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Formatted question description: https://leetcode.ca/all/2347.html

2347. Best Poker Hand

  • Difficulty: Easy.
  • Related Topics: Array, Hash Table, Counting.
  • Similar Questions: .

Problem

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:

  • "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.

Solution (Java, C++, Python)

  • class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        HashMap<Character, Integer> map = new HashMap<>();
        for (char suit : suits) {
            if (map.containsKey(suit)) {
                map.put(suit, map.get(suit) + 1);
                if (map.get(suit) == 5) {
                    return "Flush";
                }
            } else {
                map.put(suit, 1);
            }
        }
        String s = "";
        HashMap<Integer, Integer> map2 = new HashMap<>();
        for (int rank : ranks) {
            if (map2.containsKey(rank)) {
                map2.put(rank, map2.get(rank) + 1);
                if (map2.get(rank) == 2) {
                    s = "Pair";
                } else if (map2.get(rank) == 3) {
                    s = "Three of a Kind";
                    return s;
                }
            } else {
                map2.put(rank, 1);
            }
        }
        return s.isEmpty() ? "High Card" : s;
    }
}

############

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:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        if len(set(suits)) == 1:
            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'

############

# 2347. Best Poker Hand
# https://leetcode.com/problems/best-poker-hand/

class Solution:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        if len(set(suits)) == 1: return "Flush"
        
        if max(Counter(ranks).values()) >= 3: return "Three of a Kind"
        
        if max(Counter(ranks).values()) >= 2: return "Pair"
        
        return "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()
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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