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A modified game of Nim

Last Updated : 03 Aug, 2022
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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:
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:
 


 


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++
// 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;
}
Java
// 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
Python3
# 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
C#
// 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
PHP
<?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.
?>
JavaScript
<script>

// Javascript implementation of the approach

// Function to return the winner of the game
function getWinner(arr, n)
{
    let movesA = 0, movesB = 0, movesBoth = 0;

    for (let 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
    let arr = [ 1, 2, 3, 5, 6 ];
    let n = arr.length;
    document.write(getWinner(arr, n));

</script>

Output: 
A

 

Time Complexity: O(n)

Auxiliary Space: O(1)


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