LeetCode #3360 — EASY

Stone Removal Game

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

Solve on LeetCode

The Problem

Problem Statement

Alice and Bob are playing a game where they take turns removing stones from a pile, with Alice going first.

Alice starts by removing exactly 10 stones on her first turn.
For each subsequent turn, each player removes exactly 1 fewer stone than the previous opponent.

The player who cannot make a move loses the game.

Given a positive integer n, return true if Alice wins the game and false otherwise.

Example 1:

Input: n = 12

Output: true

Explanation:

Alice removes 10 stones on her first turn, leaving 2 stones for Bob.
Bob cannot remove 9 stones, so Alice wins.

Example 2:

Input: n = 1

Output: false

Explanation:

Alice cannot remove 10 stones, so Alice loses.

Constraints:

1 <= n <= 50

Step 01

Brute Force Baseline

Problem summary: Alice and Bob are playing a game where they take turns removing stones from a pile, with Alice going first. Alice starts by removing exactly 10 stones on her first turn. For each subsequent turn, each player removes exactly 1 fewer stone than the previous opponent. The player who cannot make a move loses the game. Given a positive integer n, return true if Alice wins the game and false otherwise.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

The constraints are small enough that a brute-force solution is feasible.

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 #3360: Stone Removal Game
class Solution {
    public boolean canAliceWin(int n) {
        int x = 10, k = 0;
        while (n >= x) {
            n -= x;
            --x;
            ++k;
        }
        return k % 2 == 1;
    }
}

// Accepted solution for LeetCode #3360: Stone Removal Game
func canAliceWin(n int) bool {
	x, k := 10, 0
	for n >= x {
		n -= x
		x--
		k++
	}
	return k%2 == 1
}

# Accepted solution for LeetCode #3360: Stone Removal Game
class Solution:
    def canAliceWin(self, n: int) -> bool:
        x, k = 10, 0
        while n >= x:
            n -= x
            x -= 1
            k += 1
        return k % 2 == 1

// Accepted solution for LeetCode #3360: Stone Removal Game
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3360: Stone Removal Game
// class Solution {
//     public boolean canAliceWin(int n) {
//         int x = 10, k = 0;
//         while (n >= x) {
//             n -= x;
//             --x;
//             ++k;
//         }
//         return k % 2 == 1;
//     }
// }

// Accepted solution for LeetCode #3360: Stone Removal Game
function canAliceWin(n: number): boolean {
    let [x, k] = [10, 0];
    while (n >= x) {
        n -= x;
        --x;
        ++k;
    }
    return k % 2 === 1;
}

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(log n)

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.