Stone Game V

Hard

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ArraysDynamic Programming

There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.

In each round of the game, Alice divides the row into two non-empty rows (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the values of all the stones in this row. Bob throws away the row which has the maximum value, and Alice's score increases by the value of the remaining row. If the value of the two rows are equal, Bob lets Alice decide which row will be thrown away. The next round starts with the remaining row.

The game ends when there is only one stone remaining. Alice's is initially zero.

Return the maximum score that Alice can obtain.

Example 1:

Input: stoneValue = [6,2,3,4,5,5]
Output: 18
Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11.
In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5).
The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row.

Example 2:

Input: stoneValue = [7,7,7,7,7,7,7]
Output: 28

Example 3:

Input: stoneValue = [4]
Output: 0

Constraints:

1 <= stoneValue.length <= 500
1 <= stoneValue[i] <= 10⁶

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 range of values for each stone in the input array?
Can the input array be empty or contain only one stone?
If there are multiple splits where the sums are equal, which split should I choose?
If no split leads to a non-zero score, what should I return?
Is the input array guaranteed to contain only positive integers?

Brute Force Solution

Approach

In this game, we want to find the highest score possible by repeatedly splitting a row of stones. The brute force strategy explores every single possible split and chooses the path that leads to the best final score.

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

Consider every possible place to cut the row of stones into two parts: a left part and a right part.
Calculate the sum of the stone values in the left part and the sum of the stone values in the right part.
Compare the two sums. If they are equal, we can choose either side to continue the game. We will explore both options. If they are not equal, we pick the side with the smaller sum.
Once one side is selected, we can apply the same process recursively to that selected side: considering every possible cut within that row of stones, calculating sums, and comparing them.
Keep doing this until a row is down to a single stone (which can't be split anymore), at which point the game ends on that path.
Keep track of the score for each possible game path that we explore.
Once all paths have been explored, choose the game path with the highest score as the final answer.

Code Implementation

def stone_game_v_brute_force(stones):
    def calculate_max_score(current_stones):
        number_of_stones = len(current_stones)
        if number_of_stones <= 1:
            return 0

        max_score_so_far = 0

        for split_index in range(1, number_of_stones):
            left_stones = current_stones[:split_index]
            right_stones = current_stones[split_index:]

            left_sum = sum(left_stones)
            right_sum = sum(right_stones)

            # Determine which side to continue with
            if left_sum == right_sum:
                
                left_score = calculate_max_score(left_stones)
                right_score = calculate_max_score(right_stones)
                max_score_so_far = max(max_score_so_far, left_sum + max(left_score, right_score))

            elif left_sum < right_sum:
                # Left side yields higher potential
                max_score_so_far = max(max_score_so_far, left_sum + calculate_max_score(left_stones))

            else:
                # Right side yields higher potential
                max_score_so_far = max(max_score_so_far, right_sum + calculate_max_score(right_stones))

        return max_score_so_far

    return calculate_max_score(stones)

Big(O) Analysis

Time Complexity

O(n^3) – The brute force approach considers all possible splits of the stone row. For a row of size n, there are n-1 possible split points. At each split, the sums of the left and right sub-rows need to be computed, which takes O(n) time. Recursively applying this splitting and summing process at each level of the game tree leads to a depth of at most n. Therefore, the total time complexity can be approximated as O(n * n * n), which simplifies to O(n^3).

Space Complexity

O(N^2) – The primary space complexity arises from the recursive calls described in the plain English explanation. In the worst-case scenario, each call splits the row of stones, potentially leading to a recursion depth proportional to N, where N is the initial number of stones. However, the dominant factor is the memoization needed to avoid recomputing the same subproblems repeatedly (though not explicitly stated, optimization would require it). Memoization typically involves storing the results of function calls in a table or dictionary of size O(N^2) to store the maximum achievable score for each possible subarray defined by start and end indices.

Optimal Solution

Approach

This problem involves strategically splitting a row of stones to maximize your score. The trick is to efficiently calculate the sums of different stone sections and reuse those calculations to avoid redundant work.

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

First, calculate the total value of all the stones up to each stone, which will help quickly determine the sum of any consecutive group of stones.
Consider all possible places to split the stones into two groups: a left group and a right group.
For each possible split, calculate the value of the left group and the value of the right group using the pre-calculated sums.
If the left group's value is less than the right group's value, you get the left group's value as your score, and you continue playing on the left group.
If the right group's value is less than the left group's value, you get the right group's value as your score, and you continue playing on the right group.
If both groups have equal value, you get that value as your score, and you can choose to continue playing on either the left or right group. Choose the one that yields the highest future score.
Repeat this process on the chosen group until only one stone remains. The total score accumulated across all splits is your final score. Remember to use the previous calculations of smaller ranges of stones to speed up the process.

Code Implementation

def stone_game_v(stone_values):    number_of_stones = len(stone_values)
    prefix_sums = [0] * (number_of_stones + 1)
    for i in range(number_of_stones):        prefix_sums[i + 1] = prefix_sums[i] + stone_values[i]

    memo = {}

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

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

        max_score = 0
        for split_index in range(left_index + 1, right_index + 1):
            left_group_sum = prefix_sums[split_index] - prefix_sums[left_index]
            right_group_sum = prefix_sums[right_index + 1] - prefix_sums[split_index]

            if left_group_sum < right_group_sum:
                # Explore the left side, as it yields the lower sum
                max_score = max(max_score, left_group_sum + calculate_max_score(left_index, split_index - 1))
            elif right_group_sum < left_group_sum:
                # Explore the right side, as it yields the lower sum
                max_score = max(max_score, right_group_sum + calculate_max_score(split_index, right_index))
            else:
                # Key Decision: Choose the side that yields a higher subsequent score
                max_score = max(max_score,
                                left_group_sum + max(calculate_max_score(left_index, split_index - 1), calculate_max_score(split_index, right_index)))

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

    # Initiate the process for the entire range of stones
    result = calculate_max_score(0, number_of_stones - 1)

    return result

Big(O) Analysis

Time Complexity

O(n^3) – The solution involves a recursive approach with memoization. The recursion explores all possible splits of the stone row, resulting in nested loops implicitly. For each split point (of which there are up to n), we calculate the left and right sums in O(1) time using prefix sums. In the worst case where left and right sums are equal, we explore both left and right subproblems recursively. Because each function call processes from one stone to n stones, and it has to pick every split point, this results in O(n) splits to explore recursively at each level up to n levels, leading to the time complexity being O(n^3) after considering the redundant calls removed by memoization.

Space Complexity

O(N^2) – The algorithm pre-calculates prefix sums, which requires an auxiliary array of size N. The recursive calls, in the worst case, might explore all possible subranges of the stones, and we use memoization to store the results of these subproblems. The memoization table will store the maximum scores for all possible subranges, resulting in a 2D array of size N x N. Therefore, the auxiliary space required is proportional to N^2.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there are no stones to play with.
Array with only one element	Return 0 because only one stone exists and no split is possible.
Array with two elements	Return the minimum of the two elements as that's the only move possible.
Array with all elements having the same value	The recursive calls will explore all possible splits, correctly calculating the optimal score, possibly leading to stack overflow issues for very large n with poorly optimized implementations.
Array with a highly skewed distribution of values (e.g., mostly small values and one very large value)	The optimal split might heavily favor the smaller values side, potentially causing imbalanced recursion depth, so memoization or dynamic programming is crucial for efficiency.
Maximum size array (performance and memory)	Employ memoization or dynamic programming to avoid redundant calculations and prevent stack overflow errors associated with excessive recursion, optimizing time and space complexity.
Integer overflow when calculating prefix sums	Use a data type that can accommodate larger sums, such as `long` in Java or `long long` in C++, to prevent integer overflow errors.
The optimal play results in both sides summing to the same value repeatedly	The algorithm needs to handle cases where the sums are equal and must explore both options of taking the left or right side.

Null or empty input array

How to Handle:

Return 0 immediately as there are no stones to play with.

Array with only one element

How to Handle:

Return 0 because only one stone exists and no split is possible.

Array with two elements

How to Handle:

Return the minimum of the two elements as that's the only move possible.

Array with all elements having the same value

How to Handle:

The recursive calls will explore all possible splits, correctly calculating the optimal score, possibly leading to stack overflow issues for very large n with poorly optimized implementations.

Array with a highly skewed distribution of values (e.g., mostly small values and one very large value)

How to Handle:

The optimal split might heavily favor the smaller values side, potentially causing imbalanced recursion depth, so memoization or dynamic programming is crucial for efficiency.

Maximum size array (performance and memory)

How to Handle:

Employ memoization or dynamic programming to avoid redundant calculations and prevent stack overflow errors associated with excessive recursion, optimizing time and space complexity.

Integer overflow when calculating prefix sums

How to Handle:

Use a data type that can accommodate larger sums, such as `long` in Java or `long long` in C++, to prevent integer overflow errors.

The optimal play results in both sides summing to the same value repeatedly

How to Handle:

The algorithm needs to handle cases where the sums are equal and must explore both options of taking the left or right side.