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# 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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