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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2347\. Best Poker Hand
Easy
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.