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Find the winner of the game with N piles of boxes
  • Difficulty Level : Easy
  • Last Updated : 26 May, 2021

Given N piles containing White(W) and Black(B) boxes only, two players A and B play a game. 
Player A may remove any number of boxes from the top of the pile having the topmost box White and Player B may remove any number of boxes from the top of the pile having the topmost box Black. If there is a pile with top as W, then player A has remove one or more boxes. Similarly if there is a pile with top as B, then player B has to remove one or more boxes. They play alternating, with the player A first to move. Both players plays optimally. 
The task is to find the winner of the game (who cannot make the last move). Player last to move loses the game.
Examples: 
 

Input: N = 2 
WBW, BWB 
Output: A 
Player A can remove all boxes from pile 1. Now player B has to make a choice from pile 2. Whatever choice player B makes, he/she has to make the last move.
Input: N = 3 
WWWW, WBWB, WBBW 
Output: B

 

Approach: The game theory problems, where the player who makes the last move lose the game, are called Misere Nim’s Game. 
 

  1. If there are consecutive occurrences of the same box in any pile, we can simply replace them with one copy of the same box, since it is optimal to remove all occurrences of the box present at the top. So, we can compress all piles to get the pile of form BWBWBW or WBWBWB.
  2. Now, suppose there’s a pile, which have topmost box differing from its bottom-most box, we can prove that answer remains same even without this pile, because if one player makes a move on this pile, other player makes the next move by removing box from the same pile, resulting in exactly same position for the first player.
  3. If a pile has both top and bottom-most box as same, It requires one or the player to make one extra move. So, just count the number of extra move each player has to make. The player which runs of out extra moves first wins the game.

Below is the implementation of the above approach.
 



C++




// Program to find winner of the game
#include <bits/stdc++.h>
using namespace std;
 
// Function to return find winner of game
string Winner(int n, string pile[])
{
 
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i) {
        int l = pile[i].length();
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
int main()
{
 
    int n = 2;
    string pile[n] = { "WBW", "BWB" };
 
    // function to print required answer
    cout << Winner(n, pile);
 
    return 0;
}

Java




// Program to find winner of the game
class GFG
{
 
// Function to return find winner of game
static String Winner(int n, String []pile)
{
 
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i)
    {
        int l = pile[i].length();
 
        // Piles begins and ends with White box 'W'
        if (pile[i].charAt(0) == pile[i].charAt(l - 1) &&
            pile[i].charAt(0) == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i].charAt(0) == pile[i].charAt(l - 1) &&
            pile[i].charAt(0) == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2;
    String pile[] = { "WBW", "BWB" };
 
    // function to print required answer
    System.out.println(Winner(n, pile));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 code to find winner of the game
 
# Function to return find winner of game
def Winner(n, pile):
 
    a, b = 0, 0
    for i in range(0, n):
        l = len(pile[i])
 
        # Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] and
            pile[i][0] == 'W'):
            a += 1
 
        # Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] and
            pile[i][0] == 'B'):
            b += 1
     
    if a <= b:
        return "A"
    else:
        return "B"
 
# Driver code
if __name__ == "__main__":
 
    n = 2
    pile = ["WBW", "BWB"]
 
    # function to print required answer
    print(Winner(n, pile))
 
# This code is contributed by Rituraj Jain

C#




// Program to find winner of the game
using System;
     
class GFG
{
 
// Function to return find winner of game
static String Winner(int n, String []pile)
{
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i)
    {
        int l = pile[i].Length;
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] &&
            pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] &&
            pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 2;
    String []pile = { "WBW", "BWB" };
 
    // function to print required answer
    Console.WriteLine(Winner(n, pile));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Program to find winner of the game
 
// Function to return find winner of game
function Winner(n, pile)
{
 
    var a = 0, b = 0;
    for (var i = 0; i < n; ++i) {
        var l = pile[i].length;
 
        // Piles begins and ends with White box 'W'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'W')
            a++;
 
        // Piles begins and ends with Black box 'B'
        if (pile[i][0] == pile[i][l - 1] && pile[i][0] == 'B')
            b++;
    }
    if (a <= b)
        return "A";
    else
        return "B";
}
 
// Driver code
var n = 2;
var pile = ["WBW", "BWB"];
 
// function to print required answer
document.write( Winner(n, pile));
 
// This code is contributed by noob2000.
</script>
Output: 
A

 

Time Complexity: O(N)
 

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