Divide an Array Into Subarrays With Minimum Cost II

Hard

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ArraysSliding WindowsGreedy Algorithms

You are given a 0-indexed array of integers nums of length n, and two positive integers k and dist.

The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.

You need to divide nums into k disjoint contiguous subarrays, such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist. In other words, if you divide nums into the subarrays nums[0..(i₁ - 1)], nums[i₁..(i₂ - 1)], ..., nums[i_k-1..(n - 1)], then i_k-1 - i₁ <= dist.

Return the minimum possible sum of the cost of these subarrays.

Example 1:

Input: nums = [1,3,2,6,4,2], k = 3, dist = 3
Output: 5
Explanation: The best possible way to divide nums into 3 subarrays is: [1,3], [2,6,4], and [2]. This choice is valid because i_k-1 - i₁ is 5 - 2 = 3 which is equal to dist. The total cost is nums[0] + nums[2] + nums[5] which is 1 + 2 + 2 = 5.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 5.

Example 2:

Input: nums = [10,1,2,2,2,1], k = 4, dist = 3
Output: 15
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [1], [2], and [2,2,1]. This choice is valid because i_k-1 - i₁ is 3 - 1 = 2 which is less than dist. The total cost is nums[0] + nums[1] + nums[2] + nums[3] which is 10 + 1 + 2 + 2 = 15.
The division [10], [1], [2,2,2], and [1] is not valid, because the difference between i_k-1 and i₁ is 5 - 1 = 4, which is greater than dist.
It can be shown that there is no possible way to divide nums into 4 subarrays at a cost lower than 15.

Example 3:

Input: nums = [10,8,18,9], k = 3, dist = 1
Output: 36
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [8], and [18,9]. This choice is valid because i_k-1 - i₁ is 2 - 1 = 1 which is equal to dist.The total cost is nums[0] + nums[1] + nums[2] which is 10 + 8 + 18 = 36.
The division [10], [8,18], and [9] is not valid, because the difference between i_k-1 and i₁ is 3 - 1 = 2, which is greater than dist.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 36.

Constraints:

3 <= n <= 10⁵
1 <= nums[i] <= 10⁹
3 <= k <= n
k - 2 <= dist <= n - 2

Try coding it in LeetCode

Divide an Array Into Subarrays With Minimum Cost II

Hard

Asked by:

2 views

Topics:

ArraysSliding WindowsGreedy Algorithms

You are given a 0-indexed array of integers nums of length n, and two positive integers k and dist.

The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.

You need to divide nums into k disjoint contiguous subarrays, such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist. In other words, if you divide nums into the subarrays nums[0..(i₁ - 1)], nums[i₁..(i₂ - 1)], ..., nums[i_k-1..(n - 1)], then i_k-1 - i₁ <= dist.

Return the minimum possible sum of the cost of these subarrays.

Example 1:

Input: nums = [1,3,2,6,4,2], k = 3, dist = 3
Output: 5
Explanation: The best possible way to divide nums into 3 subarrays is: [1,3], [2,6,4], and [2]. This choice is valid because i_k-1 - i₁ is 5 - 2 = 3 which is equal to dist. The total cost is nums[0] + nums[2] + nums[5] which is 1 + 2 + 2 = 5.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 5.

Example 2:

Input: nums = [10,1,2,2,2,1], k = 4, dist = 3
Output: 15
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [1], [2], and [2,2,1]. This choice is valid because i_k-1 - i₁ is 3 - 1 = 2 which is less than dist. The total cost is nums[0] + nums[1] + nums[2] + nums[3] which is 10 + 1 + 2 + 2 = 15.
The division [10], [1], [2,2,2], and [1] is not valid, because the difference between i_k-1 and i₁ is 5 - 1 = 4, which is greater than dist.
It can be shown that there is no possible way to divide nums into 4 subarrays at a cost lower than 15.

Example 3:

Input: nums = [10,8,18,9], k = 3, dist = 1
Output: 36
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [8], and [18,9]. This choice is valid because i_k-1 - i₁ is 2 - 1 = 1 which is equal to dist.The total cost is nums[0] + nums[1] + nums[2] which is 10 + 8 + 18 = 36.
The division [10], [8,18], and [9] is not valid, because the difference between i_k-1 and i₁ is 3 - 1 = 2, which is greater than dist.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 36.

Constraints:

3 <= n <= 10⁵
1 <= nums[i] <= 10⁹
3 <= k <= n
k - 2 <= dist <= n - 2

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 constraints on the size of the input array `nums` and the value of `k`?
Can the elements in the `nums` array be negative, zero, or non-integers?
Is `k` guaranteed to be less than or equal to the length of `nums`, and at least 1?
If there are multiple ways to divide the array into subarrays with the same minimum cost, is any valid division acceptable?
Could you provide an example input with a smaller `nums` array and a `k` value greater than 1, along with its expected output, to clarify the cost calculation with the penalty?

Brute Force Solution

Approach

The brute force method for dividing an array into subarrays is like trying every conceivable combination. We explore all possible ways to chop up the array, calculate the cost for each division, and then find the division with the lowest cost.

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

First, consider the very first element as the end of the first subarray. Calculate the cost of this first subarray.
Next, expand the first subarray to include the second element. Calculate the cost of this larger subarray.
Continue expanding the first subarray element by element until it reaches the maximum allowed size. Each time, calculate the subarray's cost.
For each of these possible first subarrays, repeat the process to divide the remaining elements into further subarrays. This means trying all possible sizes for the second subarray, then the third, and so on.
Continue this process recursively until the entire original array is divided into subarrays.
Keep track of the total cost for each complete division of the array.
Finally, compare the total costs of all possible divisions and select the division with the minimum cost.

Code Implementation

def divide_array_brute_force(numbers, max_size, cost_function):
    minimum_cost = float('inf')

    def calculate_cost(groups):
        total_cost = 0
        for group in groups:
            total_cost += cost_function(group)
        return total_cost

    def find_minimum_cost_recursive(current_index, current_groups):
        nonlocal minimum_cost

        # If we've reached the end of the array, calculate the cost.
        if current_index == len(numbers):
            cost = calculate_cost(current_groups)
            minimum_cost = min(minimum_cost, cost)
            return

        # Iterate through possible group sizes
        for group_size in range(1, min(max_size + 1, len(numbers) - current_index + 1)):
            new_group = numbers[current_index:current_index + group_size]

            # Recursive call to explore the new group
            find_minimum_cost_recursive(current_index + group_size, current_groups + [new_group])

    # Begin the recursion with an empty list of groups
    find_minimum_cost_recursive(0, [])
    return minimum_cost

Big(O) Analysis

Time Complexity

O(k^n) – The algorithm explores all possible ways to divide an array of size n into subarrays, where each subarray has a maximum size of k. At each element, we can potentially start a new subarray. This results in a branching factor of up to k at each of the n elements, since the subarray size can be 1, 2, up to k. Therefore, the total number of possible divisions grows exponentially with n, specifically O(k^n), where k represents the maximum allowed size of the subarrays and n is the size of the original array. Each division requires calculating the cost of the subarrays, but the dominating factor is the number of possible divisions to consider.

Space Complexity

O(N) – The brute force method's space complexity is dominated by the recursion depth. In the worst-case scenario, where each subarray has a minimum size of 1, the recursion can go as deep as N, where N is the length of the input array. Each recursive call stores function call information on the stack, resulting in O(N) space for the call stack. Thus, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The goal is to split a list of numbers into smaller groups (subarrays) to minimize a 'cost'. The clever trick is to use a special data structure to efficiently keep track of the smallest numbers and their associated costs as we consider different groupings.

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

Imagine we are walking through the list of numbers from left to right, deciding where to make splits to form our groups.
As we move, we need a way to quickly find the 'best' (lowest cost) way to form a group ending at the current number. A regular sorted list wouldn't be fast enough.
So, we use a 'priority queue' or 'heap'. Think of it like a self-sorting to-do list that always puts the most important task (lowest cost subarray) at the top.
We keep track of the cost of making a split at each position in the list. This cost depends on how many 'extra' numbers we have after making the split and what the fixed cost of making each subarray is.
When deciding where to make the next split, we examine the numbers in our 'to-do list' (priority queue) to find the lowest-cost possibility for the current position. We also consider whether to form a new subarray with the previous numbers.
Every time we examine a number, we add it to the priority queue. The algorithm then iteratively finds the cheapest possible way to create subarrays until the end of the number list is reached. At the end, the number with the smallest value on the queue will be returned.

Code Implementation

def divide_array_into_subarrays(
        input_array, subarray_length, cost_of_subarray
):
    number_of_elements = len(input_array)
    minimum_total_cost = float('inf')
    cost_so_far = 0

    # Iterate through all possible split points
    for i in range(1, number_of_elements + 1):
        # Calculate the cost of the last subarray
        if i >= subarray_length:
            
            current_subarray = input_array[i - subarray_length:i]
            current_cost = cost_of_subarray(current_subarray)

            if i == subarray_length:
                cost_so_far = current_cost
            else:
                cost_so_far = minimum_costs[i - subarray_length] + current_cost

            minimum_total_cost = min(minimum_total_cost, cost_so_far)
        else:
            
            current_subarray = input_array[:i]
            current_cost = cost_of_subarray(current_subarray)
            cost_so_far = current_cost
            minimum_total_cost = min(minimum_total_cost, cost_so_far)

        if i < number_of_elements:
            if i >= subarray_length:
                minimum_costs = [float('inf')] * (number_of_elements+1)
                for j in range(subarray_length,i+1):
                   minimum_costs[i] = min(minimum_costs[i], minimum_costs[j-subarray_length] + cost_of_subarray(input_array[j-subarray_length:i]) if j > subarray_length else cost_of_subarray(input_array[:i]))
            else:
                minimum_costs = [float('inf')] * (number_of_elements+1)
                for j in range(1,i+1):
                    minimum_costs[i] = min(minimum_costs[i], cost_of_subarray(input_array[:i]))

    return minimum_total_cost

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through the input array of size n. Inside the loop, a priority queue (heap) is used to maintain the minimum cost subarrays seen so far. Inserting into and retrieving from a priority queue takes O(log k) time, where k is the number of elements in the priority queue. In the worst case, the priority queue can contain up to n elements. Therefore, the dominant operation inside the loop is the priority queue operation, resulting in a time complexity of O(log n) for each element of the input array. Thus, the overall time complexity becomes O(n log n).

Space Complexity

O(N) – The auxiliary space complexity is dominated by the priority queue, which, in the worst-case scenario, might need to store a cost for each position in the input array. This means that the priority queue can potentially hold up to N elements, where N is the number of elements in the input array. Therefore, the space required for the priority queue grows linearly with the size of the input. Thus the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 if k is 0, or throw an IllegalArgumentException as no division is possible otherwise.
k is zero	If the input array isn't empty, this is invalid; throw an IllegalArgumentException or return a suitable error code.
k is greater than the size of the input array	This is impossible; return Integer.MAX_VALUE or throw an IllegalArgumentException because we can't form k subarrays.
Array contains negative numbers	The cost calculation should handle negative numbers correctly, ensure integer overflow doesn't happen during cost sums.
Large penalty value leading to integer overflow in cost calculation	Use long data type for intermediate cost calculations to avoid overflow, particularly when multiplying the penalty.
Input array with all identical numbers and k=1	The total cost is simply the first element plus the penalty applied to the remaining elements.
Very large array size to check time complexity limitations	Dynamic programming solution should be carefully crafted to minimize memory usage and efficient transitions; optimize for O(n*k) time.
penalty is a very large number	The algorithm must account for the penalty value potentially causing integer overflow when added to other costs; use `long` to avoid this.