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A modified game of Nim
  • Last Updated : 26 Apr, 2019

Given an array arr[] of integers, two players A and B are playing a game where A can remove any number of non-zero elements from the array that are multiples of 3. Similarly, B can remove multiples of 5. The player who can’t remove any element loses the game. The task is to find the winner of the game if A starts first and both play optimally.

Examples:

Input: arr[] = {1, 2, 3, 5, 6}
Output: A
3 and 6 are the elements that A can remove.
5 is the only element that B can remove.
A can remove 3 in his first move then B will have to remove 5. In the next turn, A will remove 6 and B will be left with no more moves to make.

Input: arr[] = {3, 5, 15, 20, 6, 9}
Output: A

Approach: Store the count of elements only divisible by 3 in movesA, count of elements only divisible by 5 in movesB and the elements divisible by both in movesBoth. Now,



  • If movesBoth = 0 then both the players can remove only the elements which are divisible by their respective number and A will win the game only when movesA > movesB.
  • If movesBoth > 0 then in order to play optimally, A will remove all the elements that are divisible by both 3 and 5 so that B is left with no elements to remove from the common elements then A will be the winner only if movesA + 1 > movesB

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the winner of the game
string getWinner(int arr[], int n)
{
    int movesA = 0, movesB = 0, movesBoth = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Increment common moves
        if (arr[i] % 3 == 0 && arr[i] % 5 == 0)
            movesBoth++;
  
        // Increment A's moves
        else if (arr[i] % 3 == 0)
            movesA++;
  
        // Increment B's moves
        else if (arr[i] % 5 == 0)
            movesB++;
    }
  
    // If there are no common moves
    if (movesBoth == 0) {
        if (movesA > movesB)
            return "A";
        return "B";
    }
  
    // 1 is added because A can remove all the elements
    // that are part of the common moves in a single move
    if (movesA + 1 > movesB)
        return "A";
    return "B";
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << getWinner(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GfG
{
  
    // Function to return the winner of the game 
    static String getWinner(int arr[], int n) 
    
        int movesA = 0, movesB = 0, movesBoth = 0
      
        for (int i = 0; i < n; i++) 
        
      
            // Increment common moves 
            if (arr[i] % 3 == 0 && arr[i] % 5 == 0
                movesBoth++; 
      
            // Increment A's moves 
            else if (arr[i] % 3 == 0
                movesA++; 
      
            // Increment B's moves 
            else if (arr[i] % 5 == 0
                movesB++; 
        
      
        // If there are no common moves 
        if (movesBoth == 0)
        
            if (movesA > movesB) 
                return "A"
            return "B"
        
      
        // 1 is added because A can remove 
        // all the elements that are part
        // of the common moves in a single move 
        if (movesA + 1 > movesB) 
            return "A"
        return "B"
    }
  
    // Driver code 
    public static void main(String []args)
    {
          
        int arr[] = { 1, 2, 3, 5, 6 }; 
        int n = arr.length; 
        System.out.println(getWinner(arr, n));
    }
}
  
// This code is contributed by Rituraj Jain

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Python3

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# Python3 implementation of the approach 
  
# Function to return the winner of the game 
def getWinner(arr, n): 
  
    movesA, movesB, movesBoth = 0, 0, 0
    for i in range(0, n): 
  
        # Increment common moves 
        if arr[i] % 3 == 0 and arr[i] % 5 == 0
            movesBoth += 1
  
        # Increment A's moves 
        elif arr[i] % 3 == 0
            movesA += 1
  
        # Increment B's moves 
        elif arr[i] % 5 == 0
            movesB += 1
  
    # If there are no common moves 
    if movesBoth == 0
        if movesA > movesB: 
            return "A"
        return "B"
  
    # 1 is added because A can 
    # remove all the elements 
    # that are part of the common
    # moves in a single move 
    if movesA + 1 > movesB: 
        return "A"
    return "B"
  
# Driver code 
if __name__ == "__main__":
  
    arr = [1, 2, 3, 5, 6
    n = len(arr) 
    print(getWinner(arr, n)) 
  
# This code is contributed by Rituraj Jain

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C#

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// C# implementation of the approach
using System;
  
class GfG
{
  
    // Function to return the winner of the game 
    static String getWinner(int []arr, int n) 
    
        int movesA = 0, movesB = 0, movesBoth = 0; 
      
        for (int i = 0; i < n; i++) 
        
      
            // Increment common moves 
            if (arr[i] % 3 == 0 && arr[i] % 5 == 0) 
                movesBoth++; 
      
            // Increment A's moves 
            else if (arr[i] % 3 == 0) 
                movesA++; 
      
            // Increment B's moves 
            else if (arr[i] % 5 == 0) 
                movesB++; 
        
      
        // If there are no common moves 
        if (movesBoth == 0)
        
            if (movesA > movesB) 
                return "A"
            return "B"
        
      
        // 1 is added because A can remove 
        // all the elements that are part
        // of the common moves in a single move 
        if (movesA + 1 > movesB) 
            return "A"
        return "B"
    }
  
    // Driver code 
    public static void Main(String []args)
    {
          
        int []arr = { 1, 2, 3, 5, 6 }; 
        int n = arr.Length; 
        Console.WriteLine(getWinner(arr, n));
    }
}
  
// This code is contributed by
// Rajput-Ji

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the winner of the game
function getWinner($arr, $n)
{
    $movesA = 0;
    $movesB = 0;
    $movesBoth = 0;
  
    for ($i = 0; $i < $n; $i++) 
    {
  
        // Increment common moves
        if ($arr[$i] % 3 == 0 && $arr[$i] % 5 == 0)
            $movesBoth++;
  
        // Increment A's moves
        else if ($arr[$i] % 3 == 0)
            $movesA++;
  
        // Increment B's moves
        else if ($arr[$i] % 5 == 0)
            $movesB++;
    }
  
    // If there are no common moves
    if ($movesBoth == 0)
    {
        if ($movesA > $movesB)
            return "A";
        return "B";
    }
  
    // 1 is added because A can remove all the elements
    // that are part of the common moves in a single move
    if ($movesA + 1 > $movesB)
        return "A";
    return "B";
}
  
    // Driver code
    $arr = array( 1, 2, 3, 5, 6 );
    $n = sizeof($arr) / sizeof($arr[0]);
    echo getWinner($arr, $n);
  
// This code is contributed by ajit.
?>

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Output:

A

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