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: B
Explanation:
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: A
Explanation:
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 Bvoid 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the winner// of game between A and Bstatic 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 Codepublic 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 Bdef 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 Codeif __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 approachusing System;class GFG{// Function to find the winner// of game between A and Bstatic 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 Codepublic 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 Bfunction 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 stonelet arr=[1, 1, 1, 2];let N = arr.length;// Function callfindWinner(arr, N);// This code is contributed by unknown2108</script> |
Output:
B
Time Complexity: O(N)
Auxiliary Space: O(1)
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