Maximum Sum of Almost Unique Subarray

Medium

Amazon

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Topics:

ArraysSliding Windows

You are given an integer array nums and two positive integers m and k.

Return the maximum sum out of all almost unique subarrays of length k of nums. If no such subarray exists, return 0.

A subarray of nums is almost unique if it contains at least m distinct elements.

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

For example:

nums = [2,6,7,3,1,7], m = 3, k = 4

Output: 18

Explanation: There are 3 almost unique subarrays of size k = 4. These subarrays are [2, 6, 7, 3], [6, 7, 3, 1], and [7, 3, 1, 7]. Among these subarrays, the one with the maximum sum is [2, 6, 7, 3] which has a sum of 18.

As another example:

nums = [1,2,1,2,1,2,1], m = 3, k = 3

Output: 0

Explanation: There are no subarrays of size k = 3 that contain at least m = 3 distinct elements in the given array [1,2,1,2,1,2,1]. Therefore, no almost unique subarrays exist, and the maximum sum is 0.

Describe a solution to efficiently find the maximum sum, considering time and space complexity. Consider edge cases where no valid subarray exists. Provide code.

Maximum Sum of Almost Unique Subarray

Medium

Amazon

3 views

Topics:

ArraysSliding Windows

You are given an integer array nums and two positive integers m and k.

Return the maximum sum out of all almost unique subarrays of length k of nums. If no such subarray exists, return 0.

A subarray of nums is almost unique if it contains at least m distinct elements.

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

For example:

nums = [2,6,7,3,1,7], m = 3, k = 4

Output: 18

Explanation: There are 3 almost unique subarrays of size k = 4. These subarrays are [2, 6, 7, 3], [6, 7, 3, 1], and [7, 3, 1, 7]. Among these subarrays, the one with the maximum sum is [2, 6, 7, 3] which has a sum of 18.

As another example:

nums = [1,2,1,2,1,2,1], m = 3, k = 3

Output: 0

Explanation: There are no subarrays of size k = 3 that contain at least m = 3 distinct elements in the given array [1,2,1,2,1,2,1]. Therefore, no almost unique subarrays exist, and the maximum sum is 0.

Describe a solution to efficiently find the maximum sum, considering time and space complexity. Consider edge cases where no valid subarray exists. Provide code.

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 `nums` be negative, zero, or floating-point numbers?
If no almost unique subarray of size `k` exists, what should the function return?
What is the definition of 'almost unique' in this context? Is it 'at most one duplicate' within the subarray, or something else?
If there are multiple almost unique subarrays of size `k` with the same maximum sum, is it acceptable to return any one of them?

Brute Force Solution

Approach

The brute force approach to finding the maximum sum of an almost unique section is to check every possible section. We will consider every starting point and every ending point within the list of numbers. Then for each of these sections, we check if the section meets the 'almost unique' criteria.

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

Start by considering the first number in the list as a section by itself.
Then consider the first two numbers as a section, then the first three, and so on, until you've included the entire list in a section starting from the beginning.
Now, move to the second number in the list and do the same thing: consider the second number alone, then the second and third, then the second, third, and fourth, and so on, until you've included the entire list in a section starting from the second number.
Keep doing this, shifting the starting point one number at a time, until you've considered every number in the list as a possible starting point for a section.
For each of these sections you've created, check if it meets the 'almost unique' condition (meaning the number of distinct values is close to the section's length).
If a section is 'almost unique', calculate the sum of the numbers in that section.
Compare the sums of all the 'almost unique' sections you found, and keep track of the largest sum.
The largest sum you found is the maximum sum of an almost unique section.

Code Implementation

def find_max_almost_unique_subarray_sum(number_list):
    max_sum = 0
    list_length = len(number_list)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            subarray = number_list[start_index : end_index + 1]
            subarray_length = len(subarray)

            # Check if the subarray is almost unique.
            unique_elements = set(subarray)

            # We use length-1 to check for almost unique condition as described
            if len(unique_elements) >= subarray_length - 1:
                current_sum = sum(subarray)

                # Keep track of largest sum
                if current_sum > max_sum:
                    max_sum = current_sum

    return max_sum

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through all possible subarrays of the input array of size n. There are approximately n * (n+1) / 2 such subarrays, which is O(n²). For each subarray, the algorithm checks if it's almost unique, which requires iterating through the subarray to count distinct elements, taking O(n) time. Therefore, the overall time complexity is O(n²) * O(n) = O(n³).

Space Complexity

O(N) – The described brute force approach requires auxiliary space primarily for checking the 'almost unique' condition of each subarray. For each subarray, we would need to store the distinct elements, which in the worst case, could be all the elements in the subarray. Thus, we need a data structure (like a set or hash map) that could potentially store up to N elements for the longest subarray being checked, where N is the size of the input list. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The key is to efficiently maintain a 'window' of numbers and track unique elements within it. We slide this window along the sequence, adding and removing numbers to ensure almost uniqueness, and calculate the sum within the window.

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

Start with an empty window at the beginning of the sequence.
Expand the window to the right by adding numbers one by one.
Keep track of how many times each number appears inside the window.
If adding a number causes the window to have too many duplicate numbers, shrink the window from the left until the 'almost unique' condition is met again.
Each time the window satisfies the 'almost unique' condition, calculate the sum of the numbers within the window.
Remember the largest sum you've seen so far.
Continue sliding the window until you reach the end of the sequence.
The largest sum you remembered is the answer.

Code Implementation

def find_maximum_sum_of_almost_unique_subarray(sequence, maximum_allowed_duplicates):
    window_start = 0
    maximum_sum = 0
    current_sum = 0
    element_counts = {}

    for window_end in range(len(sequence)):
        right_element = sequence[window_end]

        element_counts[right_element] = element_counts.get(right_element, 0) + 1
        current_sum += right_element

        # Shrink the window if too many duplicates
        while element_counts[right_element] > maximum_allowed_duplicates:

            left_element = sequence[window_start]
            element_counts[left_element] -= 1

            # Reduce the current sum as the left element is removed
            current_sum -= left_element
            window_start += 1

        # Update maximum sum after ensuring almost unique condition
        maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n) – The outer loop iterates through the input sequence of size n, extending the window to the right. The inner loop shrinks the window from the left. Each element can be added to the window and removed from the window at most once. Therefore, the number of operations is bounded by a constant factor times n, resulting in O(n) time complexity.

Space Complexity

O(N) – The provided solution maintains a window and tracks the frequency of each number within it. This frequency tracking is typically done using a hash map or an array. In the worst-case scenario, all the numbers in the input sequence are unique, meaning the hash map or array needs to store information for each of the N numbers. Therefore, the auxiliary space required is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return 0 if the array is empty or null, as there's no subarray to consider.
k (subarray length) is larger than the array length	Return 0, as no subarray of length k can be formed.
k is equal to 1	Iterate through array and return the largest unique element, or 0 if none exist meeting uniqueness criteria.
all elements in the array are the same	Return 0 if k > 1, otherwise return the element if k == 1 and its frequency in array is 1.
large array with only a few unique elements	The sliding window and frequency counter might take longer if many elements are checked repeatedly, so ensure the frequency counter is efficient.
Input array contains negative numbers	The sum calculations and frequency counting should work correctly with negative numbers.
Input array contains 0	The presence of zero should not affect the algorithm as long as the uniqueness condition is met.
no subarray exists meeting the almost unique condition	Return 0 in case no subarray fulfills the almost unique criteria, ensuring the function has a valid return even with unfavorable input.