Maximum Score From Removing Stones

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Greedy Algorithms

You are playing a solitaire game with three piles of stones of sizes a, b, and c respectively. Each turn you choose two different non-empty piles, take one stone from each, and add 1 point to your score. The game stops when there are fewer than two non-empty piles (meaning there are no more available moves).

Given three integers a, b, and c, return the maximum score you can get.

Example 1:

Input: a = 2, b = 4, c = 6
Output: 6
Explanation: The starting state is (2, 4, 6). One optimal set of moves is:
- Take from 1st and 3rd piles, state is now (1, 4, 5)
- Take from 1st and 3rd piles, state is now (0, 4, 4)
- Take from 2nd and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 6 points.

Example 2:

Input: a = 4, b = 4, c = 6
Output: 7
Explanation: The starting state is (4, 4, 6). One optimal set of moves is:
- Take from 1st and 2nd piles, state is now (3, 3, 6)
- Take from 1st and 3rd piles, state is now (2, 3, 5)
- Take from 1st and 3rd piles, state is now (1, 3, 4)
- Take from 1st and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 7 points.

Example 3:

Input: a = 1, b = 8, c = 8
Output: 8
Explanation: One optimal set of moves is to take from the 2nd and 3rd piles for 8 turns until they are empty.
After that, there are fewer than two non-empty piles, so the game ends.

Constraints:

1 <= a, b, c <= 10⁵

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 are the possible ranges for the number of stones in each pile? Can any pile initially have zero stones?
Can the number of stones in any pile ever be negative during the removal process, and if so, how should that be handled?
If I can't remove stones from two piles in a particular turn (because one or both piles are empty), should I return 0 immediately, or is there some other defined behavior?
Are the stone counts guaranteed to be integers, or could they be floating-point numbers?
If multiple sequences of moves lead to the same maximum score, is any of them acceptable, or should I try to optimize for a specific one?

Brute Force Solution

Approach

The brute force method for maximizing the score from removing stones involves exploring every possible sequence of stone removals. We essentially try every combination to find the one that yields the highest score. This means checking all orders of removals, regardless of whether they seem optimal at first.

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

Consider that you have three piles of stones.
Start by trying to remove one stone from the first and second piles.
Calculate the score from this removal.
Then, from the remaining stones, again remove one stone from the first and second piles or from any other two piles.
Calculate the score from this removal and add it to the previous score.
Keep repeating the process of removing stones from two different piles in all available combinations until one or two piles are empty.
Keep track of the score obtained in each of these removal sequences.
After trying every possible sequence of removing stones, select the sequence that gave the maximum total score.

Code Implementation

def maximum_score_from_removing_stones_brute_force(pile_one, pile_two, pile_three):
    maximum_score = 0

    def solve(current_pile_one, current_pile_two, current_pile_three, current_score):
        nonlocal maximum_score

        # Update maximum score if current score is higher
        maximum_score = max(maximum_score, current_score)

        if current_pile_one <= 0 or current_pile_two <= 0:
            if current_pile_one <= 0 and current_pile_two <= 0:
                pass
            elif current_pile_one <= 0 and current_pile_three <= 0:
                pass
            elif current_pile_two <= 0 and current_pile_three <= 0:
                pass
            else:
                pass
        else:

            # Try removing from pile one and pile two
            solve(current_pile_one - 1, current_pile_two - 1, current_pile_three, current_score + 1)

        if current_pile_one <= 0 or current_pile_three <= 0:
            if current_pile_one <= 0 and current_pile_two <= 0:
                pass
            elif current_pile_one <= 0 and current_pile_three <= 0:
                pass
            elif current_pile_two <= 0 and current_pile_three <= 0:
                pass
            else:
                pass
        else:

            # Try removing from pile one and pile three
            solve(current_pile_one - 1, current_pile_two, current_pile_three - 1, current_score + 1)

        if current_pile_two <= 0 or current_pile_three <= 0:
            if current_pile_one <= 0 and current_pile_two <= 0:
                pass
            elif current_pile_one <= 0 and current_pile_three <= 0:
                pass
            elif current_pile_two <= 0 and current_pile_three <= 0:
                pass
            else:
                pass
        else:
            # Try removing from pile two and pile three
            solve(current_pile_one, current_pile_two - 1, current_pile_three - 1, current_score + 1)

    # Initiate the recursive process
    solve(pile_one, pile_two, pile_three, 0)

    return maximum_score

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach explores every possible sequence of stone removals from three piles. In each step, we have three choices: remove from piles 1 and 2, piles 1 and 3, or piles 2 and 3. Since we continue until piles are exhausted, and in the worst case, the number of steps is proportional to the initial number of stones which we can consider to be 'n', we effectively have a recursion with three branches at each step, leading to a runtime complexity of approximately O(3^n). This exponential growth comes from exploring every combination of stone removals.

Space Complexity

O(N) – The brute force approach described explores every possible sequence of stone removals, which can be visualized as a decision tree. The depth of this tree can be at most N, where N is the total number of stones initially, representing the number of removals performed. In the worst-case scenario, where recursion is used to implement this exploration, the recursion stack can grow to a depth of N, storing function call contexts for each level of the tree. Therefore, the auxiliary space complexity is O(N) due to the maximum depth of the recursion stack.

Optimal Solution

Approach

The key is to realize that always taking the two largest piles is the best way to maximize your score. The core idea is that repeatedly removing the largest two piles guarantees the most efficient reduction of stones.

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

Identify the two largest piles of stones.
Remove one stone from each of these two largest piles.
Add the number of stones you removed (which is always 1 + 1 = 2) to your total score.
If there are still at least two piles with stones remaining, repeat the process from the start.
If you end up with only one pile remaining, or no piles, you're done. You've maximized your score.

Code Implementation

def maximum_score_from_removing_stones(a, b, c):
    stones = [a, b, c]
    total_score = 0

    while True:
        stones.sort()
        
        # If the two largest piles are empty, end the simulation.
        if stones[1] == 0:
            break

        # Choose the two largest piles
        first_largest_pile = stones[2]
        second_largest_pile = stones[1]

        # Reduce stones from largest piles
        stones[2] -= 1
        stones[1] -= 1

        # Update score - removing one stone from each.
        total_score += 1

    return total_score

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is finding the two largest piles repeatedly. If we use an efficient data structure like a max-heap (priority queue), finding the two largest elements takes O(1) time. However, removing elements and re-heapifying takes O(log n) time, where n is the number of stone piles. We perform this operation until there is at most one pile left, which takes at most n/2 iterations (or simply n iterations in the worst case). Thus, the overall time complexity is O(n log n) since each iteration involves heap operations.

Space Complexity

O(1) – The provided description suggests identifying and manipulating the two largest piles. While the *identification* process might implicitly use temporary variables to track indices or values of the largest piles, their count remains constant (two variables). No auxiliary data structures like arrays or hash maps are created to store intermediate results related to the number of piles. Thus, the auxiliary space used is independent of the number of piles, N, and remains constant. The space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input arrays (a = [], b = [], c = [])	Return 0 immediately as there are no stones to remove.
One or two piles are empty (e.g., a = [5], b = [], c = [])	Return 0 as we need three piles to perform an operation.
Arrays with a single element each (a = [5], b = [3], c = [10])	Return the minimum of (a[0] + b[0], b[0] + c[0], a[0] + c[0]) if possible, otherwise return 0 if any sum is less than zero.
Arrays with extremely large values, causing integer overflow during summation.	Use 64-bit integers or appropriate data types to prevent overflow during calculations.
Arrays with a large number of elements - potential for stack overflow with recursive solutions.	Implement an iterative approach using a priority queue or equivalent data structure for efficiency.
Values in one pile are significantly larger than values in others.	The greedy approach of always removing from the two largest piles is optimal and handles such scenarios correctly.
The sum of the three piles does not change after each removal, so if the sum is odd, no moves can happen	If the sum of all piles is odd from the start, then return -1 because this state can never lead to all piles becoming zero simultaneously.
One pile's initial value is larger than or equal to the sum of other two.	The total number of moves is restricted by the sum of the smaller piles, which is the maximum number of moves possible.

Empty input arrays (a = [], b = [], c = [])

How to Handle:

Return 0 immediately as there are no stones to remove.

One or two piles are empty (e.g., a = [5], b = [], c = [])

How to Handle:

Return 0 as we need three piles to perform an operation.

Arrays with a single element each (a = [5], b = [3], c = [10])

How to Handle:

Return the minimum of (a[0] + b[0], b[0] + c[0], a[0] + c[0]) if possible, otherwise return 0 if any sum is less than zero.

Arrays with extremely large values, causing integer overflow during summation.

How to Handle:

Use 64-bit integers or appropriate data types to prevent overflow during calculations.

Arrays with a large number of elements - potential for stack overflow with recursive solutions.

How to Handle:

Implement an iterative approach using a priority queue or equivalent data structure for efficiency.

Values in one pile are significantly larger than values in others.

How to Handle:

The greedy approach of always removing from the two largest piles is optimal and handles such scenarios correctly.

The sum of the three piles does not change after each removal, so if the sum is odd, no moves can happen

How to Handle:

If the sum of all piles is odd from the start, then return -1 because this state can never lead to all piles becoming zero simultaneously.

One pile's initial value is larger than or equal to the sum of other two.

How to Handle:

The total number of moves is restricted by the sum of the smaller piles, which is the maximum number of moves possible.