Mahjong Winning Hand Detection (14 Tiles)

Mountain View onsite coding round 1: given a hand of 14 tiles, decide whether it is a winning mahjong hand. The candidate must clarify the rule set (typically: 4 melds of three identical tiles or three consecutive ranks + 1 pair). Requirements Input: array of 14 integers representing tile values (clarify suit / range during the interview). A winning hand consists of: One pair (two identical tiles), and Four melds, each being either: A triplet (3 identical tiles), or A run (3 consecutive ranks in the same suit). Minority variant: A newer version instead specifies one pair plus two melds while retaining a 14-tile input. Because that accounts for only eight tiles, clarify the intended meld count and how the remaining six tiles should be handled before coding. Output: boolean. Notes Clarify the rules — the prompt is intentionally sparse. Confirm: are honor tiles (winds/dragons) included? Do flowers count? Multi-suit? Greedy + backtracking is the standard approach: Try every possible pair (tile, tile); remove the pair. On the remaining 12 tiles, repeatedly take the smallest remaining tile and try to form a triplet of it or a run starting at it; backtrack on failure. For 14 tiles this runs in well under 1 ms; the interviewer cares more about the case analysis and clean recursion than micro-optimizations. The same algorithm extends to special hands (seven pairs / 国士无双) if the interviewer adds them as a follow-up. Preparation Practice writing the backtracking decomposer for tiles / counters; the canonical "expression add operators" recursion trains the same backtracking skeleton. Have a tile-frequency-map mental model — sorting is the alternative but harder to reason about for runs across suits. Pre-plan for the inevitable rules clarification: pull out at least 3 questions before coding.