Although the solution is one way, still it is a great hand with a great solution. Let's see again, what had happened:
We play 3NT from South hand, the system unfortunately allowed to East to double 3♣ Stayman. North's 3♦ bid denies 4 card major, promises stopper, so our 3NT bid is obligatory.
The lead is the two of clubs. We have to duck it 2 times, so we win the 3rd round of club, discarding spade from the dummy and we start counting our tricks and possibilities.
We have 7 top tricks and we can have 2 more tricks in the red suits without losing a trick.
We have to try our red suits carefully, because we can win 4 heart tricks in case one of the opponent has doubleton Jack of hearts. The diamond suit can come later, and we have communication in this suit.
Heart is lying 3-3, so we can cash our 4th heart trick later.. Both opps are discarding spades, West has to put the 9.
Now we cash 2 diamond tricks and we have to recognize, that East had only the singleton ten of diamonds in his hand.
Now we have to put West to have the Ace of spades, because if East has it, we are lost: we have only 8 tricks, and East can cash 2 more club tricks after the Ace of spades, and we cannot squeeze him, because West has the rest of diamonds.
West has 3 clubs, 3 hearts, 5 diamonds and 2 spades. So now, when we cash our 4th heart trick from the dummy, West will be squeezed to make a nice striptease squeeze and an elimination play, while we can hold King and a small card in spades and King and a small card in diamonds.
If he discards diamond, we cash our last diamond trick and we give him the trick with diamond in trick 11, so he has to play back from Ax in spades.
If he discards spades, (which is a far better play), we have to put him to hold single Ace in spades and 3 diamonds with J97, so we have to play small spade from both sides, because if we cash our diamond tricks first, we lose the contract anyway: both opponents have 2 high minor tricks in his suit.
At the table, West discarded a diamond, so Zöld could make an easy endplay in trick 11.
The complete distribution was: