LeetCode #3222 — EASY

Find the Winning Player in Coin Game

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

You are given two positive integers x and y, denoting the number of coins with values 75 and 10 respectively.

Alice and Bob are playing a game. Each turn, starting with Alice, the player must pick up coins with a total value 115. If the player is unable to do so, they lose the game.

Return the name of the player who wins the game if both players play optimally.

Example 1:

Input: x = 2, y = 7

Output: "Alice"

Explanation:

The game ends in a single turn:

Alice picks 1 coin with a value of 75 and 4 coins with a value of 10.

Example 2:

Input: x = 4, y = 11

Output: "Bob"

Explanation:

The game ends in 2 turns:

Alice picks 1 coin with a value of 75 and 4 coins with a value of 10.
Bob picks 1 coin with a value of 75 and 4 coins with a value of 10.

Constraints:

1 <= x, y <= 100

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers x and y, denoting the number of coins with values 75 and 10 respectively. Alice and Bob are playing a game. Each turn, starting with Alice, the player must pick up coins with a total value 115. If the player is unable to do so, they lose the game. Return the name of the player who wins the game if both players play optimally.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

2
7

Example 2

4
11

Core Insight

What unlocks the optimal approach

The only way to make 115 is to use one coin of value 75 and four coins of value 10. Each turn uses up these many coins.
Hence the number of turns is <code>min(x, y / 4)</code>.
Determine the winner from its parity.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3222: Find the Winning Player in Coin Game
class Solution {
    public String losingPlayer(int x, int y) {
        int k = Math.min(x / 2, y / 8);
        x -= k * 2;
        y -= k * 8;
        return x > 0 && y >= 4 ? "Alice" : "Bob";
    }
}

// Accepted solution for LeetCode #3222: Find the Winning Player in Coin Game
func losingPlayer(x int, y int) string {
	k := min(x/2, y/8)
	x -= 2 * k
	y -= 8 * k
	if x > 0 && y >= 4 {
		return "Alice"
	}
	return "Bob"
}

# Accepted solution for LeetCode #3222: Find the Winning Player in Coin Game
class Solution:
    def losingPlayer(self, x: int, y: int) -> str:
        k = min(x // 2, y // 8)
        x -= k * 2
        y -= k * 8
        return "Alice" if x and y >= 4 else "Bob"

// Accepted solution for LeetCode #3222: Find the Winning Player in Coin Game
/**
 * [3222] Find the Winning Player in Coin Game
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn losing_player(x: i32, y: i32) -> String {
        let count = x.min(y / 4);

        if count % 2 == 1 {
            "Alice".to_owned()
        } else {
            "Bob".to_owned()
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3222() {
        assert_eq!("Alice".to_owned(), Solution::losing_player(2, 7));
        assert_eq!("Bob".to_owned(), Solution::losing_player(4, 11));
    }
}

// Accepted solution for LeetCode #3222: Find the Winning Player in Coin Game
function losingPlayer(x: number, y: number): string {
    const k = Math.min((x / 2) | 0, (y / 8) | 0);
    x -= k * 2;
    y -= k * 8;
    return x && y >= 4 ? 'Alice' : 'Bob';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.