++i costs less gas compared to i++ or i += 1
++i costs less gas compared to i++ or i += 1 for unsigned integer, as pre-increment is
cheaper (about 5 gas iteration). This statement is true even with the optimizer enabled.
i++ increments i and retuns the initial value of i. Which means:

But ++i returns the actual incremented value:

In the first case, the compiler has to create a temporary variable (when used) for returning 1
instead of 2

I suggest using ++i instead of i++ to increment the value of a uint variable.
Same thing for --i and i--