LeetCode 3360 - Stone Removal Game

LeetCode Problem 3360

Difficulty: 🟢 Easy
Topics: Math, Simulation

Solution

Problem Understanding

In this game, Alice and Bob take turns removing stones from a pile. The rules are very strict because the number of stones removed on each turn is predetermined.

Alice always goes first and must remove exactly 10 stones on her first move. After that, each move removes exactly 1 fewer stone than the previous move. This means the sequence of required removals is:

• Alice removes 10
• Bob removes 9
• Alice removes 8
• Bob removes 7
• Alice removes 6
• and so on

A player loses immediately if they cannot remove the required number of stones on their turn.

The input n represents the initial number of stones in the pile. We must return:

• true if Alice eventually wins
• false if Alice loses

The constraints are very small:

• 1 <= n <= 50

This tells us that performance is not a concern. Even a direct simulation of the game is easily fast enough.

There are several important edge cases to think about:

• If n < 10, Alice cannot even make the first move, so she loses immediately.
• If the pile becomes too small for the next required move, the current player loses.
• Since the move size decreases every turn, eventually the required removal reaches smaller values like 1.
• The game is deterministic, there are no choices to make. Every move amount is fixed.

Because there are no decisions involved, we only need to simulate the process and determine who gets stuck first.

Approaches

Brute Force Approach

A brute-force style solution would recursively simulate every turn of the game.

We could define a recursive function with:

• the remaining number of stones
• the number of stones required for the current move
• whose turn it is

At each step:

• If the current player cannot remove the required number of stones, they lose.
• Otherwise, subtract the stones and continue recursively with the next smaller move size.

This approach is correct because it exactly models the rules of the game.

However, recursion is unnecessary here because there are no branching decisions. Every move is predetermined. The recursive overhead adds complexity without any benefit.

Optimal Approach

The key observation is that the game is completely deterministic.

There is only one possible sequence of moves:

10, 9, 8, 7, ...

So instead of exploring possibilities, we simply simulate the game turn by turn.

We maintain:

• the remaining number of stones
• the current required move amount
• whose turn it is

At each turn:

1. Check whether the current player can remove the required number of stones.
2. If not, that player loses.
3. Otherwise, subtract the stones.
4. Decrease the required move amount by 1.
5. Switch turns.

Since the maximum move count is tiny, this simulation is extremely efficient.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(k)	O(k)	Recursive simulation of all turns
Optimal	O(k)	O(1)	Iterative deterministic simulation

Here, k is the number of turns played, which is at most 10 because the move sizes decrease from 10 down to 1.

Algorithm Walkthrough

1. Initialize the required move size to 10 because Alice must remove exactly 10 stones first.
2. Initialize a boolean variable to track whose turn it is. We can use alice_turn = True initially.
3. Start a loop while the required move size is greater than 0.
4. For each turn, check whether the remaining number of stones is less than the required move size.

• If yes, the current player cannot make the move.
• If it is Alice's turn, return False.
• Otherwise, return True because Bob loses.

5. If the move is possible, subtract the required move size from n.
6. Decrease the required move size by 1 because every move must remove exactly one fewer stone than the previous move.
7. Flip the turn indicator so the other player moves next.
8. Continue until a player cannot move.

Why it works

The game rules completely determine every move amount. There are no strategic choices or alternate paths. Therefore, simulating the moves exactly as described always produces the correct winner. The algorithm maintains the invariant that n always represents the remaining stones before the current player's turn, and move always represents the exact number of stones that must be removed next. This problem describes a two-player deterministic game between Alice and Bob who alternately remove stones from a single pile. Alice always starts first and must remove exactly 10 stones on her first move. After that, the number of stones each player removes decreases by exactly one compared to the previous move. So Bob removes 9 on his first turn, Alice removes 8 on her second turn, Bob removes 7, and so on.

The game continues until a player is unable to remove the required number of stones on their turn. That player immediately loses. The task is to determine whether Alice wins given an initial number of stones n.

The input is a single integer n representing the number of stones in the pile at the start. The output is a boolean indicating whether Alice can force a win under optimal play, although in this problem there is no decision-making since moves are fixed.

The constraints are very small, with 1 <= n <= 50, which strongly suggests that either a simulation or a direct mathematical observation will be sufficient.

The most important edge cases arise when n is less than 10, since Alice cannot even make her first move, and when n is exactly equal to cumulative sums of the decreasing sequence of moves. Because the moves are deterministic and not strategic, the game reduces to checking whether the sequence of required removals can be satisfied until a player fails.

Approaches

Brute Force Simulation

The brute-force idea is to simulate the game step by step. We maintain a variable representing the current required removal amount, starting at 10 for Alice. We also track whose turn it is, although it does not affect the logic since the sequence is fixed: 10, 9, 8, 7, ...

At each step, we check if the remaining stones n are at least the required amount. If yes, we subtract and continue. If not, the current player loses immediately. This simulation continues until failure occurs or stones are exhausted.

This approach is correct because it directly mirrors the rules of the game. However, it is also more general than necessary. Given the constraints, it is already efficient enough, but it is still conceptually heavier than required since we can reason about the total sum of required moves instead of simulating each turn.

Key Insight (Optimal Approach)

The key observation is that the game is fully deterministic and independent of player strategy. The sequence of moves is fixed:

10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 stops when the next required move becomes non-positive.

So the only question is whether Alice can complete her first move of 10 stones. If n < 10, she immediately loses. If n >= 10, she performs her move, leaving n - 10 stones.

Now Bob needs 9 stones. If fewer than 9 remain, Bob loses and Alice wins. Otherwise, Bob continues, and so on.

However, because the sequence decreases quickly and n <= 50, we can simply simulate or precompute the full consumption pattern. The total maximum sum of the sequence is:

10 + 9 + 8 + ... + 1 = 55

So the entire game is bounded and small. We can simulate until stones run out or a player fails.

The optimal solution is therefore a simple loop that decrements the required move size each turn and alternates players implicitly.

Comparison Table

Approach	Time Complexity	Space Complexity	Notes
Brute Force Simulation	O(k) where k ≤ 10	O(1)	Simulate each move until failure
Optimal Simulation	O(k) where k ≤ 10	O(1)	Same simulation but directly structured around decreasing moves

Algorithm Walkthrough

1. Initialize a variable move = 10 representing Alice’s first required removal. This encodes the deterministic rule that each player must remove one fewer stone than the previous move.
2. While there are still stones remaining and the required move is greater than zero, check whether the current player can make their move. If n < move, the current player cannot move and loses immediately.
3. If the move is possible, subtract move from n, then decrement move by 1 to reflect the rule that the next move is one less.
4. Repeat this process until either n becomes insufficient for the next move or move reaches zero.
5. If the loop finishes without Alice being blocked on her required move, Alice wins; otherwise, she loses.

Why it works

The correctness relies on the fact that the game has no branching decisions. The sequence of required removals is fixed and strictly decreasing, meaning the entire game is a deterministic consumption process. Therefore, the winner is determined solely by whether the sequence of required moves can be fully executed until Alice survives all turns where she is forced to act.

Python Solution

class Solution:
    def canAliceWin(self, n: int) -> bool:
        move = 10
        alice_turn = True

        while move > 0:
            if n < move:
                return not alice_turn

            n -= move
            move -= 1
            alice_turn = not alice_turn

        return False

The implementation directly follows the simulation described in the algorithm walkthrough.

The variable move stores the exact number of stones that must be removed on the current turn. It starts at 10 because Alice's first move is fixed.

The boolean alice_turn tracks whose turn it currently is. When it is True, it means Alice is attempting the current move. After each successful move, we flip the boolean using not.

Inside the loop, the first operation checks whether the current player can legally make the required move. If n < move, the current player loses immediately. Since the function must return whether Alice wins, we return not alice_turn.

If the move is possible, we subtract the stones, decrease the required move amount by 1, and switch turns.

The loop eventually terminates because move continuously decreases. PythonRun

### Implementation Explanation

The code models the decreasing sequence of required moves starting at 10. The variable `move` tracks how many stones must be removed on the current turn. We alternate turns using `alice_turn`, although it is not strictly necessary for correctness because failure immediately determines the loser.

At each iteration, we check whether the remaining stones are sufficient. If not, the current player cannot move and loses, so we return whether Alice is the winner based on whose turn it is. Otherwise, we subtract the move and continue with the next smaller move requirement.

Once all moves down to 1 are processed, Alice is guaranteed to win, since she survives the deterministic sequence.

## Go Solution

```go
func canAliceWin(n int) bool {
    move := 10
    aliceTurn := true

    for move > 0 {
        if n < move {
            return !aliceTurn
        }

        n -= move
        move--
        aliceTurn = !aliceTurn
    }

    return false
}

The Go implementation mirrors the Python solution almost exactly.

Go uses explicit variable declarations with :=, and boolean toggling is handled with !aliceTurn.

There are no overflow concerns because the constraints are extremely small. The implementation also does not require slices, arrays, or additional data structures because the game state consists of only a few integers and a boolean.

Worked Examples

Example 1

Input:

n = 12

Simulation:

Turn	Player	Required Move	Stones Before	Stones After
1	Alice	10	12	2
2	Bob	9	2	Cannot move

Bob cannot remove 9 stones, so Bob loses.

Result:

true

Example 2

Input:

n = 1

Simulation:

Turn	Player	Required Move	Stones Before	Result
1	Alice	10	1	Cannot move

Alice cannot make the first move.

Result:

false

Go

### Go-Specific Notes

The Go implementation mirrors the Python logic closely. The main differences are syntactic: Go uses a `for` loop instead of `while`, and booleans are explicitly toggled using `!aliceTurn`. Integer behavior is safe because values remain small and well within bounds. There are no slice or reference considerations here since the algorithm is purely arithmetic.

## Worked Examples

### Example 1: n = 12

| Turn | Player | Move Required | Remaining n Before | Action | Remaining n After |
| --- | --- | --- | --- | --- | --- |
| 1 | Alice | 10 | 12 | Alice removes 10 | 2 |
| 2 | Bob | 9 | 2 | Bob cannot move | Bob loses |

Alice wins immediately because Bob cannot perform his required move of 9.

### Example 2: n = 1

| Turn | Player | Move Required | Remaining n Before | Action |
| --- | --- | --- | --- | --- |
| 1 | Alice | 10 | 1 | Alice cannot move |

Alice immediately loses because she cannot satisfy her first required move.

## Complexity Analysis

| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | O(1) | At most 10 turns are simulated |
| Space | O(1) | Only a few variables are used |

Although we describe the runtime as constant time, another valid interpretation is `O(k)` where `k` is the number of turns. Since `k` is bounded by 10, the runtime is effectively constant for all valid inputs.

The algorithm uses only integer variables and a boolean flag, so the memory usage remains constant.

## Test Cases

```sql
sol = Solution()

assert sol.canAliceWin(12) == True   # Example where Alice wins immediately
assert sol.canAliceWin(1) == False   # Alice cannot make first move

assert sol.canAliceWin(10) == True   # Alice removes all stones exactly
assert sol.canAliceWin(9) == False   # Just below first required move
assert sol.canAliceWin(19) == False  # Alice removes 10, Bob removes 9 exactly
assert sol.canAliceWin(20) == True   # Bob cannot remove next required amount

assert sol.canAliceWin(27) == False  # Sequence reaches Alice failing later
assert sol.canAliceWin(28) == True   # One more stone changes winner

assert sol.canAliceWin(50) == False  # Large valid input
assert sol.canAliceWin(55) == True   # Total sum from 10 to 1

Test	Why
n = 12	Verifies the provided winning example
n = 1	Verifies the provided losing example
n = 10	Alice removes all stones exactly
n = 9	Alice cannot make the opening move
n = 19	Both first moves succeed exactly
n = 20	Bob fails on the second move
n = 27	Tests a deeper sequence of turns
n = 28	Verifies neighboring values can change the outcome
n = 50	Tests a large input within constraints
n = 55	Total of all required moves from 10 to 1

Edge Cases

One important edge case occurs when n < 10. Since Alice must remove exactly 10 stones on her first turn, any smaller value means she immediately loses. A careless implementation might attempt to continue simulation even though the first move is impossible. The solution handles this naturally with the if n < move check at the start of the loop.

Another important case is when the remaining stones become exactly equal to the required move amount. For example, with n = 10, Alice can remove all 10 stones successfully. The implementation correctly allows this because it only fails when n < move, not when n == move.

A third edge case happens later in the sequence after several successful moves. For example, in n = 27, the game proceeds through multiple turns before a player eventually gets stuck. This verifies that the implementation correctly updates the move size and alternates turns after every move, instead of assuming the outcome depends only on the first one or two turns. | Time | O(10) → O(1) | The loop runs at most 10 iterations since moves decrease from 10 to 1 | | Space | O(1) | Only a few scalar variables are used |

The complexity is constant because the sequence length is fixed and independent of n. Even though we simulate, the bounded nature of the problem ensures constant runtime.

Test Cases

PythonRun

Test Summary

Test	Why
n = 12	basic winning scenario
n = 1	Alice cannot make first move
n = 10	minimal valid Alice move
n = 55	full sequence completes exactly
n = 54	just below full completion boundary
n = 50	upper-range stress case

Edge Cases

One important edge case is when n < 10. In this situation, Alice immediately loses because she cannot perform her required first move. This must be handled before any further logic.

Another edge case is when n is exactly equal to a partial prefix sum of the decreasing sequence (such as 10, 19, 27, etc.). These cases determine whether the turn transitions correctly lead to a loss for the next player. The simulation naturally handles this because it checks feasibility before each subtraction.

A final edge case is when n is close to the maximum possible total sum of all moves (55). In such cases, the entire sequence may complete successfully, and the winner is determined by whether the last valid move is completed. The loop-based simulation ensures correctness by exhausting all possible moves in order.