GATE | GATE CS 2008 | Question 74

Consider the following C program

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int f1(int n)
{
  if(n == 0 || n == 1)
    return n;
  else
    return (2*f1(n-1) + 3*f1(n-2));
}
  
int f2(int n)
{
  int i;
  int X[N], Y[N], Z[N] ;
  X[0] = Y[0] = Z[0] = 0;
  X[1] = 1; Y[1] = 2; Z[1] = 3;
  for(i = 2; i <= n; i++)
  {
    X[i] = Y[i-1] + Z[i-2];
    Y[i] = 2*X[i];
    Z[i] = 3*X[i];
  }
  return X[n] ;
}

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The running time of f1(n) and f2(n) are
(A) \theta(n) and \theta(n)
(B) \theta(2^n) and \theta(n)
(C) \theta(n) and \theta(2^n)
(D) \theta(2^n) and \theta(2^n)
(A) A
(B) B
(C) C
(D) D


Answer: (B)

Explanation: See Question 3 of https://www.geeksforgeeks.org/c-language-set-5/

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