Best Poker Hand
Problem Statement - link #
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.
Examples #
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 == 51 <= ranks[i] <= 13'a' <= suits[i] <= 'd'- No two cards have the same rank and suit.
Solutions #
class Solution {
public:
string bestHand(vector<int>& ranks, vector<char>& suits) {
unordered_set<char> set(suits.begin(), suits.end());
if(set.size() == 1) return "Flush";
unordered_map<int, int> map;
int cur = 1;
for(int &i: ranks) {
map[i]++;
cur = max(cur, map[i]);
}
if(cur >= 3) {
return "Three of a Kind";
}
else if(cur == 2) {
return "Pair";
}
return "High Card";
}
};