# HackerRank Fun Game problem solution

In this HackerRank Fun Game problem solution we have given the value of two N-sized integer arrays, A and B., and is played by two players P1 and P2. can you determine the outcome of the game? Print First if P1 will win, Second if P2 will win, or Tie if they will tie. Assume both players always move optimally.

## Problem solution in Python.

```import sys

# Process input
T = int(data[0])

def run_game(N, a, b):
c = [a[i] + b[i] for i in range(N)]  # a-b
cs = [i for i, x in sorted(
enumerate(c), key=(lambda y : y[1]))] # keys that sort c

# Play game
p1, p2 = (0, 0)  # payoffs to the two players
for i in range(N):
if (i % 2)==0:  # P1's turn
p1 += a[cs.pop(-1)]
else:  # P2's turn
p2 += b[cs.pop(-1)]

# Print results
if p1 > p2:
print("First")
elif p1==p2:
print("Tie")
else:
print("Second")

for k in range(T):
N = int(data[3*k + 1])
a = list(map(int, data[3*k + 2].split()))
b = list(map(int, data[3*k + 3].split()))
run_game(N, a, b)```

## Problem solution in Java.

```import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int T = input.nextInt();
for(int t=0; t<T; t++){
int n = input.nextInt();
int[] a = new int[n];
int[] b = new int[n];
int aScore = 0, bScore = 0;
TreeSet<Pair> c = new TreeSet<>();

for(int i=0; i<n; i++){
a[i] = input.nextInt();
}

for(int i=0; i<n; i++){
b[i] = input.nextInt();
}

int turn = 0;
for(Pair p:c){
if(turn == 0)
aScore += a[p.index];
else
bScore += b[p.index];
turn = 1 - turn;
}
if(aScore > bScore){
System.out.println("First");
}else if(aScore == bScore){
System.out.println("Tie");
}else{
System.out.println("Second");
}

}
}

public static class Pair implements Comparable<Pair>{
int sum;
int index;

Pair(int s, int i){
this.sum = s;
this.index = i;
}

@Override
public int compareTo(Pair p){
if(this.sum < p.sum){
return 1;
}else{
return -1;
}
}

public String toString(){
return "{ s = "+sum+", i = "+index+" }";
}
}
}```

## Problem solution in C++.

```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int Maxn = 1005;

int t;
int n;
int sum;
int a[Maxn];

int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
sum = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]); sum += a[i];
}
for (int i = 0; i < n; i++) {
int tmp; scanf("%d", &tmp);
a[i] += tmp; sum -= tmp;
}
sort(a, a + n);
for (int i = n - 1; i >= 0; i--)
if ((n - 1 - i) % 2 == 0) sum += a[i];
else sum -= a[i];
if (sum > 0) printf("First\n");
else if (sum == 0) printf("Tie\n");
else printf("Second\n");
}
return 0;
}```

## Problem solution in C.

```#include<stdio.h>
int solve(int *a, int *b, int n, int max)
{
int temp = 0, loss, ret = a[0], i;
loss = b[0] - ( max - a[0] );
for( i = 1 ; i < n ; i++ )
{
if( ( b[i] - ( max - a[i] ) ) > loss )
{
loss = b[i] - ( max - a[i] );
ret = a[i];
temp = i;
}
}
a[temp] = b[temp] = 0;
return ret;
}
int main()
{
int t, n, start, i;
long P1, P2;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
int a[n], b[n], max = -1;
for( i = 0 ; i < n ; i++ )
{
scanf("%d", a + i);
if( max < a[i] )
{
max = a[i];
}
}
for( i = 0 ; i < n ; i++ )
{
scanf("%d", b + i);
}
start = 1, P1 = P2 = 0;
for( i = 0 ; i < n ; i++ )
{
start > 0 ? ( P1 += solve(a, b, n, max) ): ( P2 += solve(b, a, n, max) );
start = !start;
}
puts(P1 > P2 ? "First" : P1 < P2 ? "Second" : "Tie");
}
return 0;
}```