用对数增长计算嵌套循环的时间复杂度

Question

I am learning at my own pace online. I was solving some examples but I can't wrap my mind around this one:

while(i<n)
{
     for(int j=1; j<=i; j++)
         sum = sum + 1;
     i *=2;
}

I think the answer should be 2^n but my friend says nlog(n)

Can someone find the big-O for this loop and explain to me how to do so?

Answer 1

The outer loop will enter it's body log2(n) times, because i is increasing exponentially and thereby reaches the end n faster and faster. For example, if n were 1024 , it would need only 10 iterations, with n=65536 , it were 16 iterations. The accurate count is log2(n) , but in terms of runtime complexity the logarithmic behaviour is enough. So here the complexity is O(log(n)) .

The inner loop for(int j=1; j<=i; j++) , each time when evaluated, will run to the corresponding i . It can be shown that the average run width is about n / log2(n) , since i is 1 , 2 , 4 , ... n with log2(n) steps. For example, if n is 31 , i is 1 , 2 , 4 , 8 , 16 , the sum is 31 with 5 steps. So it is permissible to take complexity O(n/log(n)) here.

The overal complexity is then O(log(n)*n/log(n)) , which is O(n) .

Answer 2

We can assume without loss of generality that N is equal to 2^k + 1 . We need to find the number of iterations of the inner loop. There will be k iterations of outer loop with 2^0, 2^1, ..., 2^k iterations of the inner loop. Let's sum up this values.

Answer 3

It's n .

if n=2^k then while complexity is k,

second loop : 2^1 + 2^2 + ... 2^k = 2^(k+1)-1 ~= 2^(k+1)

2^(k+1) = 2*n