LeetCode in Kotlin

2347. Best Poker Hand

Easy

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:

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.

Solution

class Solution {
    fun bestHand(ranks: IntArray, suits: CharArray): String {
        val map = HashMap<Char, Int>()
        for (suit in suits) {
            if (map.containsKey(suit)) {
                map[suit] = map[suit]!! + 1
                if (map[suit] == 5) {
                    return "Flush"
                }
            } else {
                map[suit] = 1
            }
        }
        var s = ""
        val map2 = HashMap<Int, Int>()
        for (rank in ranks) {
            if (map2.containsKey(rank)) {
                map2[rank] = map2[rank]!! + 1
                if (map2[rank] == 2) {
                    s = "Pair"
                } else if (map2[rank] == 3) {
                    s = "Three of a Kind"
                    return s
                }
            } else {
                map2[rank] = 1
            }
        }
        return if (s.isEmpty()) "High Card" else s
    }
}

This site is open source. Improve this page.