The difference of time Complexity between given functions can be represented by:
void fun1(int n)
{
for(int i=1;i<=n;i++) for(int j=1;j<=i*i;j++) if(j%i==0) for(int k=1;k<=j;k++) s++; return 0; }
void fun2(int n)
for(int i=1;i<=n;i++) for(int j=1;j<=i*i;j++) for(int k=1;k<=j;k++) s++; return 0; }
$i. O(n^2)$
$ii. O(n)$
$iii.O(1)$
$iv. O(n^{1.5})$
https://gateoverflow.in/230638/ace-algorithms
Shaik Masthan
For the first function number of time, the inner loop will execute will be
$\frac{n^{2}*(n+1)}{2}$
So, the complexity will be O(n^3).
@Shaik MasthanCan u please tell me what is the meaning of difference of time complexity between functions here?
For f1 time complexity see this : https://gateoverflow.in/230638/ace-algorithms