# Solving Leetcode Interviews in Seconds with AI: Subarrays Distinct Element Sum of Squares II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2916" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's an efficient solution to calculate the sum of squares of distinct counts of all subarrays, along with a complexity analysis.

*   **High-Level Approach:** The core idea is to iterate through all possible subarrays of the input array `nums`. For each subarray, determine the count of distinct elements. Accumulate the square of this count into a running sum, taking the modulo at each step to prevent overflow. The distinct count is calculated efficiently using a `set` to avoid recounting duplicates within each subarray.

*   **Complexity:**
    *   Runtime: O(n<sup>2</sup>), where n is the length of the input array `nums`.
    *   Storage: O(n) in the worst case, due to the space used by the `distinct_elements` set.

	
	# Code
	```python
	def sum_of_squares_of_distinct_counts(nums):
    """
    Calculates the sum of the squares of distinct counts of all subarrays of nums.

    Args:
        nums: A 0-indexed integer array.

    Returns:
        The sum of the squares of distinct counts modulo 10^9 + 7.
    """

    n = len(nums)
    total_sum = 0
    modulo = 10**9 + 7

    for i in range(n):
        for j in range(i, n):
            subarray = nums[i : j+1]
            distinct_elements = set(subarray)
            distinct_count = len(distinct_elements)
            total_sum = (total_sum + (distinct_count * distinct_count)) % modulo

    return total_sum
	```