Date: 2026-06-06
Time: 15:21
This file solves LeetCode 2347 — Best Poker Hand. Given exactly 5 cards (as parallel arrays of ranks and suits), it returns the best hand classification. It's a straightforward classification problem — no game simulation, no hand comparison.
bestpokerhand(ranks, suits) -> strTakes two parallel lists of length 5 and returns one of four strings in a fixed priority order:
| Priority | Return Value | Condition |
|----------|-------------|-----------|
| 1 (best) | "Flush " | All 5 suits identical |
| 2 | "Three of a Kind " | Any rank appears 3+ times |
| 3 | "Pair " | Any rank appears exactly 2 times |
| 4 (worst) | "High Card " | None of the above |
The trailing space in each return value matches the LeetCode expected output format.
Early-return cascade: The function checks hands from strongest to weakest and returns the first match. This avoids nested conditionals and makes the priority ordering explicit.
Frequency analysis via Counter: Rather than writing manual loops to count duplicates, the solution reduces the rank list to a frequency map and inspects only the maximum frequency. This collapses "Three of a Kind" and "Full House" (which the problem treats the same since maxfreq >= 3 catches both), and similarly "Pair" and "Two Pair" (both have maxfreq == 2, and the problem doesn't distinguish them).
Set cardinality for uniformity check: len(set(suits)) == 1 is the idiomatic Python way to test if all elements are identical.
Imports: collections.Counter — the only dependency, used for rank frequency counting.
Imported by: best-poker-hand/test_solution.py consumes this function. The large "Imported By" list in the prompt is an artifact of the cross-reference tool — those are unrelated test files, not actual consumers of this function.
1. Convert suits to a set. If the set has exactly one element, all suits match — return "Flush ".
2. Build a Counter over ranks, extract the maximum frequency value.
3. If max_freq >= 3: return "Three of a Kind ".
4. If max_freq == 2: return "Pair ".
5. Otherwise (max_freq == 1, all ranks distinct): return "High Card ".
The flush check comes first because a flush beats all other hands in this simplified ranking. The rank-frequency checks are mutually exclusive by the >=3 / ==2 / implicit ==1 boundaries.
[1, 13] and suits are single characters from {'a', 'b', 'c', 'd'}.>= 3 threshold means a hand with 4 or 5 of the same rank still returns "Three of a Kind " — correct per the problem statement, which doesn't define "Four of a Kind" as a distinct category.None. Invalid input (empty lists, wrong types) would propagate as unhandled exceptions from set(), Counter(), or max(). This is appropriate for a LeetCode solution where inputs are guaranteed valid.
best-poker-hand/test_solution.py — See what edge cases are covered (all same rank, flush + pair overlap, etc.)best-poker-hand/plan.md — The approach reasoning before implementationbest-poker-hand/review.md — Post-implementation review notescounter-max-frequency-pattern — This max(Counter(...).values()) idiom recurs across many LeetCode solutions in this repo for duplicate-detection problemsx-of-a-kind-in-a-deck-of-cards/solution.py — Related card/frequency problem that uses GCD instead of max frequencyflush-beats-all — Flush is checked before rank-based hands and takes priority over Three of a Kind, matching the problem's hand rankingno-full-house-distinction — The >= 3 threshold treats Full House (3+2) identically to Three of a Kind, which is correct per the problem constraintsno-two-pair-distinction — Two Pair and One Pair both return "Pair " because the check uses only max_freq == 2, not the count of pairstrailing-space-in-output — All return strings end with a trailing space, matching LeetCode's expected output formatfive-card-assumption — The function assumes exactly 5 cards with no length validation; correctness depends on the caller providing valid input