Partition a Card Deck into Fifteens

Given the visible values of an entire card deck, decide whether every card can be consumed by partitioning the deck into disjoint groups of exactly three cards whose values sum to 15.
Cards are distinct physical cards even when values repeat. Every card must appear in exactly one group.

Function
canPartitionIntoFifteens(cards: int[]) → boolean

Examples
Example 1
cards = [1,5,9,2,6,7]
return = true
The deck can be grouped as (1,5,9) and (2,6,7).
Example 2
cards = [1,1,1]
return = false
The only triple sums to 3.
Example 3
cards = [5,5,5,5,5,5]
return = true
Two identical triples each sum to 15.

Constraints
0 <= cards.length <= 36 and its length is divisible by 3.
1 <= cards[i] <= 9.

The trick is that values are only 1 to 9, so don't track individual cards. Track a count array of size 10. Take the smallest value still remaining, try every pair (b, c) with b <= c and b >= a such that a+b+c = 15 and counts allow it, decrement, recurse. Always anchoring on the smallest remaining value kills duplicate orderings. Memoize on the count tuple, since at most 36 cards across 9 values gives a small state space. Quick prunes: length not divisible by 3 returns false, total sum must equal 5 times the number of triples, and the empty deck returns true. The pitfall is picking triples in arbitrary order, which explodes the search, or using greedy on largest first, which gives wrong answers. Valid triples are few: (1,5,9), (2,4,9), (2,5,8), (3,3,9), (5,5,5) and so on. If the recursion gets tangled live, StealthCoder is the hedge that gives you a clean version.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.