Find the winner of a game of removing any number of stones from the least indexed non-empty pile from given N piles Last Updated : 17 Jun, 2021 Comments Improve Suggest changes Like Article Like Report Given an array arr[] consisting of N integers, each representing size of a pile of stones. The task is to determine the winner of the game when two players, A and B, play a game optimally as per the following conditions: Player A always starts the game.In one move, a player may remove any number of stones (at least 1), from the first non-empty pile with minimal index.The first player who cannot make a move loses the game. Print “A” if player A wins the game. Otherwise, print “B”. Examples: Input: arr[] = {1, 1, 1, 1, 1, 1}Output: BExplanation:Here, each pile has only one stone so A and B will alternatively remove one stone each.A removes 1 stone from 1st pile, B removes 1 from 2nd pile and so on.Last stone in 6th pile is removed by B.Since A has no choice left, B wins the game. Input: arr[] = {1, 1, 2, 1}Output: AExplanation:Here, A removes 1 stone from 1st pile, B removes 1 from 2nd pile, A removes only 1 from 3rd pile, and now B is forced to remove 1 from 3rd pile. Last stone in 4th pile is removed by A.Since B has no choice left, A wins the game. Approach: The idea is to figure out the optimal ways that players should follow to win the game. Below are two ways to play optimally: For all piles except the last one, if there are K stones on a pile then the current player will pick only (K - 1) stones (only if K > 1) so that another player is forced to pick the remaining 1 stone. This guarantees the current player to get a chance to pick stones from the next pile and eventually win.For the last pile, if it has K stones, then all K stones will be picked by the current player so that the other player does not get a chance to pick stones. This eventually guarantees the current player to win. Follow the steps below to solve this problem: Initially, assuming that the player will remove all stones from the current pile, A will be the winner if the size of the array is odd and B if the size is even.Now, iterate over the range [0, N - 2] and check the current index and number of stones present on the current pile to decide which player will win.While traversing, if the index is even, A will get a chance to pick a stone. Otherwise, B will get a chance to pick stones.If the current index is even and the number of stones is greater than 1, then the winner is set to B. Update winner to A.If the current index is odd and the number of stones exceeds 1, the winner is set to be A. Then, update the winner to B.Finally, print the winner of the game. Below is the implementation of the above approach: C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the winner // of game between A and B void findWinner(int a[], int n) { // win = 1 means B is winner // win = 0 means A is winner int win = 0; // If size is even, winner is B if (n % 2 == 0) win = 1; // If size is odd, winner is A else win = 0; for (int i = n - 2; i >= 0; i--) { // Stone will be removed by B if (i % 2 == 1) { // If B wants to win // B will take n-1 stones // from current pile having // n stones and force A to // pick 1 stone if (win == 0 && a[i] > 1) win = 1; } // Stone will be removed by A else { // If A wants to win // A will take n-1 stones from // current pile having n stones // and force B to pick 1 stone if (win == 1 && a[i] > 1) win = 0; } } // Print the winner accordingly if (win == 0) cout << "A"; else cout << "B"; } // Driver Code int main() { // Given piles of stone int arr[] = { 1, 1, 1, 2 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call findWinner(arr, N); return 0; } Java // Java program for the above approach import java.util.*; class GFG{ // Function to find the winner // of game between A and B static void findWinner(int a[], int n) { // win = 1 means B is winner // win = 0 means A is winner int win = 0; // If size is even, winner is B if (n % 2 == 0) win = 1; // If size is odd, winner is A else win = 0; for(int i = n - 2; i >= 0; i--) { // Stone will be removed by B if (i % 2 == 1) { // If B wants to win // B will take n-1 stones // from current pile having // n stones and force A to // pick 1 stone if (win == 0 && a[i] > 1) win = 1; } // Stone will be removed by A else { // If A wants to win // A will take n-1 stones from // current pile having n stones // and force B to pick 1 stone if (win == 1 && a[i] > 1) win = 0; } } // Print the winner accordingly if (win == 0) System.out.print("A"); else System.out.print("B"); } // Driver Code public static void main(String[] args) { // Given piles of stone int arr[] = { 1, 1, 1, 2 }; int N = arr.length; // Function call findWinner(arr, N); } } // This code is contributed by Amit Katiyar Python3 # Python3 program for the above approach # Function to find the winner # of game between A and B def findWinner(a, n): # win = 1 means B is winner # win = 0 means A is winner win = 0 # If size is even, winner is B if (n % 2 == 0): win = 1 # If size is odd, winner is A else: win = 0 for i in range(n - 2, -1, -1): # Stone will be removed by B if (i % 2 == 1): # If B wants to win # B will take n-1 stones # from current pile having # n stones and force A to # pick 1 stone if (win == 0 and a[i] > 1): win = 1 # Stone will be removed by A else: # If A wants to win # A will take n-1 stones from # current pile having n stones # and force B to pick 1 stone if (win == 1 and a[i] > 1): win = 0 # Print the winner accordingly if (win == 0): print("A") else: print("B") # Driver Code if __name__ == '__main__': # Given piles of stone arr = [ 1, 1, 1, 2 ] N = len(arr) # Function call findWinner(arr, N) # This code is contributed by mohit kumar 29 C# // C# program for the // above approach using System; class GFG{ // Function to find the winner // of game between A and B static void findWinner(int []a, int n) { // win = 1 means B is winner // win = 0 means A is winner int win = 0; // If size is even, winner is B if (n % 2 == 0) win = 1; // If size is odd, winner is A else win = 0; for(int i = n - 2; i >= 0; i--) { // Stone will be removed by B if (i % 2 == 1) { // If B wants to win // B will take n-1 stones // from current pile having // n stones and force A to // pick 1 stone if (win == 0 && a[i] > 1) win = 1; } // Stone will be removed by A else { // If A wants to win // A will take n-1 stones from // current pile having n stones // and force B to pick 1 stone if (win == 1 && a[i] > 1) win = 0; } } // Print the winner accordingly if (win == 0) Console.Write("A"); else Console.Write("B"); } // Driver Code public static void Main(String[] args) { // Given piles of stone int []arr = {1, 1, 1, 2}; int N = arr.Length; // Function call findWinner(arr, N); } } // This code is contributed by Rajput-Ji JavaScript <script> // Javascript program for the above approach // Function to find the winner // of game between A and B function findWinner(a, n) { // win = 1 means B is winner // win = 0 means A is winner let win = 0; // If size is even, winner is B if (n % 2 == 0) win = 1; // If size is odd, winner is A else win = 0; for(let i = n - 2; i >= 0; i--) { // Stone will be removed by B if (i % 2 == 1) { // If B wants to win // B will take n-1 stones // from current pile having // n stones and force A to // pick 1 stone if (win == 0 && a[i] > 1) win = 1; } // Stone will be removed by A else { // If A wants to win // A will take n-1 stones from // current pile having n stones // and force B to pick 1 stone if (win == 1 && a[i] > 1) win = 0; } } // Print the winner accordingly if (win == 0) document.write("A"); else document.write("B"); } // Driver Code // Given piles of stone let arr=[1, 1, 1, 2]; let N = arr.length; // Function call findWinner(arr, N); // This code is contributed by unknown2108 </script> Output: B Time Complexity: O(N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Asymptotic Notations for Analysis of Algorithms S Shivamthakur77 Follow Improve Article Tags : DSA Similar Reads Basics & PrerequisitesTime and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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