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++)
{
}
}
}
|
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++)
{
}
}
}
|
Java
static void fun( int n, int k)
{
for ( int i = 1 ; i <= n; i++)
{
int p = Math.pow(i, k);
for ( int j = 1 ; j <= p; j++)
{
}
}
}
|
Python3
def fun(n, k):
for i in range ( 1 , n + 1 ):
p = pow (i, k)
for j in range ( 1 , p + 1 ):
|
C#
static void fun( int n, int k)
{
for ( int i = 1; i <= n; i++)
{
int p = Math.Pow(i, k);
for ( int j = 1; j <= p; j++)
{
}
}
}
|
Javascript
<script>
function fun(n, k)
{
for (let i = 1; i <= n; i++)
{
int p = Math.pow(i, k);
for (let j = 1; j <= p; j++)
{
}
}
}
</script>
|
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
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!
Last Updated :
30 Oct, 2023
Like Article
Save Article