Time Complexity of Loop with Powers
Last Updated :
30 Oct, 2023
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) + ?(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)
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