How do you solve Mean of Array After Removing Some Elements on LeetCode?

Mean of Array After Removing Some Elements (LeetCode #1619) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting is essential to identify the smallest and largest elements.

What is the brute force solution for Mean of Array After Removing Some Elements?

The brute force approach for Mean of Array After Removing Some Elements is Brute Force, with O(n log n) time complexity and O(1) space complexity. The brute force approach involves sorting the array and then manually removing the smallest and largest 5% of the elements. After that, we calculate the mean of the remaining elements. This is straightforward but not efficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Mean of Array After Removing Some Elements?

Mean of Array After Removing Some Elements uses the following concepts: Array, Sorting. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Mean of Array After Removing Some Elements?

Always clarify edge cases, such as very small arrays. Explain your thought process while coding to demonstrate understanding. Practice explaining your code to ensure you can communicate effectively in an interview.

What are common mistakes when solving Mean of Array After Removing Some Elements?

Not sorting the array before removing elements. Failing to correctly calculate the number of elements to remove.

#1619

Mean of Array After Removing Some Elements

Easy

ArraySortingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting but focuses on calculating the mean directly after determining the indices of the elements to remove. This avoids unnecessary storage and calculations.

⚙️

Algorithm

4 steps

1Step 1: Sort the array.
2Step 2: Calculate the number of elements to remove (5% of the length of the array).
3Step 3: Calculate the sum of the remaining elements directly using a loop.
4Step 4: Calculate the mean using the sum and the count of remaining elements.

solution.py11 lines

1# Full working Python code
2
3def mean_after_removal(arr):
4    arr.sort()
5    n = len(arr)
6    remove_count = n // 20  # 5% of the array
7    total_sum = sum(arr[remove_count:n-remove_count])
8    return total_sum / (n - 2 * remove_count)
9
10# Example usage
11print(mean_after_removal([1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]))

ℹ

Complexity note: The time complexity remains O(n log n) due to sorting, but we optimize the calculation of the mean by summing directly during the loop. The space complexity is O(1) as we are not using extra space proportional to input size.

1Sorting is essential to identify the smallest and largest elements.
2Removing a fixed percentage of elements simplifies the calculation of the mean.

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