Find the winner of the Game of removing odd or replacing even array elements

Given an array arr[] consisting of N integers. Two players, Player 1 and Player 2, play turn-by-turn in which one player can may either of the following two moves:

• Convert even array element to any other integer.
• Remove odd array element.

The player who is not able to make any move loses the game. The task is to print the winner of the game. Print -1 if the game may go on forever.

Examples:

Input: arr[] = {3, 1, 9, 7}
Output: Player 2
Explanation: Since all array elements are odd, no conversion is possible.
Turn 1: Player 1 deletes 3.
Turn 2: Player 2 deletes 1.
Turn 3: Player 1 deletes 9.
Turn 4: Player 2 deletes 7. Now, Player 1 has no moves left. Therefore, Player 2 wins the game.

Input: arr[]={4, 8}
Output: -1

Approach: Follow the steps below to solve the problem:

• Traverse the array.
• Count the number of even and odd elements present in the array.
• If the number of even elements is zero, perform the following operations: 
  • If the count of odd is even, then Player 2 will win the game.
  • Otherwise, Player 1 will win the game.
• If the count of odd is odd and only one even element is present in the array, then Player 1 will win the game.
• Otherwise, there will be a draw every time.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to evaluate the
// winner of the game
void findWinner(int arr[], int N)
{
    // Stores count of odd
    // array elements
    int odd = 0;

    // Stores count of even
    // array elements
    int even = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // If array element is odd
        if (arr[i] % 2 == 1) {

            odd++;
        }
        // Otherwise
        else {

            even++;
        }
    }

    // If count of even is zero
    if (even == 0) {

        // If count of odd is even
        if (odd % 2 == 0) {

            cout << "Player 2" << endl;
        }

        // If count of odd is odd
        else if (odd % 2 == 1) {

            cout << "Player 1" << endl;
        }
    }

    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1) {

        cout << "Player 1" << endl;
    }

    // Otherwise
    else {

        cout << -1 << endl;
    }
}

// Driver Code
int main()
{
    int arr[] = { 3, 1, 9, 7 };
    int N = sizeof(arr)
            / sizeof(arr[0]);

    findWinner(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to evaluate the
// winner of the game
static void findWinner(int arr[], int N)
{

    // Stores count of odd
    // array elements
    int odd = 0;

    // Stores count of even
    // array elements
    int even = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If array element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }

        // Otherwise
        else
        {
            even++;
        }
    }

    // If count of even is zero
    if (even == 0)
    {

        // If count of odd is even
        if (odd % 2 == 0)
        {
            System.out.println("Player 2");
        }

        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            System.out.println("Player 1");
        }
    }

    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        System.out.println("Player 1");
    }

    // Otherwise
    else
    {
        System.out.println(-1);
    }
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { 3, 1, 9, 7 };
    int N = arr.length;

    findWinner(arr, N);
}
}

// This code is contributed by ipg2016107

Python3

# Python3 program for the above approach

# Function to evaluate the
# winner of the game
def findWinner(arr, N):

    # Stores count of odd
    # array elements
    odd = 0

    # Stores count of even
    # array elements
    even = 0

    # Traverse the array
    for i in range(N):

        # If array element is odd
        if (arr[i] % 2 == 1):
            odd += 1

        # Otherwise
        else:
            even += 1

    # If count of even is zero
    if (even == 0):

        # If count of odd is even
        if (odd % 2 == 0):
            print("Player 2")

        # If count of odd is odd
        elif (odd % 2 == 1):
            print("Player 1")

    # If count of odd is odd and
    # count of even is one
    elif (even == 1 and odd % 2 == 1):
        print("Player 1")

    # Otherwise
    else:
        print(-1)

# Driver code
if __name__ == '__main__':

    arr = [ 3, 1, 9, 7 ]
    N = len(arr)

    findWinner(arr, N)

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;

class GFG{

// Function to evaluate the
// winner of the game
static void findWinner(int []arr, int N)
{

    // Stores count of odd
    // array elements
    int odd = 0;

    // Stores count of even
    // array elements
    int even = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If array element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }

        // Otherwise
        else
        {
            even++;
        }
    }

    // If count of even is zero
    if (even == 0)
    {

        // If count of odd is even
        if (odd % 2 == 0)
        {
            Console.WriteLine("Player 2");
        }

        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            Console.WriteLine("Player 1");
        }
    }

    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        Console.WriteLine("Player 1");
    }

    // Otherwise
    else
    {
        Console.WriteLine(-1);
    }
}

// Driver Code
public static void Main()
{
    int []arr = { 3, 1, 9, 7 };
    int N = arr.Length;

    findWinner(arr, N);
}
}

// This code is contributed by bgangwar59

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to evaluate the
// winner of the game
function findWinner(arr, N)
{

    // Stores count of odd
    // array elements
    let odd = 0;

    // Stores count of even
    // array elements
    let even = 0;

    // Traverse the array
    for(let i = 0; i < N; i++)
    {

        // If array element is odd
        if (arr[i] % 2 == 1)
        {
            odd++;
        }

        // Otherwise
        else
        {
            even++;
        }
    }

    // If count of even is zero
    if (even == 0)
    {

        // If count of odd is even
        if (odd % 2 == 0)
        {
            document.write("Player 2");
        }

        // If count of odd is odd
        else if (odd % 2 == 1)
        {
            document.write("Player 1");
        }
    }

    // If count of odd is odd and
    // count of even is one
    else if (even == 1 && odd % 2 == 1)
    {
        document.write("Player 1");
    }

    // Otherwise
    else
    {
        document.write(-1);
    }
}

// Driver Code

    let arr = [ 3, 1, 9, 7 ];
    let N = arr.length;

    findWinner(arr, N);

</script>

Output:

Player 2

Time Complexity: O(N)  
Auxiliary Space: O(1)