Question

for(i=0;i< m; i++)
{

   for(j=i+1; j < m; j++)
   {

      for(k=0; k < n;k++)
      {
         for(l=0;l< n;l++)
         {if(condition) do something}
      }
   }

} 

Answer 1

In details:

The two first loops will result in (m-1) + (m-2) + (m-3) + … + 1 repetitions, which is equal to m*(m-1)/2. As for the second two loops, they basically run from 0 to n-1 so they need n^2 iterations.

As you have no clue whether the condition will be fulfilled or not, then you take the worst case, which is it being always fulfilled.

Then the number of iterations is:

m*(m+1)/2*n^2*NumberOfIterations(Something)

In O notation, the constants and lower degrees are not necessary, so the complexity is:

O(m^2*n^2)*O(Something)