Find the winner of a game of donating i candies in every i-th move

Given two integers X and Y representing the number of candies allocated to players A and B respectively, where both the players indulge in a game of donating i candies to the opponent in every i^th move. Starting with player A, the game continues with alternate turns until a player is unable to donate the required amount of candies and loses the game, the task is to find the winner of the game.

Examples:

Input: X = 2, Y = 3
Output: A
Explanation: The game turns out in the following manner:

Step	X	Y
0	2	3
1	1	4
2	3	2
3	0	5
4	4	1

Since A fails to donate 5 candies in the 5^th step, B wins the game.

 Input: X = 2, Y = 1
Output: B
Explanation: The game turns out in the following manner:

Step	X	Y
0	2	1
1	1	2
2	3	0
3	0	3

Since B fails to give 4 candies in the 4^th step, A wins the game.

Approach: The idea is to solve the problem based on the following observation:

Step	Number of candies Possessed by A	Number of candies Possessed by B
0	X	Y
1	X - 1	Y + 1
2	X - 1 + 2 = X + 1	Y + 1 - 2 = Y - 1
3	X - 2	Y + 2
4	X + 2	Y - 2

• The player whose candies reduce to 0 first, will not be having enough candies to give in the next move.
• Count of candies of player A decreases by 1 in odd moves.
• Count of candies of player B decreases by 1 in even moves.

Follow the steps below to solve the problem:

1. Initialize two variables, say chanceA and chanceB, representing the number of moves in which the number of candies possessed by a player reduces to 0.
2. Since count of candies of player A decreases by 1 in odd moves, chanceA = 2 * (X - 1) + 1
3. Since count of candies of player B decreases by 1 in even moves, chanceB = 2 * Y
4. If chanceA < chanceB, then B will be the winning player.
5. Otherwise, A will be the winning player.
6. Print the winning player.

Below is the implementation of the above approach:

// C++ Program for the
// above approach

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

// Function to find the winning
// player in a game of donating i
// candies to opponent in i-th move
int stepscount(int a, int b)
{
    // Steps in which number of
    // candies of player A finishes
    int chanceA = 2 * a - 1;

    // Steps in which number of
    // candies of player B finishes
    int chanceB = 2 * b;

    // If A's candies finishes first
    if (chanceA < chanceB) {
        cout << "B";
    }

    // Otherwise
    else if (chanceB < chanceA) {
        cout << "A";
    }

    return 0;
}

// Driver Code
int main()
{
    // Input

    // Candies possessed
    // by player A
    int A = 2;

    // Candies possessed
    // by player B
    int B = 3;

    stepscount(A, B);

    return 0;
}

// Java Program for above approach
class GFG{
    
// Function to find the winning
// player in a game of donating i
// candies to opponent in i-th move
static void stepscount(int a, int b)
{
    
    // Steps in which number of
    // candies of player A finishes
    int chanceA = 2 * a - 1;

    // Steps in which number of
    // candies of player B finishes
    int chanceB = 2 * b;

    // If A's candies finishes first
    if (chanceA < chanceB)
    {
        System.out.print("B");
    }

    // Otherwise
    else if (chanceB < chanceA) 
    {
        System.out.print("A");
    }
}

// Driver code
public static void main(String[] args)
{
    
    // Input
    // Candies possessed by player A
    int A = 2;

    // Candies possessed by player B
    int B = 3;

    stepscount(A, B);
}
}

// This code is contributed by abhinavjain194

# Python3 program for the above approach

# Function to find the winning
# player in a game of donating i
# candies to opponent in i-th move
def stepscount(a, b):
    
    # Steps in which number of
    # candies of player A finishes
    chance_A = 2 * a - 1
    
    # Steps in which number of
    # candies of player B finishes
    chance_B = 2 * b
    
    # If A's candies finishes first
    if (chance_A < chance_B):
        return 'B'
        
    # Otherwise    
    else:
        return "A"

# Driver code

# Candies possessed by player A        
A = 2

# Candies possessed by player B
B = 3

print(stepscount(A, B))

# This code is contributed by abhinavjain194

// C# program for the above approach
using System;

class GFG{
    
// Function to find the winning
// player in a game of donating i
// candies to opponent in i-th move
static void stepscount(int a, int b)
{
    
    // Steps in which number of
    // candies of player A finishes
    int chanceA = 2 * a - 1;

    // Steps in which number of
    // candies of player B finishes
    int chanceB = 2 * b;

    // If A's candies finishes first
    if (chanceA < chanceB)
    {
        Console.Write("B");
    }

    // Otherwise
    else if (chanceB < chanceA) 
    {
        Console.Write("A");
    }
}

// Driver code
static void Main()
{
    
    // Input
    // Candies possessed by player A
    int A = 2;

    // Candies possessed by player B
    int B = 3;

    stepscount(A, B);
}
}

// This code is contributed by abhinavjain194

<script>

// Javascript program for the above approach


// Function to find the winning
// player in a game of donating i
// candies to opponent in i-th move
function stepscount( a, b)
{
    // Steps in which number of
    // candies of player A finishes
    let chanceA = 2 * a - 1;

    // Steps in which number of
    // candies of player B finishes
    ;let chanceB = 2 * b;

    // If A's candies finishes first
    if (chanceA < chanceB) {
        document.write("B");
    }

    // Otherwise
    else if (chanceB < chanceA) {
        document.write("A");
    }

    return 0;
}

// Driver Code

// Input

// Candies possessed
// by player A
let A = 2;

// Candies possessed
// by player B
let B = 3;

stepscount(A, B);

</script>

B

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