Intermediate.
Put your skills to the test.

Problem summary

Four players receive five cards each. Each rank from 2 to 14 occurs four times in the deck. Score hands by their repeated-rank pattern: four of a kind 6, full house 5, three of a kind 4, two pairs 3, one pair 2, all distinct 1. Find the unique winner.

Constraints

Ranks are between 2 and 14; the deal is valid and exactly one player has the highest score.

Worked example

Input and output for the worked example discussed below.

5 2 11 11 4
7 6 9 8 8
8 13 8 13 13
2 9 9 2 14

3 5

Solution analysis

1. Count how many times each rank appears in a hand. The actual rank values do not break ties: this problem compares only the six specified multiplicity patterns, not ordinary poker straights or suits.
2. Sort the nonzero frequencies in descending order. The largest count distinguishes four-of-a-kind, three-of-a-kind, pairs, and distinct cards. If the largest is 3, inspect whether the next is 2; if it is 2, inspect whether the next is also 2.
3. In the example player 3 has frequencies 3 and 2, so scores 5. Player 4 has 2,2,1 and scores 3, while players 1 and 2 score 2. Keep the player whenever its score exceeds the current best.

Why it works

Each valid five-card hand has exactly one of the six integer partitions listed in the statement. Frequency counting recovers that partition exactly, so the decision tree assigns its required score. Taking the maximum identifies the promised unique winner.

Watch out. Do not add ordinary-poker rules. The winner is 1-based. Count distinct ranks, not adjacent equal cards in unsorted input.

An independently written complete program with standard headers.

#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
int main(){
    int winner=0;
    int best=0;
    for(int player=1;player<=4;player++){
        int frequency[15]={};
        for(int card=0;card<5;card++){
            int rank;
            cin>>rank;
            frequency[rank]++;
        }
        vector<int>groups;
        for(int rank=2;rank<=14;rank++){
            if(frequency[rank]>0){
                groups.push_back(frequency[rank]);
            }
        }
        sort(groups.rbegin(),groups.rend());
        int score=1;
        // The multiplicity partition completely determines this game's score.
        if(groups[0]==4){
            score=6;
        }
        else if(groups[0]==3){
            score=groups[1]==2?5:4;
        }
        else if(groups[0]==2){
            score=groups[1]==2?3:2;
        }
        if(score>best){
            best=score;
            winner=player;
        }
    }
    cout<<winner<<' '<<best<<'\n';
}