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Kotlin-based LeetCode algorithm problem solutions, regularly updated

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2347\. Best Poker Hand

Easy

You are given an integer array `ranks` and a character array `suits`. You have `5` cards where the i^th 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.