# Time Complexity of a Loop when Loop variable “Expands or Shrinks” exponentially

For such cases, time complexity of the loop is O(log(log(n))).The following cases analyse different aspects of the problem.

Case 1 :

```for (int i = 2; i <=n; i = pow(i, k))
{
// some O(1) expressions or statements
}
```

In this case, i takes values 2, 2k, (2k)k = 2k2, (2k2)k = 2k3, …, 2klogk(log(n)). The last term must be less than or equal to n, and we have 2klogk(log(n)) = 2log(n) = n, which completely agrees with the value of our last term. So there are in total logk(log(n)) many iterations, and each iteration takes a constant amount of time to run, therefore the total time complexity is O(log(log(n))).

Case 2 :

```// func() is any constant root function
for (int i = n; i > 0; i = func(i))
{
// some O(1) expressions or statements
}

```

In this case, i takes values n, n1/k, (n1/k)1/k = n1/k2, n1/k3, …, n1/klogk(log(n)), so there are in total logk(log(n)) iterations and each iteration takes time O(1), so the total time complexity is O(log(log(n))).

Refer below article for analysis of different types of loops.
http://www.geeksforgeeks.org/analysis-of-algorithms-set-4-analysis-of-loops/

This article is contributed by Rishav Raj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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