Best Poker Hand LeetCode Solution

Problem – Best Poker Hand LeetCode Solution

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.

Best Poker Hand LeetCode Solution in C++

class Solution {
public:
    string bestHand(vector<int>& ranks, vector<char>& suits) {
        map<char,int> m ; // Stores count of same suit
        map<int,int> m2 ; // Stores count of same rank
        for(auto i:suits)
        {
            m[i]++ ;
        }
        for(auto i:ranks)
        {
            m2[i]++ ;
        }
        string ans = "High Card" ;
        for(auto i:m) // Checking same suit
        {
            if(i.second == 5)
            {
                ans = "Flush" ;
				break ;
            }
        }
        for(auto i:m2) // Checking same rank
        {
            if(i.second >= 3)
            {
                ans = "Three of a Kind" ;
                break ;
            }
            else if(i.second == 2)
            {
                ans = "Pair" ;
				break ;
            }
        }
        return ans ;
    }
};

Best Poker Hand LeetCode Solution in Python

class Solution(object):
    def bestHand(self, ranks, suits):
        """
        :type ranks: List[int]
        :type suits: List[str]
        :rtype: str
        """
        cnt = max(Counter(ranks).values()) # Counts maximum number of ranks of the same kind
        ans = "High Card"   
        if(len(set(suits)) == 1): # 5 of the same
            ans = "Flush"
        elif cnt >= 3:
            ans = "Three of a Kind"
        elif cnt == 2:
            ans = "Pair"
        return ans 

Best Poker Hand LeetCode Solution in Java

class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        int []hashr = new int[15] ; // Check Constraints, Stores count of ranks
        int []hashs = new int[5] ; // Check Constraints, Stores count of suits
        for(int i = 0 ; i < 5 ; i++)
        {
            hashr[ranks[i]] += 1 ;
            hashs[suits[i] - 'a'] += 1 ;
        }
        String ans = "High Card" ;
        for(int i = 0 ; i < 5 ; i++) // Checking same suit
        {
            if(hashs[i] == 5)
            {
                ans = "Flush" ;
				break ;
            }
        }
        for(int i = 0 ; i < 14 ; i++) // Checking same rank
        {
            if(hashr[i] >= 3)
            {
                ans = "Three of a Kind" ;
                break ;
            }
            else if(hashr[i] == 2)
            {
                ans = "Pair" ;
				break ;
            }
        }
        return ans ;
    }
}
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