Subarrays Distinct Element Sum of Squares II

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ArraysSliding Windows

You are given a 0-indexed integer array nums.

The distinct count of a subarray of nums is defined as:

Let nums[i..j] be a subarray of nums consisting of all the indices from i to j such that 0 <= i <= j < nums.length. Then the number of distinct values in nums[i..j] is called the distinct count of nums[i..j].

Return the sum of the squares of distinct counts of all subarrays of nums.

Since the answer may be very large, return it modulo 10⁹ + 7.

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

Example 1:

Input: nums = [1,2,1]
Output: 15
Explanation: Six possible subarrays are:
[1]: 1 distinct value
[2]: 1 distinct value
[1]: 1 distinct value
[1,2]: 2 distinct values
[2,1]: 2 distinct values
[1,2,1]: 2 distinct values
The sum of the squares of the distinct counts in all subarrays is equal to 1² + 1² + 1² + 2² + 2² + 2² = 15.

Example 2:

Input: nums = [2,2]
Output: 3
Explanation: Three possible subarrays are:
[2]: 1 distinct value
[2]: 1 distinct value
[2,2]: 1 distinct value
The sum of the squares of the distinct counts in all subarrays is equal to 1² + 1² + 1² = 3.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

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 maximum size of the input array `nums`? What is the range of values within `nums`? Are they all non-negative integers?
By 'distinct element', do you mean unique element within a given subarray, or across the entire input array?
If a subarray is empty, should I consider its sum of squares to be 0?
Are there any specific constraints on the data type to be used for calculating the sum of squares to avoid overflow issues with large sums or values?
Could you provide an example of the expected output given a small input array?

Brute Force Solution

Approach

The brute force approach to this problem is all about checking every single possibility. We're going to look at every possible chunk of numbers within the larger list, figure out what's unique in each chunk, and then do some math to see if it's the biggest result we've found so far.

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

Consider every possible starting point for a chunk of numbers within the main list.
For each starting point, consider every possible ending point, creating a chunk of numbers.
Within that chunk, identify the unique numbers.
For each unique number, multiply it by itself (square it).
Add up all the squared unique numbers from that chunk.
Compare the sum you just calculated to the largest sum you've found so far. If it's bigger, remember it.
Once you've considered all possible chunks, the biggest sum you've remembered is the answer.

Code Implementation

def subarrays_distinct_element_sum_of_squares_brute_force(numbers):
    maximum_sum_of_squares = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            current_subarray = numbers[start_index:end_index + 1]

            # Find the distinct numbers in the subarray
            distinct_numbers = set(current_subarray)

            sum_of_squares = 0

            # Calculate the sum of squares
            for number in distinct_numbers:
                sum_of_squares += number * number

            # Update the maximum if necessary.
            if sum_of_squares > maximum_sum_of_squares:

                maximum_sum_of_squares = sum_of_squares

    return maximum_sum_of_squares

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarrays using nested loops, where the outer loop iterates from 0 to n-1 and the inner loop iterates from the current outer loop index to n-1. For each subarray, it identifies the unique elements, which, in the worst case, could take O(n) time if all elements in the subarray are distinct. Consequently, the overall time complexity becomes O(n * n * n) = O(n^3).

Space Complexity

O(N) – The brute force approach iterates through all possible subarrays of the input list. Inside the inner loop, for each subarray, we identify the unique numbers. To accomplish this, we might use a set or a dictionary to store the elements encountered in the subarray. In the worst case, all elements in the subarray are distinct, resulting in a set/dictionary of size proportional to the length of the subarray, which could be at most N, where N is the size of the original input list. Therefore, the auxiliary space used by this approach is O(N).

Optimal Solution

Approach

The key is to efficiently maintain the count of each unique number within all the possible sub-sections. We do this by cleverly tracking how these counts change as we grow and shrink these sub-sections. This allows us to quickly calculate the sum of squares for each sub-section without recomputing from scratch every time.

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

Start by going through each number one by one, treating each one as the beginning of a potential sub-section.
For each starting number, expand the sub-section to the right, keeping track of how many times each number appears within that sub-section.
As you expand, update a running total of the sum of squares of the distinct numbers in the current sub-section.
Once you have the sum of squares for the expanding sub-section, shrink it from the left, removing numbers and updating the counts and sum of squares accordingly. Do this as long as the starting number is the same.
The trick is to efficiently update the counts and sum of squares by only considering the numbers that are entering or leaving the sub-section as it grows and shrinks, instead of recalculating for the entire sub-section each time.
Since we're only looking for the *distinct* numbers, keeping track of each numbers count within the subarray, makes this a sliding window problem.
Continue this process for each starting number, making sure to handle duplicate starting numbers correctly to avoid redundant calculations.

Code Implementation

def sub_arrays_distinct_element_sum_of_squares_ii(numbers):
    total_sum_of_squares = 0
    number_of_elements = len(numbers)

    for start_index in range(number_of_elements):
        element_counts = {}
        distinct_elements_sum_of_squares = 0

        for end_index in range(start_index, number_of_elements):
            current_number = numbers[end_index]

            if current_number not in element_counts:
                element_counts[current_number] = 0

            element_counts[current_number] += 1

            # Only update the sum if the number appeared for the first time.
            if element_counts[current_number] == 1:
                distinct_elements_sum_of_squares += current_number * current_number

            total_sum_of_squares += distinct_elements_sum_of_squares

    return total_sum_of_squares

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates through each of the n elements of the input array. The inner loop expands the subarray to the right, potentially iterating through the remaining elements in the worst case. The shrinking of the subarray also takes at most n steps for each starting position. Thus, in the worst-case scenario where each starting number is unique, we have nested loops each running up to n times resulting in O(n²) time complexity.

Space Complexity

O(N) – The plain English explanation describes keeping track of how many times each number appears within a subsection. This implies using a data structure, such as a hash map (or dictionary), to store the counts of distinct numbers. In the worst-case scenario, all N numbers in the input array are distinct, so the hash map could potentially store counts for all N numbers. Therefore, the auxiliary space used is proportional to the number of distinct elements in the input array, which is at most N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there are no subarrays.
Array with a single element	Return the square of that element.
Array with all identical elements	The distinct element count in each subarray will be 1, so sum squares accordingly.
Array with very large numbers that can lead to integer overflow when squared	Use 64-bit integers (long long in C++, long in Java) to store the sum of squares.
Array with a large range of numbers (e.g., very small to very large)	The algorithm should efficiently handle the range without causing memory issues (consider using a hash table).
Maximum sized input array (constraints limit, e.g., n = 10^5)	Ensure the solution's time complexity is at most O(n^2) or better to avoid exceeding time limit, potentially using a sliding window approach.
Array containing negative numbers	The solution should correctly handle negative numbers as they contribute to the distinct count and their squares are positive.
Array containing zeros	Zeros should be treated as any other number and included in the distinct count if they appear in a subarray.

Empty input array

How to Handle:

Return 0 immediately as there are no subarrays.

Array with a single element

How to Handle:

Return the square of that element.

Array with all identical elements

How to Handle:

The distinct element count in each subarray will be 1, so sum squares accordingly.

Array with very large numbers that can lead to integer overflow when squared

How to Handle:

Use 64-bit integers (long long in C++, long in Java) to store the sum of squares.

Array with a large range of numbers (e.g., very small to very large)

How to Handle:

The algorithm should efficiently handle the range without causing memory issues (consider using a hash table).

Maximum sized input array (constraints limit, e.g., n = 10^5)

How to Handle:

Ensure the solution's time complexity is at most O(n^2) or better to avoid exceeding time limit, potentially using a sliding window approach.

Array containing negative numbers

How to Handle:

The solution should correctly handle negative numbers as they contribute to the distinct count and their squares are positive.

Array containing zeros

How to Handle:

Zeros should be treated as any other number and included in the distinct count if they appear in a subarray.