Predict the winner of the game on the basis of absolute difference of sum by selecting numbers

Given an array of N numbers. Two players X and Y play a game where at every step one player selects a number. One number can be selected only once. After all the numbers have been selected, player X wins if the absolute difference between the sum of numbers collected by X and Y is divisible by 4, else Y wins.
Note: Player X starts the game and numbers are selected optimally at every step.

Examples:

Input: a[] = {4, 8, 12, 16}
Output: X
X chooses 4
Y chooses 12
X chooses 8
Y chooses 16
|(4 + 8) – (12 + 16)| = |12 – 28| = 16 which is divisible by 4.
Hence, X wins



Input: a[] = {7, 9, 1}
Output: Y

Approach: The following steps can be followed to solve the problem:

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to decide the winner
int decideWinner(int a[], int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
  
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++) {
  
        // Condition to count
  
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
  
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
  
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
  
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
  
    // Check the winning condition for X
    if (count0 % 2 == 0
        && count1 % 2 == 0
        && count2 % 2 == 0
        && count3 == 0)
        return 1;
    else
        return 2;
}
  
// Driver code
int main()
{
  
    int a[] = { 4, 8, 5, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    if (decideWinner(a, n) == 1)
        cout << "X wins";
    else
        cout << "Y wins";
  
    return 0;
}
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// Java implementation of the approach
class GFG
{
      
// Function to decide the winner
static int decideWinner(int []a, int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
  
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++)
    {
  
        // Condition to count
  
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
  
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
  
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
  
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
  
    // Check the winning condition for X
    if (count0 % 2 == 0 && count1 % 2 == 0 && 
        count2 % 2 == 0 && count3 == 0)
        return 1;
    else
        return 2;
}
  
// Driver code
public static void main(String args[])
{
    int []a = { 4, 8, 5, 9 };
    int n = a.length;
    if (decideWinner(a, n) == 1)
        System.out.print("X wins");
    else
        System.out.print("Y wins");
}
}
  
// This code is contributed by Akanksha Rai
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# Python3 implementation of the approach
  
# Function to decide the winner
def decideWinner(a, n):
    count0 = 0
    count1 = 0
    count2 = 0
    count3 = 0
  
    # Iterate for all numbers in the array
    for i in range(n):
  
        # Condition to count
  
        # If mod gives 0
        if (a[i] % 4 == 0):
            count0 += 1
  
        # If mod gives 1
        elif (a[i] % 4 == 1):
            count1 += 1
  
        # If mod gives 2
        elif (a[i] % 4 == 2):
            count2 += 1
  
        # If mod gives 3
        elif (a[i] % 4 == 3):
            count3 += 1
      
    # Check the winning condition for X
    if (count0 % 2 == 0 and count1 % 2 == 0 and 
        count2 % 2 == 0 and count3 == 0):
        return 1
    else:
        return 2
  
# Driver code
a = [4, 8, 5, 9]
n = len(a)
if (decideWinner(a, n) == 1):
    print("X wins")
else:
    print("Y wins")
  
# This code is contributed by mohit kumar
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// C# implementation of the approach
using System;
class GFG
{
      
// Function to decide the winner
static int decideWinner(int []a, int n)
{
    int count0 = 0;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
  
    // Iterate for all numbers in the array
    for (int i = 0; i < n; i++)
    {
  
        // Condition to count
  
        // If mod gives 0
        if (a[i] % 4 == 0)
            count0++;
  
        // If mod gives 1
        else if (a[i] % 4 == 1)
            count1++;
  
        // If mod gives 2
        else if (a[i] % 4 == 2)
            count2++;
  
        // If mod gives 3
        else if (a[i] % 4 == 3)
            count3++;
    }
  
    // Check the winning condition for X
    if (count0 % 2 == 0 && count1 % 2 == 0 && 
        count2 % 2 == 0 && count3 == 0)
        return 1;
    else
        return 2;
}
  
// Driver code
public static void Main()
{
    int []a = { 4, 8, 5, 9 };
    int n = a.Length;
    if (decideWinner(a, n) == 1)
        Console.Write("X wins");
    else
        Console.Write("Y wins");
}
}
  
// This code is contributed by Akanksha Rai
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<?php
// PHP implementation of the approach 
  
// Function to decide the winner 
function decideWinner($a, $n
    $count0 = 0; 
    $count1 = 0; 
    $count2 = 0; 
    $count3 = 0; 
  
    // Iterate for all numbers in the array 
    for ($i = 0; $i < $n; $i++) 
    
  
        // Condition to count 
  
        // If mod gives 0 
        if ($a[$i] % 4 == 0) 
            $count0++; 
  
        // If mod gives 1 
        else if ($a[$i] % 4 == 1) 
            $count1++; 
  
        // If mod gives 2 
        else if ($a[$i] % 4 == 2) 
            $count2++; 
  
        // If mod gives 3 
        else if ($a[$i] % 4 == 3) 
            $count3++; 
    
  
    // Check the winning condition for X 
    if ($count0 % 2 == 0 && $count1 % 2 == 0 && 
        $count2 % 2 == 0 && $count3 == 0) 
        return 1; 
    else
        return 2; 
  
// Driver code
$a = array( 4, 8, 5, 9 ); 
$n = count($a);
  
if (decideWinner($a, $n) == 1) 
    echo "X wins"
else
    echo "Y wins"
      
// This code is contributed by Ryuga
?>
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Output:
X wins



Striver(underscore)79 at Codechef and codeforces D

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