Time Complexity

#1

Here is the code:-

   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++) 
               // Print statement - O(1) operation 
          } 
    }

Please explain the time complexity of the above code.

#2

hello @Sakshi2004

the outer loop will run n times.
the second inner loop will run approx n^2 times for each iteration of outer loop
now the third loop will run only when j is multiple of i.
so basically it will run for values j=i, 2i,3i…i * i
in worst case i is n , and ii will be n^2
so this loop will run approax n^2 ,for each iteration of outer loop(2nd one).
in total we can say
n * n ^ 2 *n ^ 2=> O(n^5)

#3

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