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Best Poker Hand

Difficulty: Easy

Problem Description

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]. Return a string representing the best type of poker hand you can make with the given cards.

Key Insights

  • A "Flush" consists of five cards of the same suit.
  • "Three of a Kind" consists of three cards of the same rank.
  • A "Pair" consists of two cards of the same rank.
  • If none of the above hands are present, the hand is classified as a "High Card".

Space and Time Complexity

Time Complexity: O(1) - The algorithm processes a fixed number of cards (5), so the computation time is constant. Space Complexity: O(1) - The space used is constant regardless of input size, as we only use a few variables to store counts.

Solution

To solve the problem, we can utilize a dictionary (or hash table) to count the occurrences of each rank and a set to track the unique suits. The algorithm proceeds as follows:

  1. Initialize a dictionary to count the number of cards for each rank.
  2. Initialize a set to track unique suits.
  3. Iterate over the ranks and suits to populate the dictionary and the set.
  4. Check if the size of the set is 1 (indicating a "Flush").
  5. Check the counts in the dictionary to determine if there is a "Three of a Kind" or "Pair".
  6. Return the best hand type based on the checks.

Code Solutions

def bestPokerHand(ranks, suits):
    rank_count = {}
    for rank in ranks:
        rank_count[rank] = rank_count.get(rank, 0) + 1
    
    if len(set(suits)) == 1:
        return "Flush"
    
    for count in rank_count.values():
        if count >= 3:
            return "Three of a Kind"
        elif count == 2:
            return "Pair"
    
    return "High Card"
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