Question

time complexity of given function

fun(int n){

for(int i = 0;i<n;i++)

   for(int j=i;j<i*i;j++)

      if(j%i==0){

         for(int k =0;k<j;k++)

             printf("*");

       }

}

Comments

^That is for j loop ( j= i to i ^2).

We can also write (i² - i ) in place of i^2. 

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You have already unrolled loops right and the count submission of i terms. may I ask how you unrolled the loops. ?

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^here j is looping for (i^2- i) times for each i, irrespective of if(j%i ==0). right ??

And 2nd term is for - when if(j%i==0) -> true.