LeetCode 2347: Best Poker Hand Solution

Master LeetCode problem 2347 (Best Poker Hand), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2347. Best Poker Hand

Easy

Array Hash Table Counting

Problem Explanation

Explanation

To solve this problem, we need to determine the best possible poker hand from the given cards. We can achieve this by iterating through the ranks and suits arrays to count the occurrences of each rank and suit. Based on these counts, we can identify the best hand as per the specified order of poker hands.

Count the occurrences of each rank and suit.
Check for a "Flush" by verifying if all cards have the same suit.
Check for a "Three of a Kind" by finding if any rank occurs three times.
Check for a "Pair" by identifying if any rank occurs twice.
Return the best possible hand based on the above checks.

Time Complexity: O(1) - Since we are dealing with a fixed number of cards (5 cards) and ranks (1 to 13), the time complexity is constant. Space Complexity: O(1) - The space used is constant regardless of the input size.

Solution Code

class Solution {
    public String bestPokerHand(int[] ranks, char[] suits) {
        int[] rankCount = new int[14];
        int[] suitCount = new int[4];

        for (int i = 0; i < 5; i++) {
            rankCount[ranks[i]]++;
            suitCount[suits[i] - 'a']++;
        }

        boolean isFlush = suitCount[suits[0] - 'a'] == 5;
        boolean isThreeOfAKind = false;
        boolean isPair = false;

        for (int i = 1; i <= 13; i++) {
            if (rankCount[i] == 3) {
                isThreeOfAKind = true;
            } else if (rankCount[i] == 2) {
                isPair = true;
            }
        }

        if (isFlush) {
            return "Flush";
        } else if (isThreeOfAKind) {
            return "Three of a Kind";
        } else if (isPair) {
            return "Pair";
        } else {
            return "High Card";
        }
    }
}

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