# Solving Leetcode Interviews in Seconds with AI: Best Poker Hand


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2347" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's a breakdown of the solution:

*   **Check for Flush:** First, determine if all suits are the same. If so, it's a "Flush", and we can return immediately.
*   **Count Rank Occurrences:** If it's not a flush, count the occurrences of each rank.
*   **Determine Best Hand:** Based on the rank counts, identify the best hand: "Three of a Kind" if any rank appears three or more times, "Pair" if any rank appears twice, and "High Card" otherwise.

*   **Runtime Complexity:** O(N), where N is the number of cards (always 5 in this problem).
*   **Storage Complexity:** O(1) since we use a fixed-size dictionary to count rank occurrences.

	
	# Code
	```python
	def bestHand(ranks, suits):
    """
    Determines the best poker hand from a set of 5 cards.

    Args:
        ranks (list): A list of integers representing the ranks of the cards.
        suits (list): A list of characters representing the suits of the cards.

    Returns:
        str: A string representing the best type of poker hand.
    """

    if len(set(suits)) == 1:
        return "Flush"

    rank_counts = {}
    for rank in ranks:
        rank_counts[rank] = rank_counts.get(rank, 0) + 1

    max_count = 0
    for count in rank_counts.values():
        max_count = max(max_count, count)

    if max_count >= 3:
        return "Three of a Kind"
    elif max_count == 2:
        return "Pair"
    else:
        return "High Card"
	```