Problem Description

Alice and Bob are playing a game with piles of stones. There are several piles arranged in a row, where each pile contains a positive integer number of stones represented by piles[i].

The game follows these rules:

• Players take turns, with Alice going first
• On each turn, a player can take all the stones from the first X piles (starting from the leftmost available pile)
• The value of X must satisfy: 1 <= X <= 2M
• After taking X piles, the value of M is updated: M = max(M, X)
• Initially, M = 1
• The game continues until all stones are taken
• The player with the most stones at the end wins

For example, if M = 1 at the start of a turn, the player can take stones from the first 1 or 2 piles (X can be 1 or 2 since 2M = 2). If they choose to take 2 piles, then M becomes max(1, 2) = 2 for the next turn.

Both Alice and Bob play optimally, meaning they always make the move that maximizes their own score. The task is to determine the maximum number of stones Alice can collect when both players play their best strategy.

The solution uses dynamic programming with memoization. The function dfs(i, m) calculates the maximum stones the current player can get starting from pile index i with the current M value being m. Since both players play optimally, when one player makes a move, they consider that their opponent will also play optimally in response. The maximum stones a player can get is the total remaining stones minus what their opponent can optimally get in the next turn.

Intuition

The key insight is that this is a zero-sum game - what one player gains, the other player loses. Since both players play optimally, we need to think recursively about the best outcome for the current player.

When it's a player's turn, they want to maximize their own score. But since the total number of stones is fixed, maximizing their score is equivalent to minimizing their opponent's score. This leads us to a minimax approach.

Consider what happens when a player makes a move:

• They take X piles (where 1 <= X <= 2M)
• The opponent then plays from the remaining piles
• The opponent will also play optimally to maximize their own score

So from the current player's perspective, if they take X piles, their total score will be:

• The stones they take now (sum of first X piles)
• Plus whatever stones they can get in future turns

But here's the clever part: instead of tracking both players' scores separately, we can think of it as "the maximum stones the current player can get from the remaining piles." When we switch turns, the "current player" changes, but the logic remains the same.

This means if there are stones from index i to the end, and the current player takes X piles, they get:

• All remaining stones (s[n] - s[i])
• Minus what the opponent can optimally get from the remaining position (dfs(i + X, new_M))

The formula becomes: current_player_score = total_remaining - opponent_best_score

We use prefix sums (s) to quickly calculate the sum of any range of piles. The base case is when 2M >= remaining_piles, meaning the current player can take all remaining stones in one turn.

The memoization with @cache ensures we don't recalculate the same game state multiple times, as the same position (i, m) can be reached through different sequences of moves.

Learn more about Math, Dynamic Programming and Prefix Sum patterns.

Solution Approach

The solution uses prefix sum combined with memoization search (dynamic programming with caching) to solve this game theory problem.

Step 1: Prefix Sum Preprocessing

First, we build a prefix sum array s where s[i] represents the sum of the first i elements from the piles array:

s = list(accumulate(piles, initial=0))

This allows us to calculate the sum of any range of piles in O(1) time. For example, s[n] - s[i] gives us the sum of all stones from index i to the end.

Step 2: Define the Recursive Function

We define dfs(i, m) which returns the maximum stones the current player can get when:

• Starting from pile index i
• The current value of M is m

Step 3: Base Case

if m * 2 >= n - i:
    return s[n] - s[i]

If 2M >= remaining_piles, the current player can take all remaining stones in one turn. The number of remaining piles is n - i, so if 2m >= n - i, we return the sum of all remaining stones.

Step 4: Recursive Case

return max(
    s[n] - s[i] - dfs(i + x, max(m, x)) for x in range(1, m << 1 | 1)
)

The current player tries all possible moves where they take x piles (1 <= x <= 2m):

• range(1, m << 1 | 1) generates values from 1 to 2m (inclusive). The expression m << 1 | 1 equals 2m + 1
• For each choice of x:
  • The player takes the first x piles starting from index i
  • The new position becomes i + x for the opponent
  • The new M value becomes max(m, x)
  • The player's score is: total remaining stones minus what the opponent can optimally get
  • Formula: s[n] - s[i] - dfs(i + x, max(m, x))

Step 5: Memoization

The @cache decorator automatically memoizes the results of dfs(i, m). This prevents recalculation of the same game state, reducing time complexity from exponential to polynomial.

Step 6: Return the Answer

return dfs(0, 1)

Alice starts first from index 0 with M = 1, so the answer is dfs(0, 1).

The time complexity is O(n³) where n is the number of piles, as there are O(n²) possible states (i, m) and each state requires O(n) time to compute. The space complexity is O(n²) for the memoization cache.

Example Walkthrough

Let's walk through a small example with piles = [2, 7, 9, 4, 4].

Initial Setup:

• Build prefix sum array: s = [0, 2, 9, 18, 22, 26]
• Total stones = 26
• Alice starts with dfs(0, 1) (index 0, M=1)

Turn 1 - Alice's turn (i=0, m=1):

• Alice can take X piles where 1 ≤ X ≤ 2M = 2

• Option 1: Take 1 pile (X=1)

  • Alice gets pile[0] = 2 stones immediately
  • New state: i=1, M=max(1,1)=1
  • Bob will play optimally from dfs(1, 1)
  • Alice's total = (26-0) - dfs(1,1) = 26 - Bob's best from position 1

• Option 2: Take 2 piles (X=2)

  • Alice gets piles[0,1] = 2+7 = 9 stones immediately
  • New state: i=2, M=max(1,2)=2
  • Bob will play optimally from dfs(2, 2)
  • Alice's total = (26-0) - dfs(2,2) = 26 - Bob's best from position 2

Let's trace Option 2 deeper:

Turn 2 - Bob's turn (i=2, m=2):

• Remaining stones from index 2: s[5]-s[2] = 26-9 = 17
• Bob can take X piles where 1 ≤ X ≤ 2M = 4
• Since 2M=4 ≥ remaining piles (3), Bob can take all remaining stones!
• Base case triggered: Bob gets all 17 remaining stones

Back to Turn 1:

• Option 2 gives Alice: 26 - 17 = 9 stones

Let's check Option 1:

Turn 2 - Bob's turn (i=1, m=1):

• Remaining stones: s[5]-s[1] = 26-2 = 24
• Bob can take X piles where 1 ≤ X ≤ 2
• Bob evaluates taking 1 pile vs 2 piles
• After calculation (involving more recursive calls), Bob's optimal play gives him 15 stones

Back to Turn 1:

• Option 1 gives Alice: 26 - 15 = 11 stones

Final Decision: Alice chooses Option 1 (take 1 pile) to get 11 stones total, which is better than Option 2 (9 stones).

The key insight: Even though taking more piles initially seems better (9 stones vs 2 stones), it gives Bob a larger M value, allowing him to take more stones on his turn. By taking fewer piles, Alice limits Bob's options and ultimately gets more stones overall.

Solution Implementation

Time and Space Complexity

The time complexity is O(n^3), and the space complexity is O(n^2).

Time Complexity Analysis:

• The function dfs(i, m) has two parameters: i ranges from 0 to n-1, and m can range from 1 to at most n.
• This gives us O(n^2) possible unique states for the memoization cache.
• For each state (i, m), we iterate through x from 1 to 2*m, which in the worst case can be up to 2n iterations (when m approaches n).
• Therefore, the total time complexity is O(n^2) * O(n) = O(n^3).

Space Complexity Analysis:

• The @cache decorator stores all unique combinations of (i, m) parameters.
• Since i can be from 0 to n-1 and m can be from 1 to at most n, there are O(n^2) possible states to cache.
• The prefix sum array s takes O(n) space.
• The recursion depth is at most O(n) in the worst case.
• The dominant factor is the cache storage, giving us O(n^2) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Understanding of the Game Rules

Pitfall: A common mistake is misunderstanding how the parameter M works. Some might think:

• M is fixed throughout the game
• M only affects the current player
• The update rule M = max(M, X) applies globally rather than tracking separately for each game branch

Why it happens: The problem statement's complexity and the interaction between X and M can be confusing. Each recursive branch maintains its own M value based on the choices made in that particular game path.

Solution: Remember that M is passed as a parameter in the recursion and updated locally for each branch. When a player takes X piles, the NEW M value (max(M, X)) is what gets passed to the opponent's turn, not a global M.

2. Off-by-One Errors in Range Calculation

Pitfall: Using incorrect ranges when iterating through possible moves:

# Wrong - doesn't include 2*m
for x in range(1, 2 * m):  

# Wrong - starts from 0
for x in range(0, 2 * m + 1):  

# Correct
for x in range(1, 2 * m + 1):

Why it happens: The problem states "1 <= X <= 2M", which means both 1 and 2M are valid choices (inclusive range).

Solution: Always use range(1, 2 * m + 1) to include all valid moves from 1 to 2M inclusive.

3. Forgetting to Check Array Bounds

Pitfall: Not handling the case where taking X piles would exceed the array bounds:

# Wrong - might access invalid indices
def dfs(i, m):
    max_score = 0
    for x in range(1, 2 * m + 1):
        # This could make i + x > n
        next_score = dfs(i + x, max(m, x))
        # ...

Why it happens: When near the end of the array, 2M might be larger than the remaining piles.

Solution: Either add a bounds check in the loop or handle it in the base case:

# Option 1: Limit the range
for x in range(1, min(2 * m + 1, n - i + 1)):

# Option 2: Handle in base case (as shown in the solution)
if m * 2 >= n - i:
    return s[n] - s[i]

4. Incorrectly Calculating Opponent's Optimal Play

Pitfall: Trying to maximize both players' scores or minimize the opponent's score directly:

# Wrong - trying to minimize opponent's score
return min(dfs(i + x, max(m, x)) for x in range(1, 2 * m + 1))

# Wrong - trying to maximize current player's immediate gain
return max(sum(piles[i:i+x]) + dfs(i + x, max(m, x)) for x in range(1, 2 * m + 1))

Why it happens: Misunderstanding the minimax principle in game theory. Since both players play optimally, after the current player's move, the opponent will maximize THEIR score, not help the current player.

Solution: Use the correct formula: The current player's best score = Total remaining stones - Opponent's best score from the resulting position:

return max(s[n] - s[i] - dfs(i + x, max(m, x)) for x in range(1, 2 * m + 1))

5. Not Using Memoization or Using It Incorrectly

Pitfall: Implementing plain recursion without caching, leading to exponential time complexity:

# Wrong - no memoization
def dfs(i, m):
    # recursive logic without caching
  
# Wrong - manual memoization with incorrect key
memo = {}
def dfs(i, m):
    if i in memo:  # Should use (i, m) as key, not just i
        return memo[i]

Why it happens: Underestimating the number of repeated subproblems or incorrectly identifying the state variables.

Solution: Use Python's @cache decorator or implement manual memoization with the correct state tuple (i, m):

@cache
def dfs(i, m):
    # recursive logic

# OR manual memoization
memo = {}
def dfs(i, m):
    if (i, m) in memo:
        return memo[(i, m)]
    # ... calculate result ...
    memo[(i, m)] = result
    return result