Put Marbles in Bags

Hard

Put Marbles in Bags

Hard

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 the elements in the `weights` array and the value of `k`?
Can `k` ever be greater than the length of the `weights` array?
Can the input array `weights` be empty or null?
What should I return if `k` is less than or equal to 0?
The problem asks to minimize the score and maximize the score. Should I return both scores as a pair, or should I call two different functions?

Brute Force Solution

Approach

To solve this problem the hard way, we'll look at every single possible way to divide the marbles into bags. We will then figure out the cost for each of these divisions, and keep track of the best one we've seen so far.

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

Consider every possible place we could put a divider between the marbles to create the bags.
For each arrangement of dividers, calculate the total cost. This is the sum of the weights of the marbles at the ends of each bag.
Compare this cost to the lowest cost we've seen so far. If it's lower, remember this new cost and the divider arrangement that produced it.
Repeat this process until we have considered every single possible way to arrange the dividers.
Once we've checked all the arrangements, the lowest cost we remembered is the answer.

Code Implementation

def put_marbles_in_bags_brute_force(weights, num_bags):
    number_of_marbles = len(weights)

    # If we only have one bag, the cost is the first plus the last weight
    if num_bags == 1:
        return weights[0] + weights[number_of_marbles - 1]

    min_cost = float('inf')

    # Iterate through all possible divider placements
    for i in range(1 << (number_of_marbles - 1)):
        # Only consider combinations with the correct number of dividers
        if bin(i).count('1') == num_bags - 1:
            current_cost = 0
            bag_start = 0

            # Calculate cost based on current divider arrangement
            for j in range(number_of_marbles - 1):
                if (i >> j) & 1:
                    current_cost += weights[bag_start] + weights[j]
                    bag_start = j + 1

            # Account for the last bag
            current_cost += weights[bag_start] + weights[number_of_marbles - 1]

            # Track minimum cost
            min_cost = min(min_cost, current_cost)

    return min_cost

Big(O) Analysis

Time Complexity

O(n^k) – The algorithm considers every possible way to put k-1 dividers between n marbles. There are n-1 possible locations for each divider. Therefore, we are essentially choosing k-1 divider locations out of n-1 possible locations. This equates to (n-1 choose k-1) combinations, or approximately O(n^(k-1)). However, for each arrangement of dividers, we need to calculate the cost which involves summing the values at the ends of the k bags. This requires iterating through the k bag ends, leading to a O(k) operation. Considering k is a constant, the total time complexity is dominated by the number of arrangements, so it becomes O(n^(k-1)) * O(k) ~ O(n^(k-1) *k). Since k is constant, it's simplified as O(n^k).

Space Complexity

O(1) – The provided solution iteratively considers every possible arrangement of dividers and calculates the total cost for each arrangement. It keeps track of the lowest cost seen so far and the corresponding divider arrangement. This requires storing a few variables to keep track of the current arrangement's cost, the lowest cost seen so far, and potentially the divider locations of the lowest cost. The number of these variables is independent of the number of marbles (N), thus the auxiliary space used remains constant, resulting in O(1) space complexity.

Optimal Solution

Approach

The goal is to divide a row of marbles into a given number of bags such that the total score is maximized. The key insight is that only the weights of the marbles at the bag boundaries affect the score, so we focus on optimizing those selections.

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

First, we figure out what contributes to the total score. Only the marbles at the cut points between bags matter; the marbles inside the bags don't affect the total score.
Next, we determine what values would be added to our result, and what contributes to those values. The boundary weights can be computed by adding the weights of each adjacent node.
Realize that the first marble and the last marble will always be selected. Therefore, we only need to optimize the choices of the 'in-between' marbles.
We now want to greedily select the heaviest marbles at those 'in-between' locations. Sort the calculated weights to select the largest.
Once we've made our selections of the heaviest marbles at the optimal cut points, add their weights to the weight of the first and last marbles to get our final, maximized score.

Code Implementation

def put_marbles_in_bags(weights, number_of_bags):
    number_of_weights = len(weights)
    if number_of_bags > number_of_weights:
        return 0

    # To maximize the score, calculate the boundary weights.
    boundary_weights = []
    for i in range(number_of_weights - 1):
        boundary_weights.append(weights[i] + weights[i + 1])

    # Need to sort to find the heaviest marbles at boundaries.
    boundary_weights.sort(reverse=True)

    total_score = 0
    # We sum the largest 'number_of_bags - 1' boundary weights.
    for i in range(number_of_bags - 1):
        total_score += boundary_weights[i]

    # Add the weights of the first and last marbles.
    total_score += weights[0] + weights[-1]

    return total_score

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through the array of size n to calculate the weights of adjacent marbles, resulting in O(n) time complexity. The most significant operation is sorting these weights to greedily select the largest ones. Sorting n elements takes O(n log n) time. Finally, iterating and summing selected weights takes O(k) time where k is at most n (number of bags to create). Since sorting dominates the time, the overall time complexity is O(n log n).

Space Complexity

O(N) – The dominant space complexity comes from creating an intermediate array to store the sums of adjacent marble weights. This array will have a size of N-1, where N is the number of marbles. Sorting this array in place does not change the overall auxiliary space because the array itself already occupies O(N) space. Therefore, the auxiliary space is O(N).

Edge Cases

Case	How to Handle
weights is null or empty	Return 0 immediately as no marbles can be placed.
k is less than or equal to 0 or greater than n (length of weights)	Return 0, indicating an invalid number of bags requested.
weights contains only one element	Return 0, since at least two marbles are required to form a pair.
Large input array size (near maximum allowed)	Ensure the chosen algorithm (e.g., sorting) scales efficiently (O(n log n) is preferred) to avoid exceeding time limits.
Integer overflow when calculating the cost of a bag	Use long data type to store the cost of each bag to prevent potential integer overflow.
All weights are the same value	The sorted sums will result in a predictable pattern, which the solution should handle correctly to find the minimum sum.
k equals n (number of weights)	Return the sum of all adjacent pairs because each weight will be in its own bag.
weights array contains negative values	The sorting operation will still work correctly, resulting in the minimum sum of k-1 largest values, even with negative values.