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 :



title-img


                            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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