Given an array arr[] of positive distinct integers, two players A and B are playing a game. At each move, a player selects two numbers x and y from the array and if |x – y| is not present in the array then the player adds this number to the array (size of the array increases by 1). The player who can’t make the move loses the game. The task is to find the winner of the game if player A always starts the game.
Examples:
Input: arr[] = {2, 3}
Output: A
After A’s move, array will be {2, 3, 1} and B can’t make any move.
Input: arr[] = {5, 6, 7}
Output: B
Approach: Observe here that at the end of the game (when there are no more moves to make), the resultant array will contain all the multiples of the gcd of the original array upto the maximum element of the original array.
For example, arr[] = {8, 10}
Since, gcd(8, 10) = 2. So the resultant array at the end of the game will contain all the multiples of 2 ? max(arr) i.e. 10.
Hence, arr[] = {2, 4, 6, 8, 10}
From the above observation, the number of moves that can be performed on the original array can be found which will determine the winner of the game, if the number of moves is even then B will be the winner of the game else A wins the game.
Number of moves can be found as, (max(arr) / gcd) – n where gcd is the gcd of the original array elements and max(arr) / gcd gives the total number of elements in the resultant array. Subtracting original count of elements from the count of elements in the resultant array will give the number of moves.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the winner of the gamechar getWinner(int arr[], int n){ // To store the gcd of the original array int gcd = arr[0]; // To store the maximum element // from the original array int maxEle = arr[0]; for (int i = 1; i < n; i++) { gcd = __gcd(gcd, arr[i]); maxEle = max(maxEle, arr[i]); } int totalMoves = (maxEle / gcd) - n; // If number of moves are odd if (totalMoves % 2 == 1) return 'A'; return 'B';}// Driver Codeint main(){ int arr[] = { 5, 6, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << getWinner(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to calculate gcdstatic int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to return the winner// of the gamestatic char getWinner(int []arr, int n){ // To store the gcd of the // original array int gcd = arr[0]; // To store the maximum element // from the original array int maxEle = arr[0]; for (int i = 1; i < n; i++) { gcd = __gcd(gcd, arr[i]); maxEle = Math.max(maxEle, arr[i]); } int totalMoves = (maxEle / gcd) - n; // If number of moves are odd if (totalMoves % 2 == 1) return 'A'; return 'B';}// Driver Codepublic static void main(String args[]){ int []arr = { 5, 6, 7 }; int n = arr.length; System.out.print(getWinner(arr, n));}}// This code is contributed// by Akanksha Rai |
Python3
# Python3 implementation of the approach from math import gcd# Function to return the winner # of the game def getWinner(arr, n) : # To store the gcd of the # original array __gcd = arr[0]; # To store the maximum element # from the original array maxEle = arr[0]; for i in range(1, n) : __gcd = gcd(__gcd, arr[i]); maxEle = max(maxEle, arr[i]); totalMoves = (maxEle / __gcd) - n; # If number of moves are odd if (totalMoves % 2 == 1) : return 'A'; return 'B'; # Driver Code if __name__ == "__main__" : arr = [ 5, 6, 7 ]; n = len(arr) print(getWinner(arr, n)) # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;class GFG{ // Function to calculate gcdstatic int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to return the winner// of the gamestatic char getWinner(int []arr, int n){ // To store the gcd of the // original array int gcd = arr[0]; // To store the maximum element // from the original array int maxEle = arr[0]; for (int i = 1; i < n; i++) { gcd = __gcd(gcd, arr[i]); maxEle = Math.Max(maxEle, arr[i]); } int totalMoves = (maxEle / gcd) - n; // If number of moves are odd if (totalMoves % 2 == 1) return 'A'; return 'B';}// Driver Codepublic static void Main(){ int []arr = { 5, 6, 7 }; int n = arr.Length; Console.Write(getWinner(arr, n));}}// This code is contributed// by Akanksha Rai |
PHP
<?php// PHP implementation of the approach// Function to calculate gcdfunction __gcd($a, $b) { if ($b == 0) return $a; return __gcd($b, $a % $b); } // Function to return the winner// of the gamefunction getWinner($arr, $n){ // To store the gcd of the // original array $gcd = $arr[0]; // To store the maximum element // from the original array $maxEle = $arr[0]; for ($i = 1; $i < $n; $i++) { $gcd = __gcd($gcd, $arr[$i]); $maxEle = max($maxEle, $arr[$i]); } $totalMoves = ($maxEle / $gcd) - $n; // If number of moves are odd if ($totalMoves % 2 == 1) return 'A'; return 'B';}// Driver Code$arr = array(5, 6, 7);$n = sizeof($arr);echo getWinner($arr, $n);// This code is contributed// by Akanksha Rai?> |
Javascript
<script> // JavaScript implementation of the approach // Function to calculate gcd function __gcd(a, b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to return the winner // of the game function getWinner(arr, n) { // To store the gcd of the // original array let gcd = arr[0]; // To store the maximum element // from the original array let maxEle = arr[0]; for (let i = 1; i < n; i++) { gcd = __gcd(gcd, arr[i]); maxEle = Math.max(maxEle, arr[i]); } let totalMoves = parseInt(maxEle / gcd, 10) - n; // If number of moves are odd if (totalMoves % 2 == 1) return 'A'; return 'B'; } let arr = [ 5, 6, 7 ]; let n = arr.length; document.write(getWinner(arr, n)); </script> |
B
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
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