Find the player who wins the game by removing the last of given N cards (original) (raw)

Last Updated : 4 May, 2021

Given two integers N and K, where N represents the total number of cards present when game begins and K denotes the maximum number of cards that can be removed in a single turn. Two players A and B get turns to remove at most K cards, one by one starting from player A. The player to remove the last card is the winner. The task is to check if A can win the game or not. If found to be true, print 'A' as the answer. Otherwise, print 'B'.

Examples:

Input: N = 14, K = 10
Output: Yes
Explanation:
Turn 1: A removes 3 cards in his first turn.
Turn 2: B removes any number of cards from the range [1 - 10]
Finally, A can remove all remaining cards and wins the game, as the number of remaining cards after turn 2 will be ≤ 10

Input: N= 11, K=10
Output: No

Approach: The idea here is to observe that whenever the value of N % (K + 1) = 0, then A will never be able to win the game. Otherwise A always win the game.
Proof:

1. If N ≤ K: The person who has the first turn will win the game, i.e. A.
2. If N = K + 1: A can remove any number of cards in the range [1, K]. So, the total number of cards left after the first turn are also in the range [1, K]. Now B gets the turn and number of cards left are in the range [1, K]. So, B will win the game.
3. If K + 2 ≤ N ≤ 2K + 1: A removes N - (K + 1) cards in his first turn. B can remove any number of cards in the range [1, K] in the next turn. Therefore, the total number of cards left now are in the range [1, K].Now, since the remaining cards left are in the range [1, K], so A can remove all the cards and win the game.

Therefore the idea is to check if N % (K + 1) is equal to 0 or not. If found to be true, print B as the winner. Otherwise, print A as the winner.
Below is the implementation of the above approach:

C++ `

// C++ Program to implement // the above approach

#include <bits/stdc++.h> using namespace std;

// Function to check which // player can win the game void checkWinner(int N, int K) { if (N % (K + 1)) { cout << "A"; } else { cout << "B"; } }

// Driver code int main() {

int N = 50;
int K = 10;
checkWinner(N, K);

}

Java

// Java program to implement // the above approach import java.util.*;

class GFG{

// Function to check which // player can win the game static void checkWinner(int N, int K) { if (N % (K + 1) > 0) { System.out.print("A"); } else { System.out.print("B"); } }

// Driver code public static void main(String[] args) { int N = 50; int K = 10;

checkWinner(N, K);

} }

// This code is contributed by Amit Katiyar

Python3

Python3 program to implement

the above approach

Function to check which

player can win the game

def checkWinner(N, K):

if(N % (K + 1)):
    print("A")
else:
    print("B")

Driver Code

N = 50 K = 10

Function call

checkWinner(N, K)

This code is contributed by Shivam Singh

C#

// C# program to implement // the above approach using System;

class GFG{

// Function to check which // player can win the game static void checkWinner(int N, int K) { if (N % (K + 1) > 0) { Console.Write("A"); } else { Console.Write("B"); } }

// Driver code public static void Main(String[] args) { int N = 50; int K = 10;

checkWinner(N, K);

} }

// This code is contributed by Amit Katiyar

JavaScript

`

Time Complexity: O(1)
Auxiliary Space: O(1)