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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2301_2400.s2347_best_poker_hand;

// #Easy #Array #Hash_Table #Counting #2022_07_26_Time_1_ms_(76.92%)_Space_41.6_MB_(30.77%)

import java.util.HashMap;

/**
 * 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.
**/
public class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        HashMap map = new HashMap<>();
        for (char suit : suits) {
            if (map.containsKey(suit)) {
                map.put(suit, map.get(suit) + 1);
                if (map.get(suit) == 5) {
                    return "Flush";
                }
            } else {
                map.put(suit, 1);
            }
        }
        String s = "";
        HashMap map2 = new HashMap<>();
        for (int rank : ranks) {
            if (map2.containsKey(rank)) {
                map2.put(rank, map2.get(rank) + 1);
                if (map2.get(rank) == 2) {
                    s = "Pair";
                } else if (map2.get(rank) == 3) {
                    s = "Three of a Kind";
                    return s;
                }
            } else {
                map2.put(rank, 1);
            }
        }
        return s.isEmpty() ? "High Card" : s;
    }
}