Algo:Time complexity

Question

void fun(intn)
{
    int s=0;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i*i;j++)
        {
            if(j%i==0)
            {
                for(k=1;k<=j;K++)
                s++
            }
         }
    }
}

Comments

Okay I made a small mistake I got $O(n^4)$

for i  = 1, j =1, k =1

for i = 2, j = 4  k = 2+4 = 2(1+2)

for i = 3, j = 9  k = 3+6+9 = 3(1+2+3)

..

..

for i = n, j = n^2  k = n+2n+3n...n^2 = n(1+2+3...n)

 

hence observe last term it is n(1+2+3...n) = n(n(n+1)/2) = O(n^3)

Now second last term will be (n-1)(1+2+3...(n-1)) = O(n^3)

Hence all n^3 upto n will sum up and it will be $O(n^4)$

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n*O(n³) = O(n⁴); n is coz of outer loop na ?

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Yes $n$ times $O(n^3)$ will be $O(n^4)$