Given an integer array arr, return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.
Answers within 10-5 of the actual answer will be considered accepted.
Example 1:
Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3] Output: 2.00000 Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.
Example 2:
Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0] Output: 4.00000
Example 3:
Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4] Output: 4.77778
Constraints:
20 <= arr.length <= 1000arr.length is a multiple of 20.0 <= arr[i] <= 105The simplest way to tackle this problem is to first sort the array. Once sorted, it's easy to remove the smallest and largest 5% of the elements. After erasing these entries, compute the mean of the remaining numbers.
This C program sorts the array using qsort, then calculates the number of elements to remove (5% from both ends). It then calculates the sum of the middle elements (those not removed) and computes the mean.
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The time complexity is O(n log n) due to the sorting step, and the space complexity is O(1) since we are sorting in-place.
Another approach involves modifying partitioning techniques from Quickselect to find and remove the smallest and largest 5% of elements without fully sorting the array. This can reduce the average time complexity in certain scenarios compared to always sorting.
The C solution uses a variation of quickselect to partition the array around the smallest and largest 5% elements. It shifts elements only around the necessary indices and computes the mean of the remaining central segment.
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Average time complexity is O(n) for partitioning, though it degrades to O(n^2) in the worst case. Space complexity is O(1), as we modify the array in place.
| Approach | Complexity |
|---|---|
| Sorting Approach | The time complexity is O(n log n) due to the sorting step, and the space complexity is O(1) since we are sorting in-place. |
| Partition Approach Using Variation of Quickselect | Average time complexity is O(n) for partitioning, though it degrades to O(n^2) in the worst case. Space complexity is O(1), as we modify the array in place. |
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