If I understand the game correctly, it takes a minimum of 5 cards to clear the array: a K, a J, two tens, and an 8.  The K J 10 10 eliminate everything above a 9, and with an 8 you can made 3 sets of 15.  For instance, [4,5,6], [7,8],[A,2,3,9]

Part II)

The K J 10 10 8 add to 52.  That leaves 63 other points in the stack.  This must be 4 Queens (48) plus a set of 15.

So, the stack could have 

  K J 10 10 Q Q Q Q + cards adding to 23.

  While many solutions are possible, the cards that add to 23 could be as few as 3 (6,8,9) or as many as 6 (1,2,3,4,5,8) or even (1,2,3,4,6,7) because we don't necessarily need an 8.

Or may I have misunderstood the question (it happens a lot).