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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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ``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

 ` ``\$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.``?>`
Output:
```A
```

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