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

Last Updated : 08 Apr, 2021

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:

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.
Since count of candies of player A decreases by 1 in odd moves, chanceA = 2 * (X – 1) + 1
Since count of candies of player B decreases by 1 in even moves, chanceB = 2 * Y
If chanceA < chanceB, then B will be the winning player.
Otherwise, A will be the winning player.
Print the winning player.

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 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 << "B";
    }
  
    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;
}

Output:

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

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My Personal Notes

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

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