Divide an Array Into Subarrays With Minimum Cost I

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ArraysTwo PointersSliding Windows

You are given an array of integers nums of length n.

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 3 disjoint contiguous subarrays.

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

Example 1:

Input: nums = [1,2,3,12]
Output: 6
Explanation: The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6.
The other possible ways to form 3 subarrays are:
- [1], [2,3], and [12] at a total cost of 1 + 2 + 12 = 15.
- [1,2], [3], and [12] at a total cost of 1 + 3 + 12 = 16.

Example 2:

Input: nums = [5,4,3]
Output: 12
Explanation: The best possible way to form 3 subarrays is: [5], [4], and [3] at a total cost of 5 + 4 + 3 = 12.
It can be shown that 12 is the minimum cost achievable.

Example 3:

Input: nums = [10,3,1,1]
Output: 12
Explanation: The best possible way to form 3 subarrays is: [10,3], [1], and [1] at a total cost of 10 + 1 + 1 = 12.
It can be shown that 12 is the minimum cost achievable.

Constraints:

3 <= n <= 50
1 <= nums[i] <= 50

Try coding it in LeetCode

Divide an Array Into Subarrays With Minimum Cost I

Easy

Asked by:

1 view

Topics:

ArraysTwo PointersSliding Windows

You are given an array of integers nums of length n.

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 3 disjoint contiguous subarrays.

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

Example 1:

Input: nums = [1,2,3,12]
Output: 6
Explanation: The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6.
The other possible ways to form 3 subarrays are:
- [1], [2,3], and [12] at a total cost of 1 + 2 + 12 = 15.
- [1,2], [3], and [12] at a total cost of 1 + 3 + 12 = 16.

Example 2:

Input: nums = [5,4,3]
Output: 12
Explanation: The best possible way to form 3 subarrays is: [5], [4], and [3] at a total cost of 5 + 4 + 3 = 12.
It can be shown that 12 is the minimum cost achievable.

Example 3:

Input: nums = [10,3,1,1]
Output: 12
Explanation: The best possible way to form 3 subarrays is: [10,3], [1], and [1] at a total cost of 10 + 1 + 1 = 12.
It can be shown that 12 is the minimum cost achievable.

Constraints:

3 <= n <= 50
1 <= nums[i] <= 50

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 `nums` array and the value of `k`?
Can the elements in the `nums` array be negative, zero, or only positive integers?
Is `k` always guaranteed to be less than or equal to the length of the `nums` array? If `k` is greater than the length of `nums`, what should the function return?
If there are multiple ways to divide the array into `k` subarrays with the same minimum cost, is any one of them acceptable, or is there a specific division I should prioritize?
Could you provide an example where the minimum cost is achieved with a specific division to ensure my understanding?

Brute Force Solution

Approach

The brute force approach tackles this problem by exploring every single possible way to cut the initial collection of numbers into smaller groups. We check each arrangement to see if it meets some condition, like limiting the number of groups, and then calculate a score for it. We keep the arrangement with the best score.

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

Consider dividing the collection into just one group.
Then, consider dividing it into two groups, trying every possible place to make the split.
Next, consider dividing it into three groups, again trying every single combination of split locations.
Continue this process until you've considered the maximum allowed number of groups.
For each way you divide the collection into groups, calculate a value or score based on some rule.
Compare the scores of all the different arrangements you've tried.
Select the arrangement with the best score as your answer.

Code Implementation

def divide_array_into_subarrays_with_minimum_cost_brute_force(numbers, max_subarrays):

    best_cost = float('inf')

    def calculate_cost(subarray_division):
        cost = 0
        for subarray in subarray_division:
            if subarray:
                cost += subarray[0] # Cost based on first element
        return cost

    def generate_subarrays(index, current_subarrays):
        nonlocal best_cost
        
        if index == len(numbers):
            if len(current_subarrays) <= max_subarrays and len(current_subarrays) > 0:
                cost = calculate_cost(current_subarrays)
                best_cost = min(best_cost, cost)
            return

        # Add to the last subarray
        if current_subarrays:
            generate_subarrays(index + 1, current_subarrays[:-1] + [current_subarrays[-1] + [numbers[index]]])

        # Start a new subarray
        # Core logic: explore the partition decision
        generate_subarrays(index + 1, current_subarrays + [[numbers[index]]])

    # Start with an empty set of subarrays
    generate_subarrays(0, [])

    if best_cost == float('inf'):
        return -1
    return best_cost

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach iterates through all possible ways to divide an array of size n into up to k subarrays. When dividing into i subarrays, we are essentially choosing (i-1) split points from (n-1) possible locations. This can be represented by the binomial coefficient (n-1 choose i-1). Summing this for i from 1 to k gives the total number of possible subarray divisions. In the worst case, where k is close to n, this can approach O(n^k) as we need to consider potentially all possible combinations of dividing the array into various numbers of subarrays up to k. Calculating the cost for each such division will be included in the constant factor.

Space Complexity

O(N^K) – The brute force approach explores all possible ways to divide the array into up to K subarrays. In the worst case, it might involve storing multiple subarrays or intermediate results for each possible division. Storing each division requires space proportional to the number of subarrays times the size of each subarray, resulting in a worst-case space complexity where the algorithm has to evaluate N choose (K-1) combinations and store each intermediate combination in some manner. This can be roughly approximated as O(N^K), where N is the input array size and K is the maximum number of subarrays. Therefore, the space complexity is polynomial with respect to N, given K.

Optimal Solution

Approach

The key to solving this problem efficiently is to recognize that we only need to focus on the starting elements of the subarrays. We want to minimize the cost, so we should always choose the smallest starting numbers to represent our subarrays.

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

First, sort the original list of numbers from smallest to largest.
Determine the number of subarrays we need to create, based on the problem's requirements.
Select the smallest numbers from the sorted list. The number of smallest numbers to select is equal to the required number of subarrays.
Calculate the cost. The cost is the sum of the selected smallest numbers. This sum represents the minimum cost required to divide the array.
Return the calculated cost.

Code Implementation

def divide_array_into_subarrays(numbers, subarray_count):
    numbers.sort()

    # Select the smallest elements for minimum cost
    smallest_numbers = numbers[:subarray_count]

    # Calculate the cost, summing the smallest numbers.
    total_cost = sum(smallest_numbers)

    return total_cost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is sorting the input array of size n. Common sorting algorithms like merge sort or quicksort have a time complexity of O(n log n). Selecting the k smallest elements after sorting takes O(k) time, where k is the number of subarrays, which is less than n, so does not change the overall complexity. Calculating the sum of these k elements also takes O(k) time. Therefore, the overall time complexity is dominated by the sorting step, which is O(n log n).

Space Complexity

O(N) – The dominant space usage comes from sorting the original list of numbers. Most sorting algorithms, like merge sort or quicksort (in the worst case), require auxiliary space. In the worst-case scenario, this sorting process requires a temporary array of size N, where N is the number of elements in the input array, to facilitate the sorting. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return infinity or throw an exception since a valid partition is impossible.
k is larger than the size of the input array	Return infinity or throw an exception as it is impossible to divide the array into more subarrays than elements.
k is equal to 1	The cost is simply the first element of the array.
Array contains negative numbers	The algorithm should handle negative numbers correctly, as they can affect the minimum cost.
Array contains only zeros	The algorithm should handle arrays with only zeros without issues, with the result being zero if k is valid.
Integer overflow potential when summing costs	Use a larger data type (e.g., long) to store the sum of the costs to prevent integer overflow.
Array sorted in ascending order	If k > 1, the first k elements will be optimal division points.
Large input array size for dynamic programming approach	Consider optimizing space complexity of dynamic programming solution to avoid exceeding memory limits.