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)
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