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double foo(int n)
{
    int i;
    double sum;
    if(n == 0)
    {
        return 1.0;
    }
    else
    {
        sum = 0.0;
        for(i = 0; i < n; i++)
        {
            sum += foo(i);
        }
        return sum;
    }

}

The time complexity of the above code is?

asked in Algorithms by Veteran (319k points) 577 1445 2962
recategorized by | 384 views
<n = ?> : <return value> : <number of times called>
n = 0 : 1 : 1
n = 1 : 1 : 2
n = 2 : 2 : 4
n = 3 : 4 : 8
n = 4 : 8 : 16
n = 5 : 16 : 32
n = 6 : 32 : 64
n = 7 : 64 : 128
n = 8 : 128 : 256
n = 9 : 256 : 512
n = 10 : 512 : 1024

number_of_times_called = pow(2, n-1);

thanks :)

1 Answer

+10 votes
Best answer

Answer is $O(2^n)$

T(n) = \sum_{i=0}^{n-1} T(i) + cn 

The last $cn$ is for the $n$ times the for loop is executing. 

T(0) = 1
T(1) = 2
T(2) =  5
T(3) = 11
T(4) = 23
T(5) = 47

So,  $$T(n) = 2^n + 2^{n-1} - 1 = O(2^n)$$

answered by Veteran (319k points) 577 1445 2962
selected by
explain the recurrence relation???
The sigma term is adding the complexities of all the recursive calls and the "cn" is for the $n$ no. of iterations which also adds to the complexity. If we just do fun(n-1) inside fun(n), recurrence relation will be $$T(n) = T(n-1) + c$$

Here, because of $n$ number of times, a recursive call happens (at the level of n), we have to add $cn$ to the complexity.
Still not clear... can u plz elaborate Sir

 

 

 

I think it will help u

thanks :)

@__PyuriNot clear for me .
When n=2

For ( i=0 ;i < n ; i++)
sum=sum+f(i)

so f(0) is called which returns 1
f(1) will call f(0).

Total of 3  function call and producing a sum of 2

Am i correct ?

In your diagram you showed that their are 5 function calls . How is that possible ?

@arjun Sir , i understood your recurrence .T(n)=Sigma T(i) + cn But didn't follow the solution .

Could you pls explain me how to solve it using

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