 where 2 is for comparing the minimums as well the maximums of the left and right subarrays
On solving, T(n) = 1.5n - 2.
While doing linear scan, it would take 2*(n-1) comparisons in the worst case to find both minimum as well maximum in one pass.
There are 9 questions to complete.

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