Validate a Mahjong Hand Partition

A Mahjong-like hand is represented by an unordered array tiles whose values are digits from 1 through 9.
Return whether every tile can be partitioned into:
exactly one pair of equal values (the pair of eyes); and
zero or more three-tile groups, where each group is either three equal values or three consecutive values x, x + 1, x + 2.
Every tile occurrence must belong to exactly one group. The order of tiles in the input does not matter.

Function
isValidMahjongHand(tiles: int[]) → boolean

Examples
Example 1
tiles = [1,1,1,2,3,4,5,6]
return = true
Use [1,1] as the pair, then form the runs [1,2,3] and [4,5,6].
Example 2
tiles = [1,1,1,2,2]
return = true
The hand partitions into the triple [1,1,1] and the pair [2,2].
Example 3
tiles = [1,1,1,1,2]
return = false
After choosing any pair, the remaining three tiles form neither a triple nor a consecutive run.

Constraints
2 <= tiles.length <= 20.
Every tile value is between 1 and 9.
A necessary valid-hand length has the form 3g + 2 for some g >= 0.

Build a count array of size 10. Reject early if length mod 3 isn't 2. Then loop over each digit d with count at least 2, remove two as the eyes, and check whether the rest splits into groups. The check is greedy from digit 1 upward: at the lowest digit with remaining count c, first strip triples (c >= 3 takes three equal tiles), then whatever is left must form runs starting there, so you need the next two digits to have enough tiles. Actually the cleaner rule: if count[i] >= 3, subtract 3; then if count[i] > 0, you need count[i+1] and count[i+2] at least that amount. Lowest digit is forced, which is why greedy works. The common pitfall is trying only one pair candidate, or running past digit 9. Backtracking with memoization also works at 20 tiles. If you freeze on the greedy proof, StealthCoder is the hedge during the live OA.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.