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:

  • The player’s alternate turns.
  • In each turn, the current player must choose one or more elements of the current sequence S such that the values of all chosen elements are identical, and erase these elements from S.
  • When a player cannot choose anything (the sequence S is already empty), this player loses the game.

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:

C++

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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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Python3

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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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Output:

No

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

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Improved By : chitranayal