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Subarrays Distinct Element Sum of Squares II

Difficulty: Hard

Problem Description

You are given a 0-indexed integer array nums. The distinct count of a subarray of nums is defined as the number of distinct values in that subarray. Your task is to return the sum of the squares of the distinct counts of all subarrays of nums, modulo 10^9 + 7.

Key Insights

  • The problem involves calculating the distinct elements in all possible subarrays, which can be computationally intensive if approached naively.
  • Efficiently maintaining the distinct count while iterating through subarrays can be achieved using techniques like the sliding window or two-pointer approach.
  • The result needs to be calculated modulo 10^9 + 7 to handle large numbers.

Space and Time Complexity

Time Complexity: O(n^2) in the naive approach, but can be optimized to O(n) using a sliding window technique.
Space Complexity: O(n) for storing counts of elements.

Solution

To solve the problem, we can use a sliding window approach along with a hash map (or a frequency array) to track the counts of distinct elements. The idea is to expand the window by moving the right pointer and contract it by moving the left pointer while keeping track of the distinct counts. For each valid subarray defined by the current window, we calculate the square of the distinct count and accumulate the results.

Code Solutions

def distinctElementSumOfSquares(nums):
    MOD = 10**9 + 7
    n = len(nums)
    left = 0
    count = {}
    sum_of_squares = 0

    for right in range(n):
        # Add the current element to the count
        if nums[right] in count:
            count[nums[right]] += 1
        else:
            count[nums[right]] = 1

        # Count the distinct elements in the current window
        distinct_count = len(count)

        # Calculate the sum of squares
        sum_of_squares += distinct_count * distinct_count
        sum_of_squares %= MOD

        # Move the left pointer to maintain the window
        while left <= right and count[nums[right]] > 1:
            count[nums[left]] -= 1
            if count[nums[left]] == 0:
                del count[nums[left]]
            left += 1

    return sum_of_squares
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