How do you solve Subarrays Distinct Element Sum of Squares II on LeetCode?

Subarrays Distinct Element Sum of Squares II (LeetCode #2916) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency map allows for efficient counting of distinct elements.

What is the brute force solution for Subarrays Distinct Element Sum of Squares II?

The brute force approach for Subarrays Distinct Element Sum of Squares II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subarrays and counting the distinct elements in each. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subarrays Distinct Element Sum of Squares II?

Subarrays Distinct Element Sum of Squares II uses the following concepts: Array, Dynamic Programming, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Subarrays Distinct Element Sum of Squares II?

Explain your thought process clearly, especially when transitioning from brute force to optimal. Practice writing clean and efficient code, as readability is important in interviews. Be prepared to discuss the time and space complexities of your solution.

What are common mistakes when solving Subarrays Distinct Element Sum of Squares II?

Not handling the frequency map correctly, leading to incorrect distinct counts. Failing to consider the modulo operation, which can cause overflow.

#2916

Subarrays Distinct Element Sum of Squares II

Hard

ArrayDynamic ProgrammingBinary Indexed TreeSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages a sliding window approach combined with a frequency map to efficiently count distinct elements in subarrays. This reduces the need for nested loops and allows us to compute the result in linear time.

⚙️

Algorithm

5 steps

1Step 1: Initialize a frequency map to count occurrences of elements and variables to track the total sum and the current distinct count.
2Step 2: Use a single loop to iterate through the array, treating each index as the end of the current subarray.
3Step 3: For each new element added to the subarray, update the frequency map and adjust the distinct count accordingly.
4Step 4: For each position, calculate the sum of squares of distinct counts from the current subarray and add it to the total sum.
5Step 5: Return the total sum modulo 10^9 + 7.

solution.py31 lines

1# Full working Python code
2
3def sum_of_squares(nums):
4    MOD = 10**9 + 7
5    total_sum = 0
6    freq = {}
7    distinct_count = 0
8    n = len(nums)
9
10    left = 0
11    for right in range(n):
12        if nums[right] in freq:
13            freq[nums[right]] += 1
14        else:
15            freq[nums[right]] = 1
16            distinct_count += 1
17
18        total_sum += distinct_count ** 2
19        total_sum %= MOD
20
21        while left <= right and freq[nums[left]] > 1:
22            freq[nums[left]] -= 1
23            if freq[nums[left]] == 0:
24                del freq[nums[left]]
25                distinct_count -= 1
26            left += 1
27
28    return total_sum
29
30# Example usage
31print(sum_of_squares([1, 2, 1]))  # Output: 15

ℹ

Complexity note: The time complexity is O(n) because we process each element in the array once. The space complexity is O(n) due to the frequency map that stores counts of elements.

1Using a frequency map allows for efficient counting of distinct elements.
2Sliding window techniques can reduce the time complexity significantly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.