What is the time complexity of the below function?

## C

`void` `fun(` `int` `n, ` `int` `k)` `{` ` ` `for` `(` `int` `i = 1; i <= n; i++) ` ` ` `{` ` ` `int` `p = ` `pow` `(i, k);` ` ` `for` `(` `int` `j = 1; j <= p; j++) ` ` ` `{` ` ` `// Some O(1) work` ` ` `}` ` ` `}` `}` |

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Time complexity of above function can be written as 1^{k} + 2^{k} + 3^{k} + … n1^{k}.

Let us try few examples:

k=1 Sum = 1 + 2 + 3 ... n = n(n+1)/2 = n^{2}/2 + n/2 k=2 Sum = 1^{2}+ 2^{2}+ 3^{2}+ ... n1^{2}. = n(n+1)(2n+1)/6 = n^{3}/3 + n^{2}/2 + n/6 k=3 Sum = 1^{3}+ 2^{3}+ 3^{3}+ ... n1^{3}. = n^{2}(n+1)^{2}/4 = n^{4}/4 + n^{3}/2 + n^{2}/4

In general, asymptotic value can be written as **(n ^{k+1})/(k+1) + Θ(n^{k})**

If n>=k then the time complexity will be considered in

**O((n**and if n<k, then the time complexity will be considered as in the

^{k+1})/(k+1))**O(**

**n**^{k})Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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