Before the explanation: the problem has no tricks and no corner cases, since ~H~ is guaranteed to be odd. Thus if the algorithm performs correctly, you will AC all test cases.

Now, a reasonable way to solve this problem would be to have two for loops (one for the top of the diamond, the other for the bottom). To figure out how many asterisks we have to put on ~i^\text{th}~ line, consider how far we are in the loop. The number of asterisks can be found using the equation: ~height / 2 - i~ and the number of spaces ~2i+1~. This is really all you need to do to solve this 5 point problem.

Time Complexity: ~\mathcal O(H^2)~