Find Minimum Cost to Remove Array Elements

Medium

DE Shaw

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Topics:

ArraysDynamic Programming

You are given an integer array nums. Your task is to remove all elements from the array by performing one of the following operations at each step until nums is empty:

Choose any two elements from the first three elements of nums and remove them. The cost of this operation is the maximum of the two elements removed.
If fewer than three elements remain in nums, remove all the remaining elements in a single operation. The cost of this operation is the maximum of the remaining elements.

Return the minimum cost required to remove all the elements.

Example 1:

Input: nums = [6,2,8,4]

Output: 12

Explanation:

Initially, nums = [6, 2, 8, 4].

In the first operation, remove nums[0] = 6 and nums[2] = 8 with a cost of max(6, 8) = 8. Now, nums = [2, 4].
In the second operation, remove the remaining elements with a cost of max(2, 4) = 4.

The cost to remove all elements is 8 + 4 = 12. This is the minimum cost to remove all elements in nums. Hence, the output is 12.

Example 2:

Input: nums = [2,1,3,3]

Output: 5

Explanation:

Initially, nums = [2, 1, 3, 3].

In the first operation, remove nums[0] = 2 and nums[1] = 1 with a cost of max(2, 1) = 2. Now, nums = [3, 3].
In the second operation remove the remaining elements with a cost of max(3, 3) = 3.

The cost to remove all elements is 2 + 3 = 5. This is the minimum cost to remove all elements in nums. Hence, the output is 5.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁶

Find Minimum Cost to Remove Array Elements

Medium

DE Shaw

0 views

Topics:

ArraysDynamic Programming

You are given an integer array nums. Your task is to remove all elements from the array by performing one of the following operations at each step until nums is empty:

Choose any two elements from the first three elements of nums and remove them. The cost of this operation is the maximum of the two elements removed.
If fewer than three elements remain in nums, remove all the remaining elements in a single operation. The cost of this operation is the maximum of the remaining elements.

Return the minimum cost required to remove all the elements.

Example 1:

Input: nums = [6,2,8,4]

Output: 12

Explanation:

Initially, nums = [6, 2, 8, 4].

In the first operation, remove nums[0] = 6 and nums[2] = 8 with a cost of max(6, 8) = 8. Now, nums = [2, 4].
In the second operation, remove the remaining elements with a cost of max(2, 4) = 4.

The cost to remove all elements is 8 + 4 = 12. This is the minimum cost to remove all elements in nums. Hence, the output is 12.

Example 2:

Input: nums = [2,1,3,3]

Output: 5

Explanation:

Initially, nums = [2, 1, 3, 3].

In the first operation, remove nums[0] = 2 and nums[1] = 1 with a cost of max(2, 1) = 2. Now, nums = [3, 3].
In the second operation remove the remaining elements with a cost of max(3, 3) = 3.

The cost to remove all elements is 2 + 3 = 5. This is the minimum cost to remove all elements in nums. Hence, the output is 5.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁶

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 input array, and what is the range of values for the integers within the array?
Can the input array be empty or null?
If there are multiple ways to remove elements with the same minimum cost, is any valid solution acceptable?
Are all the numbers in the array positive integers, or could there be negative numbers or zeros?
Could you provide a small example with the expected output to ensure I understand the problem correctly?

Brute Force Solution

Approach

The brute force approach to finding the minimum cost to remove elements involves checking absolutely every possible combination of element removals. It is similar to trying every single possibility until you find the best one. We will systematically go through each combination and calculate the cost.

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

Consider all possible sets of elements to remove from the input. We could remove nothing, remove one element, remove two elements, and so on, up to removing all the elements.
For each set of elements that we choose to remove, we can compute the total cost of removing those elements.
After removing the selected elements, calculate the sum of the remaining elements. If the sum of the remaining elements meets the problem constraints (e.g., it's less than or equal to a target), then store the cost of removing the selected elements.
Repeat the removal and calculation process for every possible combination of elements to remove.
Finally, compare all the stored costs to determine the smallest cost among all valid removals that result in sums that satisfy the problem's conditions. The smallest cost is our answer.

Code Implementation

def find_minimum_cost_to_remove_array_elements_brute_force(input_array, removal_costs, target_sum):
    minimum_removal_cost = float('inf')
    number_of_elements = len(input_array)

    # Iterate through all possible subsets of elements to remove.
    for i in range(2 ** number_of_elements):
        elements_to_remove = []
        removal_cost_for_subset = 0
        remaining_elements = []
        
        for j in range(number_of_elements):
            # Check if the j-th element is in the current subset to remove
            if (i >> j) & 1:
                elements_to_remove.append(j)
                removal_cost_for_subset += removal_costs[j]
            else:
                remaining_elements.append(input_array[j])

        # Calculate the sum of the remaining elements
        sum_of_remaining_elements = sum(remaining_elements)

        # Check if the sum meets the target constraint
        if sum_of_remaining_elements <= target_sum:
            # Update the minimum cost if the current cost is lower.
            minimum_removal_cost = min(minimum_removal_cost, removal_cost_for_subset)

    # If no valid removal found, return -1.
    if minimum_removal_cost == float('inf'):
        return -1
    else:
        return minimum_removal_cost

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm considers all possible subsets of the input array of size n. Generating all subsets requires iterating through all possible combinations of including or excluding each element. For each element, there are two choices: include it or exclude it. Therefore, there are 2*2*...*2 (n times) = 2^n possible subsets. For each subset, we calculate the sum of the remaining elements and the cost of removal, which takes O(n) time in the worst case. However, the dominant factor is the generation of all subsets, resulting in a time complexity of O(2^n).

Space Complexity

O(1) – The brute force approach described explores all possible subsets of elements to remove. Although many subsets are considered, the provided explanation does not explicitly state that these subsets are stored in memory. The key part of the explanation is about 'storing the cost of removing the selected elements' only if a condition is met. Because the number of these stored costs is not related to N in the explanation and no collections or list like structures are mentioned, the space complexity is likely constant, meaning no extra data structures that grow with the input size N are used beyond a few primitive variables for calculations. Therefore, the auxiliary space is O(1).

Optimal Solution

Approach

The core idea is to use a method that remembers the best solutions to smaller versions of the problem to solve the bigger problem efficiently. We break down the problem into subproblems and reuse the solutions. This prevents recalculating the same thing multiple times.

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

Think of the problem as making a series of choices: at each position in the sequence of numbers, decide whether to remove the current number or not.
Figure out the minimum cost to remove numbers starting from the beginning up to each position. Store these costs as you go.
To calculate the cost up to a certain point, consider two options: either you remove the current number, or you don't.
If you remove the current number, the cost is the cost of removing numbers up to the previous position, plus the cost of removing the current number.
If you don't remove the current number, you need to find a previous position where you *did* remove a number, and the numbers in between form a valid group (meeting certain conditions). The cost would then be the cost of removing numbers up to that previous position.
Choose the option (remove or don't remove) that gives you the smallest cost. Store this minimum cost.
Repeat this process until you reach the end of the number sequence. The final stored cost is the minimum cost to remove elements to achieve the desired result.

Code Implementation

def find_minimum_cost_to_remove_array_elements(array, cost_to_remove, group_condition):
    number_of_elements = len(array)
    minimum_removal_costs = [0] * (number_of_elements + 1)

    for i in range(1, number_of_elements + 1):
        minimum_removal_costs[i] = minimum_removal_costs[i-1] + cost_to_remove[i-1]

        # Assume we remove current element.

        for j in range(i - 1, -1, -1):
            # Check all previous removal spots to see if we *don't* remove current element.

            sub_array = array[j:i]
            if group_condition(sub_array):

                #If removing the elements from j to i is acceptable
                minimum_removal_costs[i] = min(minimum_removal_costs[i],
                                                minimum_removal_costs[j])

    return minimum_removal_costs[number_of_elements]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array of n elements. For each element, it potentially iterates backwards to find a valid previous removal point. In the worst case, for each element, it may iterate back to the beginning of the array, resulting in a nested loop structure. This nested loop leads to approximately n * n/2 operations. Thus, the overall time complexity is O(n²).

Space Complexity

O(N) – The solution utilizes dynamic programming, specifically storing the minimum cost to remove elements up to each position in the input sequence. This requires an auxiliary data structure, likely an array (or list), to store these minimum costs. The size of this array directly corresponds to the number of elements in the input sequence, denoted as N. Therefore, the auxiliary space used is proportional to the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since there are no elements to remove, therefore no cost.
Array with one element	Return the value of the single element since it must be removed to satisfy the condition.
Array with two elements	Return the minimum of the two elements, as removing the larger one will satisfy the condition with minimal cost.
Array with all elements having the same value	The solution should still function correctly and efficiently converge to the optimal solution.
Large input array size (scalability)	The solution needs to be optimized for time complexity, ideally using dynamic programming to avoid exponential time complexity.
Array with very large integer values (potential integer overflow)	Use data types like `long` or consider using a programming language with built-in arbitrary-precision arithmetic to prevent integer overflow during summation of costs.
Array with alternating large and small values	The algorithm should correctly explore all possible removal combinations to arrive at the global minimum, not getting stuck in local minima.
Array with all positive integers.	The dynamic programming should still calculate cost correctly if all integers are positive.