GATE | GATE CS 2008 | Question 74

Consider the following C program

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] ;
}

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

(A)

(B)

(C)

(D)

Answer

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