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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).