Minimum Cost to Merge Stones

Hard

Google

2 views

Topics:

Dynamic Programming

You are given n piles of stones arranged in a row. The i<sup>th</sup> pile has stones[i] stones.

A move consists of merging exactly k consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these k piles.

Your task is to write a function that returns the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], k = 2
Output: 20
Explanation: 
We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], k = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore. So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], k = 3
Output: 25
Explanation: 
We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.

Constraints:

1 <= n <= 30
1 <= stones[i] <= 100
2 <= k <= 30

Minimum Cost to Merge Stones

Hard

Google

2 views

Topics:

Dynamic Programming

You are given n piles of stones arranged in a row. The i<sup>th</sup> pile has stones[i] stones.

A move consists of merging exactly k consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these k piles.

Your task is to write a function that returns the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], k = 2
Output: 20
Explanation: 
We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], k = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore. So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], k = 3
Output: 25
Explanation: 
We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.

Constraints:

1 <= n <= 30
1 <= stones[i] <= 100
2 <= k <= 30

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, and what is the maximum value of each stone?
Is `K` always a valid number such that it's between 2 and the length of `stones` inclusive?
If it's impossible to merge all stones into one pile, what should the function return?
Are the stone values guaranteed to be non-negative?
Could you provide a small example with the input and expected output to ensure I fully understand the merging process and cost calculation?

Brute Force Solution

Approach

We need to combine piles of stones into one big pile, and each combination costs a certain amount. The brute force method will try every single way we could possibly combine the piles, no matter how silly or inefficient it seems. We'll then pick the cheapest way from all the options we explored.

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

Consider all possible ways to group the stones in the first step. For example, if we have four piles, we could combine the first two, the first three, or all four.
For each of those first-step combinations, figure out the cost of doing that combination.
Now, for each of the results from that first combination, consider all possible ways to combine the remaining piles. Again, calculate the cost of each of these combinations.
Keep doing this, combining piles step by step, until you are left with just one big pile.
Each complete way of combining the stones from start to finish is a possible solution. Keep track of the total cost of each solution.
Once you've tried every single way to combine the piles, compare the total costs of all the solutions you found.
The solution with the lowest total cost is the answer.

Code Implementation

def minimum_cost_to_merge_stones_brute_force(stones, k):
    number_of_stones = len(stones)

    if (number_of_stones - 1) % (k - 1) != 0:
        return -1

    def calculate_cost(current_stones):
        if len(current_stones) == 1:
            return 0, current_stones[0]

        minimum_cost = float('inf')

        # Try every possible merge
        for i in range(len(current_stones) - k + 1):
            
            # We can only merge k stones
            next_stones = current_stones[:i] + \
                          [sum(current_stones[i:i+k])] + \
                          current_stones[i+k:]

            cost, _ = calculate_cost(next_stones)

            minimum_cost = min(minimum_cost, \
                                sum(current_stones[i:i+k]) + cost)
        
        return minimum_cost, sum(current_stones)

    # Initiate the recursive function.
    total_cost, _ = calculate_cost(stones)
    return total_cost

Big(O) Analysis

Time Complexity

O(K^(N/K)) – The brute force approach explores all possible ways to merge N piles of stones into one, where each merge combines K piles. The number of merge steps depends on how many piles we combine at each stage. Because we are considering all possible groupings, the branching factor at each step is heavily dependent on the input values of N and K. Therefore the number of possible trees of combinations we explore is exponential with a base relating to K and with a number of levels relating to N/K. Thus the time complexity is O(K^(N/K)).

Space Complexity

O(N^2 * N) – The brute force approach explores all possible ways to combine the stones, which inherently involves storing intermediate results for each combination. The key factor affecting space is the potential size of the call stack when recursively exploring these combinations. The maximum depth of the recursion can be related to the number of piles, N, and at each level of recursion, we are potentially creating copies of subarrays or storing intermediate costs. In the worst case, the space complexity is driven by the memoization or storing intermediate results for each subproblem that can arise as the algorithm explores all possible merging scenarios which is O(N^2 * N) where N is the number of stones.

Optimal Solution

Approach

The best way to solve this problem is by combining smaller groups of stones until you have one big pile. We want to do this in a way that costs the least amount of effort overall. The key is to figure out the cheapest way to merge stones in each step, building up to the final result.

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

First, realize you need to combine the stones into a single pile by repeatedly merging groups of stones.
Figure out how much it costs to merge any group of stones together. The cost is just the sum of the weight of all the stones in that group.
Start by looking at small groups of stones next to each other and calculating the cost to combine them.
Store the cheapest costs to combine small groups of stones.
Now, consider slightly bigger groups of stones. To find the cheapest way to combine them, use the previously calculated cheapest costs for the smaller groups that make them up.
Continue to build up larger groups of stones by using the cheapest costs of the smaller groups until you eventually figure out the cost of combining ALL the stones into one pile.
The cost to combine the largest group represents the minimum cost to merge all the stones.

Code Implementation

def minimum_cost_to_merge_stones(stones, k_stones):    number_of_stones = len(stones)
    if (number_of_stones - 1) % (k_stones - 1) != 0:
        return -1

    prefix_sum = [0] * (number_of_stones + 1)
    for i in range(1, number_of_stones + 1):
        prefix_sum[i] = prefix_sum[i - 1] + stones[i - 1]

    # dp[i][j] is min cost to merge stones[i] to stones[j]
    dp = [[0] * number_of_stones for _ in range(number_of_stones)]

    for interval_length in range(k_stones, number_of_stones + 1):
        for start_index in range(number_of_stones - interval_length + 1):
            end_index = start_index + interval_length - 1
            dp[start_index][end_index] = float('inf')

            for merge_point in range(start_index, end_index, k_stones - 1):
                dp[start_index][end_index] = min(
                    dp[start_index][end_index],
                    dp[start_index][merge_point] + dp[merge_point + 1][end_index],
                )

            # Only merge when the number of piles can be merged into one
            if (interval_length - 1) % (k_stones - 1) == 0:
                # Add the cost of merging the interval
                dp[start_index][end_index] += (
                    prefix_sum[end_index + 1] - prefix_sum[start_index]
                )

    # The final value represents the min cost to merge all stones
    return dp[0][number_of_stones - 1]

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm uses dynamic programming to compute the minimum cost to merge stones. It iterates through all possible lengths of subarrays, which takes O(n) time. For each subarray length, it iterates through all possible starting positions, which also takes O(n) time. Finally, for each subarray, it iterates through all possible split positions to find the optimal merging point which takes another O(n) time. Therefore, the overall time complexity is O(n * n * n) = O(n^3).

Space Complexity

O(N^2) – The algorithm stores the cheapest costs to combine groups of stones of various sizes. Step 4 and subsequent steps imply storing intermediate results for combining stones. Since it considers all possible contiguous subarrays, and for each subarray, it calculates and stores a cost, this implies storing costs for subarrays of size 1, 2, 3, up to N, where N is the total number of stones. This typically requires a 2D array or a similar data structure to store the minimum costs, resulting in O(N^2) auxiliary space.

Edge Cases

Case	How to Handle
Null or empty stones array	Return 0 if stones is null or empty, as there is nothing to merge.
stones array with size less than K	Return -1 since it's impossible to merge less than K stones into 1.
stones array size doesn't satisfy (n - 1) % (K - 1) == 0	Return -1 because it's impossible to merge down to a single pile.
K is 1	If K is 1, merging isn't possible if there's more than 1 stone, so return 0 if array size is 1, and -1 otherwise.
stones array with all the same values	The algorithm should correctly calculate the cost even if all values are equal.
Integer overflow when calculating the sum of stones in a merge	Use long data type to store the sums to prevent overflow.
Large input array size exceeding memory limits	The dynamic programming table can become very large; consider if memory optimization techniques like rolling arrays are applicable if the entire table is not needed.
Negative stone values	The problem description doesn't specify non-negative stones, but if they exist, they don't break the algorithm, however the minimum cost could be negative as well.