2347. Best Poker Hand

Description

You are given an integer array ranks and a character array suits. You have 5 cards where the ith card has a rank of ranks[i] and a suit of suits[i].

The following are the types of poker hands you can make from best to worst:

1. "Flush": Five cards of the same suit.
2. "Three of a Kind": Three cards of the same rank.
3. "Pair": Two cards of the same rank.
4. "High Card": Any single card.

Return a string representing the best type of poker hand you can make with the given cards.

Note that the return values are case-sensitive.

 

Example 1:

Input: ranks = [13,2,3,1,9], suits = ["a","a","a","a","a"]
Output: "Flush"
Explanation: The hand with all the cards consists of 5 cards with the same suit, so we have a "Flush".

Example 2:

Input: ranks = [4,4,2,4,4], suits = ["d","a","a","b","c"]
Output: "Three of a Kind"
Explanation: The hand with the first, second, and fourth card consists of 3 cards with the same rank, so we have a "Three of a Kind".
Note that we could also make a "Pair" hand but "Three of a Kind" is a better hand.
Also note that other cards could be used to make the "Three of a Kind" hand.

Example 3:

Input: ranks = [10,10,2,12,9], suits = ["a","b","c","a","d"]
Output: "Pair"
Explanation: The hand with the first and second card consists of 2 cards with the same rank, so we have a "Pair".
Note that we cannot make a "Flush" or a "Three of a Kind".

 

Constraints:

• ranks.length == suits.length == 5
• 1 <= ranks[i] <= 13
• 'a' <= suits[i] <= 'd'
• No two cards have the same rank and suit.

Solutions

Solution 1: Counting

We first traverse the array $\textit{suits}$ to check if adjacent elements are equal. If they are, we return "Flush".

Next, we use a hash table or array $\textit{cnt}$ to count the quantity of each card:

• If any card appears $3$ times, return "Three of a Kind";
• Otherwise, if any card appears $2$ times, return "Pair";
• Otherwise, return "High Card".

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{ranks}$.

Python3JavaC++GoTypeScriptRustC

class Solution:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        # if len(set(suits)) == 1:
        if all(a == b for a, b in pairwise(suits)):
            return 'Flush'
        cnt = Counter(ranks)
        if any(v >= 3 for v in cnt.values()):
            return 'Three of a Kind'
        if any(v == 2 for v in cnt.values()):
            return 'Pair'
        return 'High Card'

class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        boolean flush = true;
        for (int i = 1; i < 5 && flush; ++i) {
            flush = suits[i] == suits[i - 1];
        }
        if (flush) {
            return "Flush";
        }
        int[] cnt = new int[14];
        boolean pair = false;
        for (int x : ranks) {
            if (++cnt[x] == 3) {
                return "Three of a Kind";
            }
            pair = pair || cnt[x] == 2;
        }
        return pair ? "Pair" : "High Card";
    }
}

class Solution {
public:
    string bestHand(vector<int>& ranks, vector<char>& suits) {
        bool flush = true;
        for (int i = 1; i < 5 && flush; ++i) {
            flush = suits[i] == suits[i - 1];
        }
        if (flush) {
            return "Flush";
        }
        int cnt[14]{};
        bool pair = false;
        for (int& x : ranks) {
            if (++cnt[x] == 3) {
                return "Three of a Kind";
            }
            pair |= cnt[x] == 2;
        }
        return pair ? "Pair" : "High Card";
    }
};

func bestHand(ranks []int, suits []byte) string {
    flush := true
    for i := 1; i < 5 && flush; i++ {
        flush = suits[i] == suits[i-1]
    }
    if flush {
        return "Flush"
    }
    cnt := [14]int{}
    pair := false
    for _, x := range ranks