# Find the last player to be able to remove a string from an array which is not already removed from other array

• Difficulty Level : Hard
• Last Updated : 13 May, 2021

Given two arrays of strings arr[] and brr[] of size N and M respectively, the task is to find the winner of the game when two players play the game optimally as per the following rules:

• Player 1 starts the game.
• Player 1 removes a string from the array arr[] if it is not already removed from the array brr[].
• Player 2 removes a string from the array brr[] if it is not already removed from the array arr[].
• The player who is not able to remove a string from the array, then the player will lose the game.

Examples:

Input: arr[] = { “geeks”, “geek” }, brr[] = { “geeks”, “geeksforgeeks” }
Output: Player 1
Explanation:
Turn 1: Player 1 removed “geeks” from arr[].
Turn 2: Player 2 removed “geeksforgeeks” from brr[]
Turn 3: Player 1 removed “geek” from brr[].
Now, player 2 cannot remove any string.
Therefore, the required output is Player 1.

Input: arr[] = { “a”, “b” }, brr[] = { “a”, “b” }
Output: Player 2
Explanation:
Turn 1: Player 1 removed “a” from arr[].
Turn 2: Player 2 removed “b” from brr[].
Therefore, the required output is Player 2

Approach: The idea to based on the fact that common strings from both the arrays can be removed only from one of the arrays. Follow the steps below to solve the problem:

• If the count of common strings from both the arrays is an odd number, then remove one string from the array brr[], as Player 1 starts the game and the first common string is removed by Player 1.
• If count of strings in arr[] is greater than the count of strings in brr[] by removing the common strings from both the arrays, then print “Player 1”.
• Otherwise, print “Player 2”.

Below is the implementation of the above approach:

## C++

 `// C++ Program for the above approach``#include``using` `namespace` `std;`` ` `// Function to find last player to be``// able to remove a string from one array``// which has not been removed from the other array``void` `lastPlayer(``int` `n, ``int` `m, vector arr,``                       ``vector brr)``{` `    ``// Stores common strings``    ``// from both the array``   ``set common;` `    ``for` `(``int` `i = 0; i < arr.size(); i++)``    ``{``        ``for` `(``int` `j = 0; j < brr.size(); j++)``        ``{``            ``if` `(arr[i] == brr[j])``            ``{` `                ``// add common elements``                ``common.insert(arr[i]);``                ``break``;``            ``}``        ``}``    ``}` `    ``// Removing common strings from arr[]``    ``set a;``    ``bool` `flag;``    ``for` `(``int` `i = 0; i < arr.size(); i++)``    ``{``        ``flag = ``false``;``        ``for` `(``auto` `value : common)``        ``{``            ``if` `(value == arr[i])``            ``{` `                ``// add common elements``                ``flag = ``true``;``                ``break``;``            ``}``        ``}``        ``if` `(flag)``            ``a.insert(arr[i]);``    ``}` `    ``// Removing common elements from B``    ``set b;``    ``for` `(``int` `i = 0; i < brr.size(); i++)``    ``{``        ``flag = ``false``;``        ``for` `(``auto` `value : common)``        ``{``            ``if` `(value == brr[i])``            ``{` `                ``// add common elements``                ``flag = ``true``;``                ``break``;``            ``}``        ``}` `        ``if` `(flag)``            ``b.insert(brr[i]);``    ``}` `    ``// Stores strings in brr[] which``    ``// is not common in arr[]``    ``int` `LenBrr = b.size();``    ``if` `((common.size()) % 2 == 1)``    ``{` `        ``// Update LenBrr``        ``LenBrr -= 1;``    ``}` `    ``if` `(a.size() > LenBrr)``    ``{``        ``cout<<(``"Player 1"``)< arr{ ``"geeks"``, ``"geek"` `};` `    ``// Set of strings for player B``    ``vector brr{ ``"geeks"``, ``"geeksforgeeks"` `};``    ``int` `n = arr.size();``    ``int` `m = brr.size();``    ``lastPlayer(n, m, arr, brr);``}` `// This code is contributed by SURENDRA_GANGWAR.`

## Java

 `// Java Program for the above approach``import` `java.io.*;``import` `java.util.*;``class` `GFG``{``    ``// Function to find last player to be``    ``// able to remove a string from one array``    ``// which has not been removed from the other array``    ``static` `void` `lastPlayer(``int` `n, ``int` `m, String[] arr,``                           ``String[] brr)``    ``{` `        ``// Stores common strings``        ``// from both the array``        ``Set common = ``new` `HashSet<>();` `        ``for` `(``int` `i = ``0``; i < arr.length; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < brr.length; j++)``            ``{``                ``if` `(arr[i] == brr[j])``                ``{` `                    ``// add common elements``                    ``common.add(arr[i]);``                    ``break``;``                ``}``            ``}``        ``}` `        ``// Removing common strings from arr[]``        ``Set a = ``new` `HashSet<>();``        ``boolean` `flag;``        ``for` `(``int` `i = ``0``; i < arr.length; i++)``        ``{``            ``flag = ``false``;``            ``for` `(String value : common)``            ``{``                ``if` `(value == arr[i])``                ``{` `                    ``// add common elements``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}``            ``if` `(flag)``                ``a.add(arr[i]);``        ``}` `        ``// Removing common elements from B``        ``Set b = ``new` `HashSet<>();``        ``for` `(``int` `i = ``0``; i < brr.length; i++)``        ``{``            ``flag = ``false``;``            ``for` `(String value : common)``            ``{``                ``if` `(value == brr[i])``                ``{` `                    ``// add common elements``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}` `            ``if` `(flag)``                ``b.add(brr[i]);``        ``}` `        ``// Stores strings in brr[] which``        ``// is not common in arr[]``        ``int` `LenBrr = b.size();``        ``if` `((common.size()) % ``2` `== ``1``)``        ``{` `            ``// Update LenBrr``            ``LenBrr -= ``1``;``        ``}` `        ``if` `(a.size() > LenBrr)``        ``{``            ``System.out.print(``"Player 1"``);``        ``}``        ``else``        ``{``            ``System.out.print(``"Player 2"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``      ` `        ``// Set of strings for player A``        ``String[] arr = { ``"geeks"``, ``"geek"` `};` `        ``// Set of strings for player B``        ``String[] brr = { ``"geeks"``, ``"geeksforgeeks"` `};``        ``int` `n = arr.length;``        ``int` `m = brr.length;``        ``lastPlayer(n, m, arr, brr);``    ``}``}` `// This code is contributed by Dharanendra L V.`

## Python

 `# Python Program for the above approach`  `# Function to find last player to be``# able to remove a string from one array``# which has not been removed from the other array``def` `lastPlayer(n, m, arr, brr):` `    ``# Stores common strings``    ``# from both the array``    ``common ``=` `list``(``set``(arr) & ``set``(brr))` `    ``# Removing common strings from arr[]``    ``a ``=` `list``(``set``(arr) ^ ``set``(common))` `    ``# Removing common elements from B``    ``b ``=` `list``(``set``(brr) ^ ``set``(common))` `    ``# Stores strings in brr[] which``    ``# is not common in arr[]``    ``LenBrr ``=` `len``(b)` `    ``if` `len``(common) ``%` `2` `=``=` `1``:` `        ``# Update LenBrr``        ``LenBrr ``-``=` `1``    ` `    ``if` `len``(a) > LenBrr:``        ``print``(``"Player 1"``)``    ``else``:``        ``print``(``"Player 2"``)`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Set of strings for player A``    ``arr ``=` `[``"geeks"``, ``"geek"``]` `    ``# Set of strings for player B``    ``brr ``=` `[``"geeks"``, ``"geeksforgeeks"``]``    ` `    ``n ``=` `len``(arr)``    ``m ``=` `len``(brr)` `    ``lastPlayer(n, m, arr, brr)`

## C#

 `// C# Program for the above approach``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG``{``  ` `    ``// Function to find last player to be``    ``// able to remove a string from one array``    ``// which has not been removed from the other array``    ``static` `void` `lastPlayer(``int` `n, ``int` `m, String[] arr,``                           ``String[] brr)``    ``{` `        ``// Stores common strings``        ``// from both the array``        ``HashSet common = ``new` `HashSet();``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``for` `(``int` `j = 0; j < brr.Length; j++)``            ``{``                ``if` `(arr[i] == brr[j])``                ``{` `                    ``// add common elements``                    ``common.Add(arr[i]);``                    ``break``;``                ``}``            ``}``        ``}` `        ``// Removing common strings from []arr``        ``HashSet a = ``new` `HashSet();``        ``bool` `flag;``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``flag = ``false``;``            ``foreach` `(String value ``in` `common)``            ``{``                ``if` `(value == arr[i])``                ``{` `                    ``// add common elements``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}``            ``if` `(flag)``                ``a.Add(arr[i]);``        ``}` `        ``// Removing common elements from B``        ``HashSet b = ``new` `HashSet();``        ``for` `(``int` `i = 0; i < brr.Length; i++)``        ``{``            ``flag = ``false``;``            ``foreach` `(String value ``in` `common)``            ``{``                ``if` `(value == brr[i])``                ``{` `                    ``// add common elements``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}` `            ``if` `(flag)``                ``b.Add(brr[i]);``        ``}` `        ``// Stores strings in brr[] which``        ``// is not common in []arr``        ``int` `LenBrr = b.Count;``        ``if` `((common.Count) % 2 == 1)``        ``{` `            ``// Update LenBrr``            ``LenBrr -= 1;``        ``}` `        ``if` `(a.Count > LenBrr)``        ``{``            ``Console.Write(``"Player 1"``);``        ``}``        ``else``        ``{``            ``Console.Write(``"Player 2"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``      ` `        ``// Set of strings for player A``        ``String[] arr = { ``"geeks"``, ``"geek"` `};` `        ``// Set of strings for player B``        ``String[] brr = { ``"geeks"``, ``"geeksforgeeks"` `};``        ``int` `n = arr.Length;``        ``int` `m = brr.Length;``        ``lastPlayer(n, m, arr, brr);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`Player 1`

Time Complexity: O(N + M)
Auxiliary Space: O(N + M)

My Personal Notes arrow_drop_up