2347. Best Poker Hand

Description

You are given an integer array ranks and a character array suits. You have 5 cards where the i^th 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:

"Flush": Five cards of the same suit.
"Three of a Kind": Three cards of the same rank.
"Pair": Two cards of the same rank.
"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 {
        cnt[x]++
        if cnt[x] == 3 {
            return "Three of a Kind"
        }
        pair = pair || cnt[x] == 2
    }
    if pair {
        return "Pair"
    }
    return "High Card"
}

function bestHand(ranks: number[], suits: string[]): string {
    if (suits.every(v => v === suits[0])) {
        return 'Flush';
    }
    const count = new Array(14).fill(0);
    let isPair = false;
    for (const v of ranks) {
        if (++count[v] === 3) {
            return 'Three of a Kind';
        }
        isPair = isPair || count[v] === 2;
    }
    if (isPair) {
        return 'Pair';
    }
    return 'High Card';
}

impl Solution {
    pub fn best_hand(ranks: Vec<i32>, suits: Vec<char>) -> String {
        if suits.iter().all(|v| *v == suits[0]) {
            return "Flush".to_string();
        }
        let mut count = [0; 14];
        let mut is_pair = false;
        for &v in ranks.iter() {
            let i = v as usize;
            count[i] += 1;
            if count[i] == 3 {
                return "Three of a Kind".to_string();
            }
            is_pair = is_pair || count[i] == 2;
        }
        (if is_pair { "Pair" } else { "High Card" }).to_string()
    }
}

char* bestHand(int* ranks, int ranksSize, char* suits, int suitsSize) {
    bool isFlush = true;
    for (int i = 1; i < suitsSize; i++) {
        if (suits[0] != suits[i]) {
            isFlush = false;
            break;
        }
    }
    if (isFlush) {
        return "Flush";
    }
    int count[14] = {0};
    bool isPair = false;
    for (int i = 0; i < ranksSize; i++) {
        if (++count[ranks[i]] == 3) {
            return "Three of a Kind";
        }
        isPair = isPair || count[ranks[i]] == 2;
    }
    if (isPair) {
        return "Pair";
    }
    return "High Card";
}

2347. Best Poker Hand

Description

Solutions

Solution 1: Counting

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