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

Subarrays Distinct Element Sum of Squares I (LeetCode #2913) 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: Understanding how to track distinct elements efficiently is crucial.

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

The brute force approach for Subarrays Distinct Element Sum of Squares I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subarrays and calculating the distinct counts for each. This is straightforward but inefficient for larger 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 I?

Subarrays Distinct Element Sum of Squares I uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

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

Always clarify the problem statement and constraints. Think about edge cases, such as arrays with all identical elements. Discuss your thought process and approach before jumping into coding.

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

Not considering all subarrays correctly. Failing to update the distinct count efficiently.

#2913

Subarrays Distinct Element Sum of Squares I

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach with a hashmap to efficiently track the frequency of elements in the current subarray, allowing us to calculate distinct counts without re-evaluating the entire subarray each time.

⚙️

Algorithm

4 steps

1Step 1: Initialize a hashmap to store the frequency of elements and two pointers for the sliding window.
2Step 2: Expand the right pointer to include new elements and update the frequency map.
3Step 3: For each new element added, calculate the distinct count and add its square to the total sum.
4Step 4: Move the left pointer to shrink the window and update the frequency map accordingly.

solution.py22 lines

1# Full working Python code
2
3def subarrayDistinctCountSumSquares(nums):
4    total_sum = 0
5    freq_map = {}
6    left = 0
7    n = len(nums)
8    for right in range(n):
9        freq_map[nums[right]] = freq_map.get(nums[right], 0) + 1
10        distinct_count = len(freq_map)
11        total_sum += distinct_count ** 2
12        while left <= right:
13            freq_map[nums[left]] -= 1
14            if freq_map[nums[left]] == 0:
15                del freq_map[nums[left]]
16            left += 1
17            distinct_count = len(freq_map)
18            total_sum += distinct_count ** 2
19    return total_sum
20
21# Example usage
22print(subarrayDistinctCountSumSquares([1, 2, 1]))  # Output: 15

ℹ

Complexity note: The time complexity is O(n) because we are using a sliding window to traverse the array only once, and the space complexity is O(n) due to the hashmap storing the frequency of elements.

1Understanding how to track distinct elements efficiently is crucial.
2Using a sliding window can significantly reduce time complexity.

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