Two player game in which a player can remove all occurrences of a number

Two players player1 and player2 are playing a game on a given number sequence S where player1 starts first and both of them play optimally. The task is to find whether player1 wins or loses. If he wins print “Yes” otherwise print “No”. 
The rules of the game are as follows: 

Examples: 

Input: S = {2, 1, 2, 2} 
Output: Yes 
Explanation: 
The first player can choose the subset of elements with indices {0, 3} {0, 2} or {2, 3} (in the sequence) to ensure his victory in the future.

Input: S = {3, 2, 2, 3, 3, 5} 
Output:No 
Explanation: 
The first player can not win in this sequence. 
If player1 chooses 2 and remove all the 2 from sequence then player2 will choose two 3 and remove only two 3 as both players play optimally. 
Then player1 removes either 3 or 5 and player2 will remove the final element. So player1 will always lose. 

Approach: The approach to the above problem can be given as if the total number of elements in sequence is even and the number of elements that occur more than once is also even then the first player can not win. If the number of elements that occur more than once is odd and the occurrence of repeated elements is greater than or equal to 4 and multiple of 2, then the first player can not win. Otherwise, player1 can be the winner.



Below is the implementation of the above approach: 

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// C++ implementation for Two player
// game in which a player can remove
// all occurrences of a number
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that print whether
// player1 can wins or loses
void game(int v[], int n)
{
    unordered_map<int, int> m;
 
    // storing the number of occurrence
    // of elements in unordered map
    for (int i = 0; i < n; i++) {
 
        if (m.find(v[i]) == m.end())
            m[v[i]] = 1;
 
        else
            m[v[i]]++;
    }
 
    int count = 0;
 
    // variable to check if the
// occurrence of repeated
// elements is >= 4 and
    // multiple of 2 or not
    int check = 0;
 
    // count elements which
// occur more than once
    for (auto i: m) {
        if (i.second > 1) {
 
            if (i.second >= 4
&& i.second % 2 == 0)
                check++;
 
            count++;
        }
    }
 
    if (check % 2 != 0)
        bool flag = false;
 
    if (check % 2 != 0)
        cout << "Yes" << endl;
 
    else if (n % 2 == 0
&& count % 2 == 0)
        cout << "No" << endl;
 
    else
        cout << "Yes" << endl;
}
 
// Driver code
int main()
{
 
    int arr[] = { 3, 2, 2, 3, 3, 5 };
 
    int size = sizeof(arr)
/ sizeof(arr[0]);
 
    game(arr, size);
 
    return 0;
}
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// Java implementation for Two player
// game in which a player can remove
// all occurrences of a number
import java.util.*;
class GFG{
     
// Function that print whether
// player1 can wins or loses
public static void game(int[] v,
                        int n)
{
  HashMap<Integer,
          Integer> m = new HashMap<>();
 
  // Storing the number of occurrence
  // of elements in unordered map
  for (int i = 0; i < n; i++)
  {
    if (!m.containsKey(v[i]))
      m.put(v[i], 1);
    else
      m.replace(v[i], m.get(v[i]) + 1);
  }
 
  int count = 0;
 
  // variable to check if the
  // occurrence of repeated
  // elements is >= 4 and
  // multiple of 2 or not
  int check = 0;
 
  // count elements which
  // occur more than once
  for (Map.Entry<Integer,
                 Integer> i : m.entrySet())
  {
    if(i.getValue() > 1)
    {
      if(i.getValue() >= 4 &&
         i.getValue() % 2 == 0)
        check++;
 
      count++;
    }
  }
   
  boolean flag;
   
  if (check % 2 != 0)
    flag = false;
  if (check % 2 != 0)
    System.out.println("Yes");
  else if (n % 2 == 0  &&
           count % 2 == 0)
    System.out.println("No");
  else
    System.out.println("Yes");
}
 
public static void main(String[] args)
{
  int[] arr = {3, 2, 2, 3, 3, 5};
  int size = arr.length;
  game(arr, size);
}
}
 
// This code is contributed by divyeshrabadiya07
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# Python3 implementation for two player
# game in which a player can remove
# all occurrences of a number
from collections import defaultdict
 
# Function that print whether
# player1 can wins or loses
def game(v, n):
 
    m = defaultdict(int)
     
    # Storing the number of occurrence
    # of elements in unordered map
    for i in range(n):
        if (v[i] not in m):
            m[v[i]] = 1
             
        else:
            m[v[i]] += 1
             
    count = 0
 
    # Variable to check if the
    # occurrence of repeated
    # elements is >= 4 and
    # multiple of 2 or not
    check = 0
 
    # Count elements which
    # occur more than once
    for i in m.values():
        if (i > 1):
            if (i >= 4 and i % 2 == 0):
                check += 1
            count += 1
 
    if (check % 2 != 0):
        flag = False
 
    if (check % 2 != 0):
        print("Yes")
 
    elif (n % 2 == 0 and count % 2 == 0):
        print("No")
         
    else:
        print("Yes")
 
# Driver code
if __name__ == "__main__":
     
    arr = [ 3, 2, 2, 3, 3, 5 ]
    size = len(arr)
     
    game(arr, size)
 
# This code is contributed by chitranayal
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// C# implementation for Two player
// game in which a player can remove
// all occurrences of a number
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function that print whether
// player1 can wins or loses
public static void game(int[] v,
                        int n)
{
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
     
    // Storing the number of occurrence
    // of elements in unordered map
    for(int i = 0; i < n; i++)
    {
        if (!m.ContainsKey(v[i]))
            m.Add(v[i], 1);
        else
            m[v[i]]= m[v[i]] + 1;
    }
     
    int count = 0;
     
    // Variable to check if the
    // occurrence of repeated
    // elements is >= 4 and
    // multiple of 2 or not
    int check = 0;
     
    // Count elements which
    // occur more than once
    foreach(KeyValuePair<int,
                         int> i in m)
    {
        if (i.Value > 1)
        {
            if (i.Value >= 4 &&
                i.Value % 2 == 0)
                check++;
             
            count++;
        }
    }
     
    bool flag;
     
    if (check % 2 != 0)
        flag = false;
    if (check % 2 != 0)
        Console.WriteLine("Yes");
    else if (n % 2 == 0 &&
         count % 2 == 0)
        Console.WriteLine("No");
    else
        Console.WriteLine("Yes");
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 3, 2, 2, 3, 3, 5 };
    int size = arr.Length;
     
    game(arr, size);
}
}
 
// This code is contributed by Amit Katiyar
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Output: 
No








 

Time complexity: The complexity of the above approach is O(N).

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