Problem Description

Alice and Bob are playing a strategic game with stones. The game setup consists of an even number of piles arranged in a row, where each pile contains a positive integer number of stones represented by piles[i].

The game rules are:

• Players take turns, with Alice going first
• On each turn, a player must take an entire pile of stones from either the beginning or the end of the row
• Players continue taking piles until no piles remain
• The winner is the player who collects the most stones in total
• Since the total number of stones is odd, there cannot be a tie

Both players play optimally, meaning they always make the best possible move to maximize their chances of winning.

The task is to determine whether Alice wins the game. Return true if Alice wins, or false if Bob wins.

For example, if piles = [5, 3, 4, 5]:

• Alice could take 5 from the beginning, leaving [3, 4, 5]
• Bob then takes from either end
• The game continues with both players making optimal choices
• The function should return whether Alice ends up with more stones than Bob

The solution uses dynamic programming with memoization to calculate the maximum advantage Alice can achieve. The dfs(i, j) function computes the maximum score difference (current player's score minus opponent's score) when choosing optimally from piles between indices i and j. At each state, the player chooses between taking the left pile (piles[i]) or the right pile (piles[j]), considering that the opponent will also play optimally in subsequent turns.

Intuition

When facing this problem, we need to think about how to model optimal play for both players. The key insight is that each player wants to maximize their own score while minimizing their opponent's score.

Let's think about what happens when a player faces a subarray of piles from index i to j. The player has two choices:

1. Take the left pile (piles[i]) and leave piles from i+1 to j for the opponent
2. Take the right pile (piles[j]) and leave piles from i to j-1 for the opponent

Here's the crucial observation: instead of tracking both players' scores separately, we can track the score difference between the current player and their opponent. This simplifies our state representation significantly.

When we take piles[i], we gain piles[i] points, but then our opponent plays optimally on the remaining piles [i+1, j]. If we denote the optimal score difference for the remaining subproblem as dfs(i+1, j), then our opponent will achieve that difference. From our perspective, this means we get piles[i] - dfs(i+1, j).

Why the subtraction? Because dfs(i+1, j) represents how much better the player (now our opponent) does compared to their opponent (us) in that subgame. So we subtract it from our gain.

Similarly, if we take piles[j], our score difference becomes piles[j] - dfs(i, j-1).

Since we play optimally, we choose the option that gives us the maximum score difference: dfs(i, j) = max(piles[i] - dfs(i+1, j), piles[j] - dfs(i, j-1))

The base case occurs when i > j, meaning no piles are left, so the score difference is 0.

Finally, Alice wins if dfs(0, len(piles)-1) > 0, meaning she can achieve a positive score difference when playing optimally from the beginning.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization using Python's @cache decorator.

Implementation Details:

1. Main Function Structure:

   • The stoneGame function takes the list of piles as input
   • It defines an inner recursive function dfs(i, j) that calculates the maximum score difference for the current player

2. Recursive Function dfs(i, j):

   • Parameters: i and j represent the left and right boundaries of the remaining piles
   • Base Case: When i > j, no piles remain, so return 0
   • Recursive Case: Calculate the maximum of two choices:
     • Take left pile: piles[i] - dfs(i + 1, j)
     • Take right pile: piles[j] - dfs(i, j - 1)

3. Memoization:

   • The @cache decorator automatically memoizes the results of dfs(i, j)
   • This prevents redundant calculations for the same subproblems
   • Without memoization, the time complexity would be exponential; with it, we achieve O(n²) time complexity

4. State Space:

   • There are O(n²) possible states where n is the number of piles
   • Each state (i, j) represents a unique subarray of piles

5. Final Decision:

   • Call dfs(0, len(piles) - 1) to get Alice's maximum advantage when starting with all piles
   • Return True if this value is positive (Alice wins), otherwise False

Time and Space Complexity:

• Time Complexity: O(n²) where n is the number of piles, as we compute each state (i, j) at most once
• Space Complexity: O(n²) for the memoization cache plus O(n) for the recursion stack

The beauty of this approach is that it naturally handles the alternating turns by recursively computing score differences, eliminating the need to track whose turn it is or maintain separate scores for each player.

Example Walkthrough

Let's walk through a small example with piles = [5, 3, 4, 5] to illustrate how the dynamic programming solution works.

We'll use dfs(i, j) to represent the maximum score difference (current player's score - opponent's score) when choosing optimally from piles between indices i and j.

Initial call: dfs(0, 3) - Alice plays first with all 4 piles available.

At dfs(0, 3), Alice has two choices:

• Take left pile (5): Score difference = 5 - dfs(1, 3)
• Take right pile (5): Score difference = 5 - dfs(0, 2)

Let's evaluate dfs(1, 3) (piles [3, 4, 5]):

• Take left (3): 3 - dfs(2, 3)
• Take right (5): 5 - dfs(1, 2)

For dfs(2, 3) (piles [4, 5]):

• Take left (4): 4 - dfs(3, 3) = 4 - 5 = -1
• Take right (5): 5 - dfs(2, 2) = 5 - 4 = 1
• Maximum: 1

For dfs(1, 2) (piles [3, 4]):

• Take left (3): 3 - dfs(2, 2) = 3 - 4 = -1
• Take right (4): 4 - dfs(1, 1) = 4 - 3 = 1
• Maximum: 1

So dfs(1, 3) = max(3 - 1, 5 - 1) = max(2, 4) = 4

Similarly, we can calculate dfs(0, 2) (piles [5, 3, 4]):

• Take left (5): 5 - dfs(1, 2) = 5 - 1 = 4
• Take right (4): 4 - dfs(0, 1) = 4 - 2 = 2
• Maximum: 4

Finally, dfs(0, 3) = max(5 - 4, 5 - 4) = max(1, 1) = 1

Since dfs(0, 3) = 1 > 0, Alice wins the game. The positive value means Alice can secure 1 more stone than Bob when both play optimally.

In this example, one optimal play sequence would be:

1. Alice takes 5 (left), leaving [3, 4, 5]
2. Bob takes 5 (right), leaving [3, 4]
3. Alice takes 4 (right), leaving [3]
4. Bob takes 3 Final scores: Alice = 9, Bob = 8

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The solution uses dynamic programming with memoization. The dfs function explores all possible game states defined by the range [i, j]. There are n possible values for i (from 0 to n-1) and n possible values for j (from 0 to n-1). However, only states where i ≤ j are valid, which gives us approximately n²/2 unique states. Each state is computed only once due to the @cache decorator, and each computation takes O(1) time (just comparing two recursive calls). Therefore, the overall time complexity is O(n²).

Space Complexity: O(n²)

The space complexity consists of two components:

1. Memoization cache: The @cache decorator stores the results of all unique (i, j) pairs where i ≤ j. This requires O(n²) space to store approximately n²/2 states.
2. Recursion call stack: In the worst case, the recursion depth can go up to n levels (when we process from one end of the array to the other). This adds O(n) to the space complexity.

Since O(n²) + O(n) = O(n²), the overall space complexity is O(n²).

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

Common Pitfalls

1. Incorrect Score Calculation Logic

Pitfall: A common mistake is trying to track both players' scores separately or incorrectly handling the alternating turns. Developers often write something like:

# INCORRECT approach
def dfs(i, j, is_alice_turn):
    if i > j:
        return 0
  
    if is_alice_turn:
        # Trying to maximize Alice's score
        return max(piles[i] + dfs(i+1, j, False), 
                  piles[j] + dfs(i, j-1, False))
    else:
        # Trying to minimize Alice's score (Bob's turn)
        return min(dfs(i+1, j, True), 
                  dfs(i, j-1, True))

Why it's wrong: This approach fails because it doesn't properly account for both players' scores. You'd need to return and track both scores, making the logic more complex.

Solution: Use the score difference approach where each recursive call returns current_player_score - opponent_score. This elegantly handles alternating turns:

def dfs(i, j):
    if i > j:
        return 0
    # The subtraction naturally handles the alternation
    return max(piles[i] - dfs(i+1, j), 
              piles[j] - dfs(i, j-1))

2. Missing or Incorrect Memoization

Pitfall: Forgetting to add memoization or implementing it incorrectly:

# INCORRECT - No memoization
def stoneGame(self, piles):
    def dfs(i, j):
        if i > j:
            return 0
        return max(piles[i] - dfs(i+1, j), 
                  piles[j] - dfs(i, j-1))
    return dfs(0, len(piles)-1) > 0

Why it's wrong: Without memoization, the time complexity becomes exponential O(2^n) because the same subproblems are calculated multiple times.

Solution: Always use memoization for overlapping subproblems:

from functools import cache

@cache  # or use lru_cache(None)
def dfs(i, j):
    # recursive logic here

3. Mathematical Insight Oversight

Pitfall: Not recognizing that Alice can always win when there's an even number of piles.

Mathematical Proof: With an even number of piles, Alice can always force a win by choosing either all odd-indexed or all even-indexed piles. She can determine which set has more stones at the start and force that outcome.

Optimal Solution: For this specific problem, you could simply return True:

def stoneGame(self, piles):
    return True

However, the dynamic programming solution is more general and works even if the rules were modified (e.g., odd number of piles or different winning conditions).

4. Index Boundary Errors

Pitfall: Incorrect base case or index handling:

# INCORRECT - Wrong base case
def dfs(i, j):
    if i == j:  # Wrong! Should be i > j
        return piles[i]  # This would double-count the last pile

Solution: The correct base case is if i > j: return 0 because when left index exceeds right index, there are no piles left to pick.

5. Forgetting to Clear Cache Between Test Cases

Pitfall: When testing multiple instances, the cache from previous test cases might interfere:

# Potential issue in testing environment
cache = {}
def dfs(i, j):
    if (i, j) in cache:
        return cache[(i, j)]
    # ... recursive logic

Solution: Either use function-local memoization with @cache decorator (which creates a new cache for each function instance) or explicitly clear the cache between test cases if using a global cache.