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 } }} |
Time complexity of above function can be written as 1k + 2k + 3k + … n1k.
Let us try few examples:
k=1
Sum = 1 + 2 + 3 ... n
= n(n+1)/2
= n2/2 + n/2
k=2
Sum = 12 + 22 + 32 + ... n12.
= n(n+1)(2n+1)/6
= n3/3 + n2/2 + n/6
k=3
Sum = 13 + 23 + 33 + ... n13.
= n2(n+1)2/4
= n4/4 + n3/2 + n2/4
In general, asymptotic value can be written as (nk+1)/(k+1) + Θ(nk)
If n>=k then the time complexity will be considered in O((nk+1)/(k+1)) and if n<k, then the time complexity will be considered as in the O(nk)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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