Stone Game VII

Medium

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ArraysDynamic Programming

Alice and Bob take turns playing a game, with Alice starting first.

There are n stones arranged in a row. On each player's turn, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.

Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference. Alice's goal is to maximize the difference in the score.

Given an array of integers stones where stones[i] represents the value of the i^th stone from the left, return the difference in Alice and Bob's score if they both play optimally.

Example 1:

Input: stones = [5,3,1,4,2]
Output: 6
Explanation: 
- Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
- Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
- Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
- Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
- Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
The score difference is 18 - 12 = 6.

Example 2:

Input: stones = [7,90,5,1,100,10,10,2]
Output: 122

Constraints:

n == stones.length
2 <= n <= 1000
1 <= stones[i] <= 1000

Try coding it in LeetCode

Stone Game VII

Medium

Asked by:

6 views

Topics:

ArraysDynamic Programming

Alice and Bob take turns playing a game, with Alice starting first.

There are n stones arranged in a row. On each player's turn, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.

Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference. Alice's goal is to maximize the difference in the score.

Given an array of integers stones where stones[i] represents the value of the i^th stone from the left, return the difference in Alice and Bob's score if they both play optimally.

Example 1:

Input: stones = [5,3,1,4,2]
Output: 6
Explanation: 
- Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
- Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
- Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
- Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
- Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
The score difference is 18 - 12 = 6.

Example 2:

Input: stones = [7,90,5,1,100,10,10,2]
Output: 122

Constraints:

n == stones.length
2 <= n <= 1000
1 <= stones[i] <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the `stones` array?
Can the values in the `stones` array be negative or zero?
If Alice and Bob play optimally and the result is a tie (same score), what value should I return?
Are the values in the stones array guaranteed to be integers?
Is the input array always non-empty?

Brute Force Solution

Approach

The brute force strategy involves exploring every possible game scenario. At each turn, we consider all possible moves for the current player. We then simulate the rest of the game for each of these moves, and choose the best outcome for the first player.

Here's how the algorithm would work step-by-step:

Consider all possible moves the current player can make, which are either removing the stone at the beginning or the stone at the end.
For each of these possible moves, calculate the score the current player would get, which is the sum of the remaining stones.
Then, recursively play the game assuming the other player makes the best possible move from their perspective after each move.
After exploring all possible moves and counter-moves, determine the best possible score the initial player can achieve.
The best score will be the largest difference between Alice and Bob's scores that Alice can guarantee for herself across all possible game plays.

Code Implementation

def stone_game_vii_brute_force(stones):
    def get_max_score_difference(current_stones, is_alice_turn):
        if not current_stones:
            return 0

        best_score = float('-inf') if is_alice_turn else float('inf')

        # Iterate through possible moves
        for i in range(2):
            if i == 0:
                new_stones = current_stones[1:]
                removed_stone = current_stones[0]
            else:
                new_stones = current_stones[:-1]
                removed_stone = current_stones[-1]

            # Calculate the score for the current move
            current_score = sum(current_stones) - removed_stone

            # Recursively calculate next player's best move
            next_score_difference = get_max_score_difference(new_stones, not is_alice_turn)

            # Determine the best score for Alice or Bob
            if is_alice_turn:
                best_score = max(best_score, current_score + next_score_difference)
            else:
                best_score = min(best_score, -current_score + next_score_difference)

        return best_score

    # Alice goes first
    return get_max_score_difference(stones, True)

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force strategy recursively explores every possible game scenario. At each step, a player has two choices: remove the stone at the beginning or the end. Since each player has two choices at each turn and there are n stones, the number of possible game plays grows exponentially. Therefore, the time complexity of exploring all possible game scenarios is O(2^n), where n is the number of stones.

Space Complexity

O(N) – The provided brute force solution uses recursion. Each recursive call explores a game state, removing either the first or last stone. The maximum depth of the recursion is proportional to the number of stones, N, because at most N stones can be removed. Each recursive call creates a new stack frame, thus the auxiliary space is determined by the maximum recursion depth, which is O(N).

Optimal Solution

Approach

The problem can be solved by thinking about who has the advantage at each turn. We can use a technique that avoids recalculating the same thing multiple times to speed things up.

Here's how the algorithm would work step-by-step:

Figure out what the total score is for all the stones at the beginning of the game.
Imagine each possible move: either removing the first stone or the last stone.
For each of those moves, consider what the other player would do to maximize their own score.
Remember the best score difference Alice can achieve from any position. If we have seen a position before, use that score rather than recalculating it.
Alice aims to maximize her score difference against Bob. Bob will, of course, try to minimize this difference.
Continue this process until only one stone is left. The final score difference at the start is the answer.

Code Implementation

def stone_game_vii(stones): 
    number_of_stones = len(stones)
    prefix_sum = [0] * (number_of_stones + 1)
    for i in range(number_of_stones):
        prefix_sum[i + 1] = prefix_sum[i] + stones[i]

    memo = {}

    def calculate_score_difference(left_index, right_index): 
        if left_index == right_index:
            return 0

        if (left_index, right_index) in memo:
            return memo[(left_index, right_index)]

        # Calculate score if Alice removes the leftmost stone
        score_removing_left = prefix_sum[right_index + 1] - prefix_sum[left_index + 1]
        
        # Calculate score if Alice removes the rightmost stone
        score_removing_right = prefix_sum[right_index] - prefix_sum[left_index]

        # Determine bob's best move after alice removes a stone
        score = max(
            score_removing_left - calculate_score_difference(left_index + 1, right_index),
            score_removing_right - calculate_score_difference(left_index, right_index - 1)
        )

        memo[(left_index, right_index)] = score
        return score

    # The game begins with Alice's first move
    result = calculate_score_difference(0, number_of_stones - 1)
    return result

Big(O) Analysis

Time Complexity

O(n²) – The provided solution employs dynamic programming with memoization to avoid redundant calculations. The core logic involves considering all possible subproblems, which are defined by the starting and ending indices of the remaining stones. Since there are approximately n * (n-1) / 2 possible subproblems (choosing any start and end index from 0 to n-1), and each subproblem's solution is computed only once due to memoization, the overall time complexity is dominated by the number of subproblems. Therefore, the algorithm exhibits a time complexity of O(n²).

Space Complexity

O(N^2) – The solution employs a memoization technique to avoid recalculating the same game positions. This is typically implemented using a data structure like a 2D array or a hash map. The size of this data structure depends on the number of possible game states, which is determined by the combinations of starting and ending indices of the remaining stones. Since there are N possible starting positions and N possible ending positions (or subproblems) in the worst case, the memoization table would require space proportional to N * N. Therefore, the space complexity is O(N^2), as we are storing intermediate results for all possible subproblems.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as there are no stones to play with, and thus no score.
Array with only two elements	Directly return the maximum of the two elements as that's the only possible move for Alice.
Array with all elements being the same	The difference in scores will always be 0 as the sum removed will always be the same regardless of choice, thus dynamic programming will converge to 0.
Array with highly skewed distribution (one very large number)	The dynamic programming approach should handle this correctly, as it considers all possible subproblems, but may increase runtime.
Array contains negative numbers	The dynamic programming approach works regardless of negative numbers as it relies on the difference in sums.
Large array size (maximum constraints)	Ensure dynamic programming table is efficiently implemented to avoid memory issues and the overall algorithm has a low time complexity, potentially O(n^2).
Integer overflow when calculating sums	Use long data type to store sums to prevent integer overflow when the stone values are large.
Recursion depth exceeding stack limit for naive recursive solutions	Utilize dynamic programming with memoization or tabulation to avoid exceeding maximum recursion depth and optimize time complexity.