Given an arr[] of positive integers of length N. Player X and Y plays a game where in each move one person chooses two elements of arr[] (say arr[i] and arr[j]) such that 0 ? i < j ? N – 1 and arri should be greater than 0 and Then decrement arri and increment arrj. When one player is unable to make such a move the player loses. The task is to find the winner of the game if Y starts the game.
Examples:
Input: N = 4, arr[] = {2, 3, 1, 3}
Output: Y
Explanation: initial arr[] is: {2, 3, 1, 3}
Player Y applied by selecting i = 0 and j = 1: {1, 4, 1, 3}
Player X applied by selecting i = 0 and j = 1: {0, 5, 1, 3}
Player Y applied by selecting i = 1 and j = 2: {0, 4, 2, 3}
Player X applied by selecting i = 1 and j = 2: {0, 3, 3, 3}
Player Y applied by selecting i = 1 and j = 2: {0, 2, 4, 3}
Player X applied by selecting i = 1 and j = 2: {0, 1, 5, 3}
Player Y applied by selecting i = 1 and j = 2: {0, 0, 6, 3}
Player X applied by selecting i = 2 and j = 3: {0, 0, 5, 4}
Player Y applied by selecting i = 2 and j = 3: {0, 0, 4, 5}
Player X applied by selecting i = 2 and j = 3: {0, 0, 3, 6}
Player Y applied by selecting i = 2 and j = 3: {0, 0, 2, 7}
Player X applied by selecting i = 2 and j = 3: {0, 0, 1, 8}
Player Y applied by selecting i = 2 and j = 3: {0, 0, 0, 9}
As there are no more indices i and j left in arr[],
Which follows 0<=i<j<N-1 and Ai > 0.
Therefore, player X can’t make its next operation.
Therefore, Player Y wins the game.
Input: N = 2, arr[] = {0, 1}
Output: X
Explanation: Player Y can’t make very first operation of the game, Therefore player X wins the game.
Approach: To solve the problem follow the below idea:
As we can see that it is clearly mentioned in the game that index i (0 ? i ? N – 1) should be greater than index j for applying operation, Thus by this observation, we can see that at any index i in input arr[] there are N – i – 1 possibles positions for subtracting one from ith index and add it to jth index. Thus the game reduced into sub-games and is similar like “Game of Nim”.
- In Game of Nim, Winner decides on the basis of Nim-Sum. If Nim-sum is not equal to zero then player who makes first move wins and if Nim-Sum is equal to zero then other player Wins.
- Same this problem is related to Combinatorial Game Theory and uses Sprague Grundy Theorem to solve the problem.
- Game of Nim depends upon only Nim-sum and we have a XOR property as (A^A) = 0.If an element is even at index i, then it will give its part as even N-i in overall Nim-sum else odd N-i in overall Nim-sum.
According to above three lines, we can conclude that to get the answer calculate XOR of N-i-1 with the ith element, where ith element is odd in arr[].
if Nim-Sum comes zero then player X wins else player Y wins.
Follow the steps to solve the problem:
- Create any long type variable let’s say Z and initialize it to 0.
- Iterate on arr[] from left to right, if an element is found odd, Then update Z as Z ? (N – i – 1).
- After completing iteration if Z = 0, Then player X wins else Y wins.
- Print X/Y as the winner based on the resultant value of Z in the above step.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void find_winner(int N, int arr[])
{
long Z = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 2 == 1) {
Z = Z ^ (N - i - 1);
}
}
if (Z == 0) {
cout<<"X";
}
else {
cout<<"Y";
}
}
int main()
{
int N = 4;
int arr[] = { 2, 3, 1, 3 };
find_winner(N, arr);
return 0;
}
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Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
public static void main(String[] args)
{
int N = 4;
int[] arr = { 2, 3, 1, 3 };
find_winner(N, arr);
}
static void find_winner(int N, int[] arr)
{
long Z = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 2 == 1) {
Z = Z ^ (N - i - 1);
}
}
if (Z == 0) {
System.out.println("X");
}
else {
System.out.println("Y");
}
}
}
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Python3
def find_winner(N, arr) :
Z = 0
for i in range(N):
if (arr[i] % 2 == 1) :
Z = Z ^ (N - i - 1)
if (Z == 0) :
print("X")
else :
print("Y")
if __name__ == "__main__":
N = 4
arr = [ 2, 3, 1, 3 ]
find_winner(N, arr)
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C#
using System;
public class GFG {
static void find_winner(int N, int[] arr)
{
long Z = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 2 == 1) {
Z = Z ^ (N - i - 1);
}
}
if (Z == 0) {
Console.WriteLine("X");
}
else {
Console.WriteLine("Y");
}
}
static public void Main()
{
int N = 4;
int[] arr = { 2, 3, 1, 3 };
find_winner(N, arr);
}
}
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Javascript
function find_winner( N, arr)
{
let Z = 0;
for (let i = 0; i < N; i++) {
if (arr[i] % 2 == 1) {
Z = Z ^ (N - i - 1);
}
}
if (Z == 0) {
console.log("X");
}
else {
console.log("Y");
}
}
let N = 4;
let arr = [ 2, 3, 1, 3 ];
find_winner(N, arr);
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Time Complexity: O(N)
Auxiliary Space: O(1)