2347. Best Poker Hand

Problem

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.

Solution (Java)

class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        HashMap<Character, Integer> 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<Integer, Integer> 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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).