# Find the player who wins the game of placing alternate + and – signs in front of array elements

Find the player who wins the game of placing alternate + and – signs in front of array elements

Given an array arr[] of length, N, the task is to find the winner of a game played by two players A and B optimally, by performing the following operations:

Player A makes the first move.

Players need to alternately place + and – sign in front of array elements in their turns.

After having placed signs in front of all array elements, player A wins if the difference of all the elements is even.

Otherwise, player B wins.

Examples:

Input: arr[] = {1, 2}Output: BExplanation:All possible ways the game can be played out are:(+1) – (+2) = -1 (−1) – (+2) = -3 (+1) – (-2) = 3 (-1) – (-2) = 1Since the differences are odd in all the possibilities, B wins.

Input: arr[] = {1, 1, 2}Output: AExplanation: All possible ways the game can be played out are:(1) – (1) – (2) = -2 (1) – (1) – (-2) = 2(1) – (-1) – (2) = 0(1) – (-1) – (-2) = 4(-1) – (1) – (2) = -4(-1) – (1) – (-2) = 0(-1) – (-1) – (2) = -2(-1) – (-1) – (-2) = 4Since the differences are even in all the possibilities, A wins.

Naive Approach: The simplest approach is to generate all the possible 2N combinations in which the signs can be placed in the array and check for each combination, check if player A can win or not. If found to be true for any permutation, then print A. Otherwise, player B wins.Time Complexity: O(2N * N)Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to optimize the above approach:Initialize a variable, say diff, to store the sum of the array elements.

Traverse the array arr[], over the range of indices [1, N], and update diff by subtracting arr[i] to it.

If diff % 2 is found to be equal to 0, then print ‘A’. Otherwise, print ‘B’.

Below is the implementation of the above approach:

C++

#include

using namespace std;

void checkWinner(int arr[], int N)

{

int diff = 0;

for (int i = 0; i < N; i++) {
diff -= arr[i];
}
if (diff % 2 == 0) {
cout