Time complexity of Q2

#1 
   void function(int n) 
   { 
     int count = 0; 

     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("*"); 
          } 
    }

How to compute time complexity of the above code. Can you list the steps.

#2

Consider the comments in the following function.

void function(int n)
{

int count = 0; 

// executes n times 
for (int i=0; i<n; i++) 

    // executes O(n*n) times. 
    for (int j=i; j< i*i; j++) 
        if (j%i == 0) 
        { 
            // executes j times = O(n*n) times 
            for (int k=0; k<j; k++) 
                printf("*"); 
        }

}

Time Complexity of the above function O(n^5)

#3

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