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 `
`} ` |

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*filter_none*

In this case, i takes values 2, 2^{k}, (2^{k})^{k} = 2^{k2}, (2^{k2})^{k} = 2^{k3}, …, 2^{k}^{logk(log(n))}. The last term must be less than or equal to n, and we have 2^{k}^{logk(log(n))} = 2^{log(n)} = n, which completely agrees with the value of our last term. So there are in total log_{k}(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 > 1; i = func(i)) `
`{ ` ` ` `// some O(1) expressions or statements `
`} ` |

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*filter_none*

In this case, i takes values n, n^{1/k}, (n^{1/k})^{1/k} = n^{1/k2}, n^{1/k3}, …, n^{1/klogk(log(n))}, so there are in total log_{k}(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.

https://www.geeksforgeeks.org/analysis-of-algorithms-set-4-analysis-of-loops/amp/

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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