# Gaming Array

### Problem Statement :

```Andy wants to play a game with his little brother, Bob. The game starts with an array of distinct integers and the rules are as follows:

Bob always plays first.
In a single move, a player chooses the maximum element in the array. He removes it and all elements to its right. For example, if the starting array arr = [2,3,5,4,1], then it becomes arr' = [2,3] after removing [5,4,1].
The two players alternate turns.
The last player who can make a move wins.
Andy and Bob play g games. Given the initial array for each game, find and print the name of the winner on a new line. If Andy wins, print ANDY; if Bob wins, print BOB.

To continue the example above, in the next move Andy will remove 3. Bob will then remove 2 and win because there are no more integers to remove.

Function Description

Complete the gamingArray function in the editor below.

gamingArray has the following parameter(s):

int arr[n]: an array of integers
Returns
- string: either ANDY or BOB

Input Format

The first line contains a single integer g, the number of games.

Each of the next g pairs of lines is as follows:

The first line contains a single integer, n, the number of elements in arr.
The second line contains n distinct space-separated integers arr[i] where 0 <= i <= n.
Constraints

Array arr contains n distinct integers.
For 35% of the maximum score:
1 <= g <= 10
1 <= n <= 1000
1<= arr[i] <= 10^5
The sum of n over all games does not exceed 1000.
For 100% of the maximum score:
1 <= g <= 100
1 <= n <= 10^5
1 <= ai <= 10^9
The sum of n over all games does not exceed 10^5.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>

using namespace std;

int main()
{
int Q;
scanf("%d", &Q);
while(Q--)
{
int N;
scanf("%d", &N);
int last=0, ans=0;
for(int i=0; i<N; i++)
{
int x;
scanf("%d", &x);
if(x>last)
last=x, ans^=1;
}
if(ans==0)
printf("ANDY\n");
else
printf("BOB\n");
}
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int g = sc.nextInt();
for(int a0 = 0; a0 < g; a0++){
int n = sc.nextInt();
int[] a = new int[n];
TreeMap<Integer, Integer> tm = new TreeMap<Integer, Integer>();
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
tm.put(a[i], i);
}
int lastIndex = n;
int ans = 0;
while (!tm.isEmpty()) {
Map.Entry<Integer, Integer> e = tm.pollLastEntry();
if (e.getValue() < lastIndex) {
lastIndex = e.getValue();
ans++;
}
}
System.out.println(ans%2==0?"ANDY":"BOB");
}
}
}

In C :

#include<stdio.h>
typedef unsigned u;
int main()
{
u t,n,p,a,r;
for(scanf("%u",&t);t--;printf(r?"BOB\n":"ANDY\n"))
for(scanf("%u",&n),p=r=0;n--;)
{
scanf("%u",&a);
if(a>p){p=a;r^=1;}
}
return 0;
}

In Python3 :

#!/bin/python3

import sys

g = int(input().strip())
for a0 in range(g):
n = int(input().strip())
a = list(map(int, input().split()))
maxtotheleft = []
curmaxidx = None
for i in range(n):
if curmaxidx == None or a[curmaxidx] < a[i]:
curmaxidx = i
maxtotheleft.append(curmaxidx)

curplayer = True
for i in reversed(range(n)):
if maxtotheleft[i] == i:
curplayer = not curplayer

if curplayer:
print("ANDY")
else:
print("BOB")```
```

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