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[] = { “neveropen”, “geek” }, brr[] = { “neveropen”, “neveropen” }
Output: Player 1
Explanation:
Turn 1: Player 1 removed “neveropen” from arr[].
Turn 2: Player 2 removed “neveropen” 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<bits/stdc++.h>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 arrayvoid lastPlayer(int n, int m, vector<string> arr, vector<string> brr){ // Stores common strings // from both the array set<string> 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<string> 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<string> 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")<<endl; } else { cout<<("Player 2")<<endl; }}// Driver Codeint main(){ // Set of strings for player A vector<string> arr{ "neveropen", "geek" }; // Set of strings for player B vector<string> brr{ "neveropen", "neveropen" }; 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 approachimport 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<String> 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<String> 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<String> 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 = { "neveropen", "geek" }; // Set of strings for player B String[] brr = { "neveropen", "neveropen" }; 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 arraydef 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 Codeif __name__ == '__main__': # Set of strings for player A arr = ["neveropen", "geek"] # Set of strings for player B brr = ["neveropen", "neveropen"] n = len(arr) m = len(brr) lastPlayer(n, m, arr, brr) |
C#
// C# Program for the above approachusing 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<String> common = new HashSet<String>(); 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<String> a = new HashSet<String>(); 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<String> b = new HashSet<String>(); 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 = { "neveropen", "geek" }; // Set of strings for player B String[] brr = { "neveropen", "neveropen" }; int n = arr.Length; int m = brr.Length; lastPlayer(n, m, arr, brr); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript 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 function lastPlayer(n, m, arr, brr) { // Stores common strings // from both the array var common = []; for (var i = 0; i < arr.length; i++) { for (var j = 0; j < brr.length; j++) { if (arr[i] === brr[j]) { // add common elements common.push(arr[i]); j = brr.length; } } } // Removing common strings from []arr var a = []; var flag; for (var i = 0; i < arr.length; i++) { flag = false; common.forEach((value) => { if (value === arr[i]) { // add common elements flag = true; i = arr.length; } }); if (flag) a.push(arr[i]); } // Removing common elements from B var b = []; for (var i = 0; i < brr.length; i++) { flag = false; common.forEach((value) => { if (value === brr[i]) { // add common elements flag = true; i = brr.length; } }); if (flag) b.push(brr[i]); } // Stores strings in brr[] which // is not common in []arr var LenBrr = b.length; if (common.length % 2 === 1) { // Update LenBrr LenBrr -= 1; } if (a.length > LenBrr) { document.write("Player 1"); } else { document.write("Player 2"); } } // Driver Code // Set of strings for player A var arr = ["neveropen", "geek"]; // Set of strings for player B var brr = ["neveropen", "neveropen"]; var n = arr.length; var m = brr.length; lastPlayer(n, m, arr, brr); // This code is contributed by rdtank. </script> |
Output
Player 1
Time Complexity: O(N + M)
Auxiliary Space: O(N + M)
Using Hash Map:
Approach:
In this approach, we can use a hash map to keep track of the removed strings in both arrays. Each time a player removes a string from one of the arrays, we mark that string as removed in the hash map. Then, we check if any string in the other array is still available to be removed. If there is no available string, the last player to remove a string wins.
Initialize an empty dictionary to keep track of the removed strings in both arrays.
Initialize a variable ‘turn’ to 1, to keep track of which player’s turn it is.
While at least one of the arrays is not empty, do the following:
If it is player 1’s turn and array ‘arr’ is not empty, remove the first string from ‘arr’ and mark it as removed in the dictionary.
If it is player 2’s turn and array ‘brr’ is not empty, remove the first string from ‘brr’ and mark it as removed in the dictionary.
Switch turns by setting ‘turn’ to 3 – ‘turn’.
Check if there are any strings left in the other array that have not been removed yet.
If there are no available strings, return the current player’s turn as the winner.
If both arrays are empty and no winner has been found, return -1 as an error code.
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;int findLastPlayer(vector<string>& arr, vector<string>& brr){ unordered_map<string, bool> removed; int turn = 1; while (!arr.empty() || !brr.empty()) { if (turn == 1) { if (arr.empty()) { return 2; } string toRemove = arr[0]; removed[toRemove] = true; arr.erase(arr.begin()); } else { if (brr.empty()) { return 1; } string toRemove = brr[0]; removed[toRemove] = true; brr.erase(brr.begin()); } turn = 3 - turn; bool foundInArr = false; for (string& s : arr) { if (removed.find(s) == removed.end()) { foundInArr = true; break; } } if (!foundInArr) { for (string& s : brr) { if (removed.find(s) == removed.end()) { return 1; } } return 2; // If both arr and brr are empty, // return 2 } } return -1; // should never reach this point}int main(){ vector<string> arr1 = { "neveropen", "geek" }; vector<string> brr1 = { "neveropen", "neveropen" }; cout << findLastPlayer(arr1, brr1) << endl; // should print 1 vector<string> arr2 = { "a", "b" }; vector<string> brr2 = { "a", "b" }; cout << findLastPlayer(arr2, brr2) << endl; // should print 2 return 0;}// This code is contributed by Abhinav Mahajan (abhinav_m22) |
Java
import java.util.ArrayList;import java.util.HashMap;public class LastPlayerFinder { static int findLastPlayer(ArrayList<String> arr, ArrayList<String> brr) { HashMap<String, Boolean> removed = new HashMap<>(); int turn = 1; while (!arr.isEmpty() || !brr.isEmpty()) { if (turn == 1) { if (arr.isEmpty()) { return 2; } String toRemove = arr.remove(0); removed.put(toRemove, true); } else { if (brr.isEmpty()) { return 1; } String toRemove = brr.remove(0); removed.put(toRemove, true); } turn = 3 - turn; boolean foundInArr = false; for (String s : arr) { if (!removed.containsKey(s)) { foundInArr = true; break; } } if (!foundInArr) { for (String s : brr) { if (!removed.containsKey(s)) { return 1; } } return 2; // If both arr and brr are empty, return 2 } } return -1; // should never reach this point } public static void main(String[] args) { ArrayList<String> arr1 = new ArrayList<>(); arr1.add("neveropen"); arr1.add("geek"); ArrayList<String> brr1 = new ArrayList<>(); brr1.add("neveropen"); brr1.add("neveropen"); System.out.println(findLastPlayer(arr1, brr1)); // should print 1 ArrayList<String> arr2 = new ArrayList<>(); arr2.add("a"); arr2.add("b"); ArrayList<String> brr2 = new ArrayList<>(); brr2.add("a"); brr2.add("b"); System.out.println(findLastPlayer(arr2, brr2)); // should print 2 }} |
Python3
def find_last_player(arr, brr): removed = {} turn = 1 while len(arr) > 0 or len(brr) > 0: if turn == 1: if len(arr) == 0: return 2 to_remove = arr.pop(0) removed[to_remove] = True else: if len(brr) == 0: return 1 to_remove = brr.pop(0) removed[to_remove] = True turn = 3 - turn for i in range(len(arr)): if arr[i] not in removed: break else: for i in range(len(brr)): if brr[i] not in removed: break else: return turn return -1 # should never reach this pointarr = ["neveropen", "geek"]brr = ["neveropen", "neveropen"]print(find_last_player(arr, brr)) # should print 1arr = ["a", "b"]brr = ["a", "b"]print(find_last_player(arr, brr)) # should print 2 |
C#
using System;using System.Collections.Generic;class Program{ static int FindLastPlayer(List<string> arr, List<string> brr) { Dictionary<string, bool> removed = new Dictionary<string, bool>(); int turn = 1; while (arr.Count > 0 || brr.Count > 0) { if (turn == 1) { if (arr.Count == 0) { return 2; } string toRemove = arr[0]; removed[toRemove] = true; arr.RemoveAt(0); } else { if (brr.Count == 0) { return 1; } string toRemove = brr[0]; removed[toRemove] = true; brr.RemoveAt(0); } turn = 3 - turn; bool foundInArr = false; foreach (string s in arr) { if (!removed.ContainsKey(s)) { foundInArr = true; break; } } if (!foundInArr) { foreach (string s in brr) { if (!removed.ContainsKey(s)) { return 1; } } return 2; // If both arr and brr are empty, return 2 } } return -1; // should never reach this point } static void Main() { List<string> arr1 = new List<string> { "neveropen", "geek" }; List<string> brr1 = new List<string> { "neveropen", "neveropen" }; Console.WriteLine(FindLastPlayer(arr1, brr1)); // should print 1 List<string> arr2 = new List<string> { "a", "b" }; List<string> brr2 = new List<string> { "a", "b" }; Console.WriteLine(FindLastPlayer(arr2, brr2)); // should print 2 }} |
Javascript
function find_last_player(arr, brr) { let removed = {}; let turn = 1; while (arr.length > 0 || brr.length > 0) { if (turn === 1) { if (arr.length === 0) { return 2; } let toRemove = arr.shift(); removed[toRemove] = true; } else { if (brr.length === 0) { return 1; } let toRemove = brr.shift(); removed[toRemove] = true; } turn = 3 - turn; let arrHasUnremovedElement = arr.some((element) => !removed[element]); if (!arrHasUnremovedElement) { let brrHasUnremovedElement = brr.some((element) => !removed[element]); if (!brrHasUnremovedElement) { return turn; } } } return -1;}let arr1 = ["neveropen", "geek"];let brr1 = ["neveropen", "neveropen"];console.log(find_last_player(arr1, brr1));let arr2 = ["a", "b"];let brr2 = ["a", "b"];console.log(find_last_player(arr2, brr2)); |
Output
1 2
Time Complexity: O(n+m), where n and m are the lengths of the two arrays.
Auxiliary Space: O(n+m), for the hash map.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Share your thoughts in the comments
