Is Subarrays Distinct Element Sum of Squares II easy or hard?

Subarrays Distinct Element Sum of Squares II is classified as a Hard problem. It requires understanding subarray contribution patterns, tracking last occurrences, and implementing advanced data structures such as Fenwick Trees or Segment Trees to achieve the optimal O(n log n) solution.

Subarrays Distinct Element Sum of Squares II Python/Java solution

Python and Java implementations typically use a Fenwick Tree or Segment Tree combined with a dictionary or array to track last occurrences. Each iteration updates contribution ranges and queries cumulative values to maintain the total sum of squared distinct counts.

How to solve Subarrays Distinct Element Sum of Squares II in O(n log n)?

Process the array from left to right while tracking the previous index of each value. When a value appears again, adjust the contribution range where it affects distinct counts. Use a Fenwick Tree or Segment Tree to apply range updates and compute prefix sums of squared distinct counts efficiently.

What is the best approach for Subarrays Distinct Element Sum of Squares II?

The most efficient solution uses a contribution-based technique with a Binary Indexed Tree (Fenwick Tree) or Segment Tree. While scanning the array, track the last occurrence of each element and update the range of subarrays whose distinct counts change. This reduces repeated work and achieves O(n log n) time complexity.

Is Subarrays Distinct Element Sum of Squares II asked at Google/Amazon/Meta?

Hard array and data structure problems involving segment trees or Fenwick trees frequently appear in interviews at companies like Google, Amazon, and Meta. This problem tests advanced subarray contribution techniques and efficient range updates, which are common interview patterns.

What data structure is used in Subarrays Distinct Element Sum of Squares II?

Efficient solutions rely on Binary Indexed Trees (Fenwick Trees) or Segment Trees to handle range updates and prefix queries. A hash map or array is also used to store the last occurrence index of each element while scanning the array.

What is the time complexity of Subarrays Distinct Element Sum of Squares II?

The brute force method runs in O(n^2) time because it enumerates every subarray and maintains a set of distinct elements. The optimized solution processes the array once and performs logarithmic range updates and queries using Fenwick or segment trees, resulting in O(n log n) time and O(n) space.

Description Hints 2 Solution Video 24:19

2916. Subarrays Distinct Element Sum of Squares II

Hard54.8% acceptance1 company

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

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

Consider the sum of the count of distinct values of subarrays ending with index i, let’s call it sum. Now if you need the sum of all subarrays ending with index i + 1 think how it can be related to sum and what extra will be needed to add to this.

Hint 2

You can find that extra sum using the segment tree.

24:19

Still stuck?Read 2 approaches, or watch a walkthrough.

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1class Solution:
2    def sumCounts(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,2,1]

Input

[2,2]

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