Coin Piles In C++

The primary challenges in algorithmic competitions often revolve around the "Coin Piles" dilemma. This guide delves into mathematical insights and optimized algorithms. Let's delve deeper into strategies for addressing this issue.

Problem statement:

You have two sets of coins, pile A and pile B, each containing a different number of coins. You have the flexibility to execute multiple actions, which can be categorized into the following options:

• ✅ Taking away 2 coins from pile A and 1 coin from pile B.
• ✅ Taking away 1 coin from pile A and 2 coins from pile B.

Your task is to determine if it is possible to clear both stacks using any of these actions.

Some major observations

To address this issue, we need to examine several important characteristics.

1. Total Coin Reduction:

When you perform a move, the total number of coins between the two piles decreases by 3, which can be achieved by taking 2 coins from pile A and 1 from pile B, or 1 coin from pile A and 2 from pile B. Therefore, if the sum of coins in pile A and pile B is divisible by 3, then one of the piles must be completely emptied.

2. Balance Between the Two:

Each turn involves taking away a greater number of coins from one stack than from the other. In order to avoid depleting one stack before the other, it is important to ensure that neither stack has more than twice the number of coins as the other.

It means:

• ✅ A <= 2 * B
• ✅ B <= 2 * A

If any of the conditions is not satisfied, it will become unfeasible to maintain equilibrium between the two stacks and proceed with valid actions.

Example:

//Program to implement the Coin piles method in C++
#include <iostream>
using namespace std;

// Function for piles entry checking
bool isEmptyPiles(int A1, int B1) {
    // The condition to check if total coins are divisible by 3
    if ((A1 + B1) % 3 != 0) {
        return false;
    }
    // If the pile has more than double the coins
    if (A1 > 2 * B1 || B1 > 2 * A1) {
        return false;
    }
    return true;
}

int main() {
    int t;  //T indicates the count of test cases
    cin >> t;
    while (t--) {
        int A1, B1;
        cin >> A1 >> B1;
        if (isEmptyPiles(A1, B1)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}

Output:

5
1 8
NO
23 56
NO
2 4
YES
1 8
NO
35 70
YES

Explanation:

The C++ script outlined tackles the "Coin Piles" challenge by checking the feasibility of reaching two coin piles, A1 and B1, using specified actions: extracting 2 coins from one pile and 1 coin from the other, or the reverse. The isEmptyPiles function assesses two critical criteria: firstly, if the total coin count (A1 + B1) is divisible by 3 since every move decreases the total by 3; secondly, ensuring neither pile holds more than double the coins of the other to allow valid moves. If either condition is not met, the function yields false with the message "NO"; if both are satisfied, it returns true with "YES". Subsequently, the main function manages multiple test scenarios, reads the input, and employs isEmptyPiles to evaluate and display outcomes for each case.

Conclusion:

In summary, "The Coin Piles" dilemma serves as a standard test for algorithm proficiency, demanding a mix of mathematical acumen and adept problem-solving skills. By leveraging the key insights that the total coin count must be divisible by 3 and that no pile can exceed double the coins of the other, it becomes more straightforward to ascertain the feasibility of emptying both piles using the specified moves. This approach streamlines the process, ensuring optimal outcomes with a constant O(1) time complexity per test scenario.

It is an excellent option for competitive programming, offering rapid solutions derived from basic arithmetic evaluations and conditional statements. Grasping the logic behind these solutions is crucial for tackling analogous combinatorial challenges and leveraging mathematical principles to design effective algorithms.