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:
- Initialize count0, count1, count2 and count3 to 0.
- Iterate for every number in the array and increase the above counters accordingly if a[i] % 4 == 0, a[i] % 4 == 1, a[i] % 4 == 2 or a[i] % 4 == 3.
- If count0, count1, count2 and count3 are all even numbers then X wins else Y will win.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Function to decide the winnerint 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 % 2 == 0)
return 1;
else
return 2;
}// Driver codeint 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;
} |
Java
// Java implementation of the approachclass GFG
{ // Function to decide the winnerstatic 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 % 2 == 0)
return 1;
else
return 2;
}// Driver codepublic 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 |
Python3
# Python3 implementation of the approach# Function to decide the winnerdef 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 % 2 == 0):
return 1
else:
return 2
# Driver codea = [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 |
C#
// C# implementation of the approachusing System;
class GFG
{ // Function to decide the winnerstatic 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 % 2 == 0)
return 1;
else
return 2;
}// Driver codepublic 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 |
PHP
<?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 && $count3 % 2 == 0)
return 1;
else
return 2;
} // Driver code$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?> |
Javascript
<script>// javascript implementation of the approach // Function to decide the winner function decideWinner(a , n) {
var count0 = 0;
var count1 = 0;
var count2 = 0;
var 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 % 2 == 0)
return 1;
else
return 2;
}
// Driver code
var a = [ 4, 8, 5, 9 ];
var n = a.length;
if (decideWinner(a, n) == 1)
document.write("X wins");
else
document.write("Y wins");
// This code contributed by Rajput-Ji</script> |
Output:
X wins
Time Complexity: O(n)
Auxiliary Space: O(1)