Maximum Sum of Distinct Subarrays With Length K

Medium

Nvidia

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Topics:

ArraysSliding Windows

You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:

The length of the subarray is k, and
All the elements of the subarray are distinct.

Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums with length 3 are:
- [1,5,4] which meets the requirements and has a sum of 10.
- [5,4,2] which meets the requirements and has a sum of 11.
- [4,2,9] which meets the requirements and has a sum of 15.
- [2,9,9] which does not meet the requirements because the element 9 is repeated.
- [9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions

Example 2:

Input: nums = [4,4,4], k = 3
Output: 0
Explanation: The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated.
We return 0 because no subarrays meet the conditions.

Constraints:

1 <= k <= nums.length <= 10^5
1 <= nums[i] <= 10^5

Maximum Sum of Distinct Subarrays With Length K

Medium

Nvidia

2 views

Topics:

ArraysSliding Windows

You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:

The length of the subarray is k, and
All the elements of the subarray are distinct.

Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums with length 3 are:
- [1,5,4] which meets the requirements and has a sum of 10.
- [5,4,2] which meets the requirements and has a sum of 11.
- [4,2,9] which meets the requirements and has a sum of 15.
- [2,9,9] which does not meet the requirements because the element 9 is repeated.
- [9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions

Example 2:

Input: nums = [4,4,4], k = 3
Output: 0
Explanation: The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated.
We return 0 because no subarrays meet the conditions.

Constraints:

1 <= k <= nums.length <= 10^5
1 <= nums[i] <= 10^5

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`? Are they small enough to allow for simpler solutions, or should I optimize for larger inputs?
Can the input array `nums` contain negative integers, zero, or only positive integers?
If no subarray of length `k` exists with all distinct elements, what value should I return?
Are the elements in the input array `nums` guaranteed to be integers, or could they be floating-point numbers?
In the problem statement, 'distinct' refers to unique values within the subarray of length `k`, is my understanding correct?

Brute Force Solution

Approach

The brute force approach for this problem is all about trying out every possible subarray of the required length. For each of these subarrays, we check if all the numbers are unique and if they are, we calculate the sum. Finally, we find the maximum sum among all valid subarrays.

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

First, consider the first 'chunk' of numbers that has the required length.
Check if all the numbers in this chunk are different from each other. If not, move on to the next chunk.
If all the numbers in the chunk *are* different, add them up to get a sum.
Remember this sum. It might be the biggest one we've seen so far.
Now, slide the chunk over by one number, and look at the next set of numbers of the required length.
Again, check if all numbers in this new chunk are different.
If they are, add them up and compare that sum to the sum you remembered earlier.
If the new sum is bigger, replace the remembered sum with this new bigger sum.
Keep sliding the chunk and repeating the process until you've checked every possible chunk of the required length in the original set of numbers.
After checking all the chunks, the sum you're remembering will be the biggest sum of a chunk with unique numbers that had the required length.

Code Implementation

def maximum_unique_subarray_brute_force(numbers, subarray_length):
    maximum_sum = 0
    
    # Iterate through all possible starting positions of the subarray
    for i in range(len(numbers) - subarray_length + 1):
        current_subarray = numbers[i:i + subarray_length]

        # Check for distinct elements
        if len(set(current_subarray)) == subarray_length:

            current_sum = sum(current_subarray)

            # Keep track of the maximum sum
            if current_sum > maximum_sum:
                maximum_sum = current_sum

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm iterates through all possible subarrays of length k within the input array of size n. For each subarray of length k, it checks for distinct elements which takes O(k) time in the worst case. Since there are approximately n such subarrays to check, the overall time complexity is O(n*k). Specifically, to check distinct elements, we may use a set that in the worst case goes up to a size of k.

Space Complexity

O(K) – The provided solution implicitly uses a sliding window approach where, in each iteration, it checks if the current subarray of length K contains distinct elements. To determine the uniqueness of the elements in the subarray, a hash set (or similar data structure) of size at most K is required to store the elements of the current subarray being examined. Therefore, the auxiliary space used depends on K, the length of the subarray. Consequently, the space complexity is O(K).

Optimal Solution

Approach

The trick to solving this problem efficiently is to avoid recomputing the sum of every subarray. Instead, we cleverly maintain a 'window' of numbers and update its sum as we slide it along. We also use a tool to quickly check if the numbers within our window are all distinct.

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

Imagine you have a magnifying glass that can see K numbers at a time in the larger list.
Start by positioning the magnifying glass at the very beginning of the list and calculate the sum of the numbers you see.
As you move the magnifying glass one position to the right, you are essentially adding one new number to the window and removing one old number from the window.
Instead of recalculating the entire sum each time you move, simply subtract the number that is leaving the window and add the number that is entering the window to the current sum.
Additionally, before moving the magnifying glass, quickly check if all the numbers currently within the window are different from each other. A technique to do this efficiently is to track the occurrences of each number. If any number appears more than once in the window, then skip the sum.
Keep track of the largest distinct sum you have seen so far.
Continue moving the magnifying glass to the right until you have covered the entire list of numbers. The largest distinct sum you tracked is your final answer.

Code Implementation

def max_sum_distinct_subarray(number_list, subarray_length):
    maximum_sum = 0
    current_sum = 0
    frequency_map = {}

    # Calculate initial window sum.
    for i in range(subarray_length):
        current_sum += number_list[i]
        frequency_map[number_list[i]] = frequency_map.get(number_list[i], 0) + 1

    # Only evaluate if initial window has distinct elements
    if len(frequency_map) == subarray_length:
        maximum_sum = current_sum

    # Slide window and update the sum and character frequency.
    for i in range(subarray_length, len(number_list)): 

        element_to_add = number_list[i]
        element_to_remove = number_list[i - subarray_length]

        current_sum += element_to_add - element_to_remove

        frequency_map[element_to_add] = frequency_map.get(element_to_add, 0) + 1
        frequency_map[element_to_remove] -= 1

        if frequency_map[element_to_remove] == 0:
            del frequency_map[element_to_remove]

        # Only consider new sum if all elements in the window are distinct.
        if len(frequency_map) == subarray_length:

            maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n) – We iterate through the input array of size n using a sliding window of size k. For each position of the window, we add the next element and remove the preceding element from the sum. Maintaining a count of elements in the window requires constant time operations. Therefore, we perform a constant amount of work for each of the n window positions, leading to a linear time complexity of O(n).

Space Complexity

O(K) – The auxiliary space is dominated by the need to track the occurrences of each number within the window of size K. This is typically done using a hash map or an array to store the frequency of each number in the window. The size of this data structure is proportional to K, the size of the window. Therefore, the space complexity is O(K).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no subarray exists.
k is larger than the array size	Return 0 since no valid subarray of length k can be formed.
Array size is equal to k, but contains duplicates	Return 0 since there is no valid subarray.
All elements in the array are the same	Return 0 if k > 1, otherwise return the value of the single element if k == 1.
Array contains negative numbers	The sliding window and distinct element check should handle negative numbers correctly during sum calculation and existence evaluation.
Integer overflow of sum calculation	Use a larger data type (long) for the sum to avoid overflow.
k is 1 and array contains duplicates	Iterate through the array and return the maximum number since each single element subarray is distinct.
Large input array to test time complexity	Sliding window with hash map ensures O(n) time complexity for a large array.