Maximum Strength of K Disjoint Subarrays

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ArraysDynamic ProgrammingSliding Windows

You are given an array of integers nums with length n, and a positive odd integer k.

Select exactly k disjoint subarrays sub₁, sub₂, ..., sub_k from nums such that the last element of sub_i appears before the first element of sub_{i+1} for all 1 <= i <= k-1. The goal is to maximize their combined strength.

The strength of the selected subarrays is defined as:

strength = k * sum(sub₁)- (k - 1) * sum(sub₂) + (k - 2) * sum(sub₃) - ... - 2 * sum(sub_{k-1}) + sum(sub_k)

where sum(sub_i) is the sum of the elements in the i-th subarray.

Return the maximum possible strength that can be obtained from selecting exactly k disjoint subarrays from nums.

Note that the chosen subarrays don't need to cover the entire array.

Example 1:

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

Output: 22

Explanation:

The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:

strength = 3 * (1 + 2 + 3) - 2 * (-1) + 2 = 22

Example 2:

Input: nums = [12,-2,-2,-2,-2], k = 5

Output: 64

Explanation:

The only possible way to select 5 disjoint subarrays is: nums[0..0], nums[1..1], nums[2..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:

strength = 5 * 12 - 4 * (-2) + 3 * (-2) - 2 * (-2) + (-2) = 64

Example 3:

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

Output: -1

Explanation:

The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.

Constraints:

1 <= n <= 10⁴
-10⁹ <= nums[i] <= 10⁹
1 <= k <= n
1 <= n * k <= 10⁶
k is odd.

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 is the expected range of values in the input array, and can the array contain negative numbers, zeros, or floating-point numbers?
What should be returned if `k` is greater than the number of elements in the array?
Are the subarrays required to be contiguous (i.e., consisting of consecutive elements)?
Is there any constraint on the minimum size of each subarray, or can a subarray consist of a single element?
If multiple sets of `k` disjoint subarrays result in the same maximum strength, is any one of them acceptable?

Brute Force Solution

Approach

The brute force method involves exploring all possible combinations of selecting K non-overlapping groups from the given numbers. We calculate a 'strength' for each group and then try to find the combination of K groups that gives us the highest overall strength. Basically, we try everything until we find the best.

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

First, think about all the ways you can pick the first group of numbers.
Then, after you've picked the first group, consider all the ways you can pick the second group from the remaining numbers.
Keep doing this until you've picked K groups. Remember that the groups can't overlap.
For each set of K groups you create, calculate its total 'strength' by adding up the strength of each individual group.
Keep track of the set of K groups that produces the highest total strength you've seen so far.
Once you've tried absolutely every possible combination of K groups, the highest strength you tracked is your answer.

Code Implementation

def maximum_strength_of_k_disjoint_subarrays_brute_force(numbers, k_subarrays):

    def calculate_strength(subarray):
        strength = 0
        for number in subarray:
            strength += number
        return strength

    def find_maximum_strength(current_index, current_subarrays):
        nonlocal maximum_total_strength

        # If we have found k subarrays, calculate and update max strength
        if len(current_subarrays) == k_subarrays:
            total_strength = 0
            for subarray in current_subarrays:
                total_strength += calculate_strength(subarray)
            maximum_total_strength = max(maximum_total_strength, total_strength)
            return

        # If we have reached the end of the array, there are no more subarrays to make, exit recursion.
        if current_index >= len(numbers):
            return

        # Explore the option of not including any subarray beginning at current_index
        find_maximum_strength(current_index + 1, current_subarrays)

        # Explore all possible subarrays that begin at current_index
        for end_index in range(current_index, len(numbers)): 
            subarray = numbers[current_index:end_index + 1]

            # Add the current subarray and proceed to find remaining subarrays
            new_subarrays = current_subarrays + [subarray]
            find_maximum_strength(end_index + 1, new_subarrays)

    maximum_total_strength = float('-inf')

    # Start the recursive process to explore all possible combinations.
    find_maximum_strength(0, [])

    return maximum_total_strength

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach iterates through all possible combinations of K disjoint subarrays within an array of size n. Selecting the first subarray has approximately n choices for its starting point. Given the first subarray, there are approximately n remaining choices for the starting point of the second subarray and this is repeated K times. Therefore, the algorithm explores roughly n * n * ... * n (K times) combinations, resulting in a time complexity of O(n^k).

Space Complexity

O(N^K) – The brute force approach explores all possible combinations of K disjoint subarrays. In the worst-case scenario, we might need to store information about each of these combinations, potentially including start and end indices for each subarray or copies of the subarrays themselves. The number of such combinations grows exponentially with K, where N is the size of the input array. Therefore, the space complexity is approximately proportional to N raised to the power of K, which is O(N^K).

Optimal Solution

Approach

The core idea is to build up the best possible result step-by-step. We avoid trying every single combination of subarrays by making optimal decisions at each stage, reusing previous results to avoid redundant calculations.

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

Imagine we're walking through the sequence of numbers, and we want to figure out the highest possible strength we can achieve using up to a certain number of subarrays at each point.
At each number, consider two choices: either include the number in a new subarray, or don't.
If we include the number in a new subarray, we need to figure out what the best possible strength we could achieve using one fewer subarray up to the point before the current subarray started. We then add the strength of the new subarray.
If we don't include the number in a new subarray, then the best possible strength is the same as the best possible strength we could achieve using the same number of subarrays up to the previous number.
To avoid recalculating the strength of subarrays repeatedly, we can precompute and store the sum of values in all possible subarrays. This is much faster than recomputing the sums every time we need them.
We'll build a table where each entry stores the maximum strength achievable using up to *k* subarrays ending at that number. We fill in the table using the choices described above, always picking the option that leads to the highest strength.
At the end, the last entry in the table (corresponding to the last number in the sequence and using all *k* subarrays) will give us the maximum strength possible.

Code Implementation

def max_strength_of_k_disjoint_subarrays(sequence, k_subarrays):
    sequence_length = len(sequence)

    #subarray_sums[i][j] stores the sum of sequence[i:j+1]
    subarray_sums = [[0] * sequence_length for _ in range(sequence_length)]
    for i in range(sequence_length):
        current_sum = 0
        for j in range(i, sequence_length):
            current_sum += sequence[j]
            subarray_sums[i][j] = current_sum

    # dp_table[i][j] is the max strength using up to i subarrays ending at index j
    dp_table = [[float('-inf')] * sequence_length for _ in range(k_subarrays + 1)]

    # Initialize: 0 subarrays have strength 0 for any ending index.
    for j in range(sequence_length):
        dp_table[0][j] = 0

    for number_of_subarrays in range(1, k_subarrays + 1):
        for current_index in range(sequence_length):
            # Option 1: Don't include sequence[current_index] in a subarray
            if current_index > 0:
                dp_table[number_of_subarrays][current_index] = dp_table[number_of_subarrays][current_index - 1]

            # Option 2: Include sequence[current_index] in a new subarray
            for start_index in range(current_index + 1):
                #Need to have at least 1 number to include
                current_subarray_strength = subarray_sums[start_index][current_index]
                if start_index > 0:
                    # The start index has to be greater than 0 in order to include the previous max value
                    dp_table[number_of_subarrays][current_index] = max(dp_table[number_of_subarrays][current_index],
                                                           dp_table[number_of_subarrays - 1][start_index - 1] + current_subarray_strength)
                else:
                    # Handle the case when start index is 0
                    dp_table[number_of_subarrays][current_index] = max(dp_table[number_of_subarrays][current_index], dp_table[number_of_subarrays - 1][0] + current_subarray_strength - subarray_sums[0][0] if number_of_subarrays>1 else current_subarray_strength)

    #The final result is the maximum strength using k_subarrays ending at the last index
    return dp_table[k_subarrays][sequence_length - 1]

Big(O) Analysis

Time Complexity

O(k*n^2) – The dominant operation is calculating the sums of all possible subarrays which takes O(n^2) time. The main dynamic programming loop iterates k times, where k is the number of subarrays. Inside this loop we iterate through the n elements of the array. For each of these n elements, in the worst case, we might look back at all the previous elements to determine the optimal start point of the last subarray. Therefore, this results in a time complexity of O(k*n^2), where k is the number of disjoint subarrays allowed and n is the length of the input array.

Space Complexity

O(N*K) – The solution uses a table to store the maximum strength achievable using up to k subarrays ending at each number. This table has dimensions dependent on both the input sequence length (N) and the maximum number of disjoint subarrays allowed (K), requiring N rows and K columns. Precomputing and storing the sum of values in all possible subarrays is mentioned, but since the table mentioned earlier already dominates the memory footprint, the space complexity is governed by the table's size. Therefore, the auxiliary space used is proportional to N multiplied by K, resulting in O(N*K) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as no subarrays can be formed from an empty array.
k is 0	Return 0, as no subarrays are needed.
k is greater than the number of positive elements in the input array	Return the sum of all positive numbers in the array, or 0 if there are no positive numbers.
Array contains only negative numbers or zero	Return 0, as the strength of any subarray containing negative numbers will be negative, and we want the maximum strength.
Array contains a mix of large positive and large negative numbers	Ensure that integer overflow is handled when calculating the sum of subarrays, possibly using long long integers or similar data types.
k is equal to the length of the array	Return the product of all positive numbers in the array (if any exist), or 0 if there are no positive numbers.
Array with only one element	If k=1 and the element is positive, return the element; otherwise, return 0.
Array contains many zero values	Skip the zeros when determining optimal subarrays, as they do not contribute to the product and can lead to incorrect results.

Null or empty input array

How to Handle:

Return 0, as no subarrays can be formed from an empty array.

k is 0

How to Handle:

Return 0, as no subarrays are needed.

k is greater than the number of positive elements in the input array

How to Handle:

Return the sum of all positive numbers in the array, or 0 if there are no positive numbers.

Array contains only negative numbers or zero

How to Handle:

Return 0, as the strength of any subarray containing negative numbers will be negative, and we want the maximum strength.

Array contains a mix of large positive and large negative numbers

How to Handle:

Ensure that integer overflow is handled when calculating the sum of subarrays, possibly using long long integers or similar data types.

k is equal to the length of the array

How to Handle:

Return the product of all positive numbers in the array (if any exist), or 0 if there are no positive numbers.

Array with only one element

How to Handle:

If k=1 and the element is positive, return the element; otherwise, return 0.

Array contains many zero values

How to Handle:

Skip the zeros when determining optimal subarrays, as they do not contribute to the product and can lead to incorrect results.