1619. Mean of Array After Removing Some Elements
Problem Description
The problem requires writing a function that calculates the trimmed mean of an array of integers. To clarify, the trimmed mean is the average obtained after removing the smallest and the largest 5% of the elements from the list. By trimming these outliers, the mean becomes a more representative measure of the central tendency for the data set, particularly when extreme values might skew the average.
For the mean to be accepted, it should be close to the actual mean within a margin of 10^-5. This means the solution should be accurate enough up to the fifth decimal place.
Intuition
The process to achieve the trimmed mean can be broken down into the following steps:
-
Sorting: Sorting the array is a crucial first step. It's necessary because to easily identify and remove the smallest and largest 5% of elements, these elements need to be at the start and end of the list, respectively.
-
Calculating Indices: Next, we calculate the indices that represent the 5% mark from both the start and the end of the sorted array. This enables us to pinpoint where to slice the array to remove the unwanted elements. These indices are derived by multiplying the total length by 0.05 and 0.95, then converting the result to an integer. In Python, this truncates the decimal, effectively implementing a floor function and ensuring we get the correct slice for the desired percentages.
-
Slicing: After sorting the array and calculating the correct indices, the next step is to slice the array to remove the smallest and largest 5% of its elements. The sliced part is the range of elements we are interested in for calculating the trimmed mean.
-
Mean Calculation and Rounding: The final step involves calculating the mean of the remaining elements and rounding it to five decimal places as to conform with the problem's requirement for precision.
Note that the solution's accuracy hinges on proper rounding, and the round() function in Python serves to round the final result to the required number of decimal places.
This approach is intuitive and efficient, taking advantage of Python's built-in functions for sorting and arithmetic operations.
Learn more about Sorting patterns.
What is the space complexity of the following code?
1int sum(int n) {
2 if (n <= 0) {
3 return 0;
4 }
5 return n + sum(n - 1);
6}
Solution Approach
The implementation of the solution uses a straightforward approach leveraging Python's list and built-in functions. Here's a breakdown of the key components of the solution:
-
Sorting (
arr.sort()): The first major step is to sort the array using the list's.sort()method. Sorting reorders the elements from the lowest to the highest values, which is essential for trimming the smallest and largest elements efficiently. -
Calculating Indices (
int(n * 0.05),int(n * 0.95)): Then * 0.05andn * 0.95calculations determine the indices for slicing the array. Givennis the total number of elements in the array, multiplying it by0.05gives the number of elements that represents the bottom 5%, and multiplying by0.95gives us the index just past the top 95% (effectively the top 5% mark). Casting the result to anintensures we're working with whole index numbers. -
Slicing (
arr[start:end]): With the start and end indices now established, we can slice the sorted array. The slicing operationarr[start:end]removes the first 5% and the last 5% of the array, which are the smallest and largest values, respectively. -
Mean Calculation (
sum(t) / len(t)) and Rounding (round(…, 5)): The trimmed arraytis then passed to thesum()function, which adds up all the remaining values. This sum is then divided bylen(t), the count of elements in the trimmed array, to get the mean. Finally, the mean is rounded to five decimal places using theround()function to satisfy the precision requirement of the problem.
The choice to use Python's list sorting and slicing is based on their efficiency and ease of use. Sorting takes O(n log n) time complexity and slicing has O(k) complexity, where k is the number of elements to be copied (the size of the trimmed array in this case). The remaining operations (calculating the sum, finding the length of the list, and dividing) have linear time complexities, resulting in an overall efficient solution.
This solution uses no additional data structures, making it space-efficient as well, with the primary space usage being the input array and the trimmed slice.
Given a sorted array of integers and an integer called target, find the element that
equals to the target and return its index. Select the correct code that fills the
___ in the given code snippet.
1def binary_search(arr, target):
2 left, right = 0, len(arr) - 1
3 while left ___ right:
4 mid = (left + right) // 2
5 if arr[mid] == target:
6 return mid
7 if arr[mid] < target:
8 ___ = mid + 1
9 else:
10 ___ = mid - 1
11 return -1
121public static int binarySearch(int[] arr, int target) {
2 int left = 0;
3 int right = arr.length - 1;
4
5 while (left ___ right) {
6 int mid = left + (right - left) / 2;
7 if (arr[mid] == target) return mid;
8 if (arr[mid] < target) {
9 ___ = mid + 1;
10 } else {
11 ___ = mid - 1;
12 }
13 }
14 return -1;
15}
161function binarySearch(arr, target) {
2 let left = 0;
3 let right = arr.length - 1;
4
5 while (left ___ right) {
6 let mid = left + Math.trunc((right - left) / 2);
7 if (arr[mid] == target) return mid;
8 if (arr[mid] < target) {
9 ___ = mid + 1;
10 } else {
11 ___ = mid - 1;
12 }
13 }
14 return -1;
15}
16Example Walkthrough
Let's use a small example to illustrate the solution approach. Suppose we are given the following array of integers:
1arr = [4, 8, 6, 5, 3, 2, 7, 9, 1, 0]
Our goal is to calculate the trimmed mean after removing the smallest and largest 5% of the elements. Since our array has 10 elements, 5% of this array corresponds to a single element at each end (since 10 * 0.05 = 0.5, which we floor to 0 elements, and we can't remove less than one whole element).
Following the steps as per the solution approach:
-
Sorting (
arr.sort()): First, we sort the array to get:1arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]This puts the elements in ascending order.
-
Calculating Indices (
int(n * 0.05),int(n * 0.95)): For our 10-element array,n * 0.05is0.5. As an integer, it's rounded down to0, representing the first element to remove. The second index isn * 0.95, which is9.5, and as an integer, it gets rounded down to9. -
Slicing (
arr[start:end]): According to the calculated indices, we slice the sorted array from index1to index9:1t = arr[1:9] # Sliced array is [1, 2, 3, 4, 5, 6, 7, 8]This removes the smallest and largest elements,
0and9. -
Mean Calculation (
sum(t) / len(t)) and Rounding (round(…, 5)): We then calculate the sum of the trimmed arraytand divide by its length to get the mean:1mean = sum([1, 2, 3, 4, 5, 6, 7, 8]) / len([1, 2, 3, 4, 5, 6, 7, 8]) # mean is 4.5Then we round the mean to five decimal places:
1trimmed_mean = round(mean, 5) # trimmed_mean is 4.5, since there are no additional decimal places
Thus, for our example array, the trimmed mean after removing the smallest and largest 5% of the elements is 4.5. Please note that in cases where there are additional decimal places, the round() function will adjust the trimmed mean to five decimal places to meet the accuracy requirements.
Solution Implementation
1from typing import List
2
3class Solution:
4 def trimMean(self, arr: List[int]) -> float:
5 # Calculate the number of elements in the array
6 num_elements = len(arr)
7
8 # Determine the indices to trim 5% of elements from both ends
9 start_index = int(num_elements * 0.05)
10 end_index = int(num_elements * 0.95)
11
12 # Sort the array in non-decreasing order
13 arr.sort()
14
15 # Trim 5% of elements from each end of the sorted array
16 trimmed_arr = arr[start_index:end_index]
17
18 # Calculate the mean of the trimmed array
19 trimmed_mean = sum(trimmed_arr) / len(trimmed_arr)
20
21 # Round the mean to five decimal places and return it
22 return round(trimmed_mean, 5)
231import java.util.Arrays; // Import Arrays class from the java.util package for sorting
2
3class Solution {
4
5 // Function to calculate the trimmed mean of an array after removing the smallest
6 // and largest 5% of elements
7 public double trimMean(int[] arr) {
8 // Sort the input array
9 Arrays.sort(arr);
10
11 // Calculate the total number of elements in the array
12 int totalElements = arr.length;
13
14 // Calculate the number of elements to trim from each end (5% from both ends)
15 int elementsToTrim = (int) (totalElements * 0.05);
16
17 // Initialize the sum of the remaining elements
18 double sum = 0;
19
20 // Loop through the array from the first element after trimming
21 // to the last element before trimming
22 for (int index = elementsToTrim; index < totalElements - elementsToTrim; ++index) {
23 // Add the current element to the sum
24 sum += arr[index];
25 }
26
27 // Calculate the trimmed mean by dividing the sum of the remaining elements
28 // by the number of elements after trimming (which is 90% of the total)
29 return sum / (totalElements * 0.9);
30 }
31}
321#include <vector>
2#include <algorithm> // Include algorithm header for sorting
3
4class Solution {
5public:
6 // Function to calculate the trimmed mean of an array
7 double trimMean(vector<int>& arr) {
8 // Sort the array in non-decreasing order
9 sort(arr.begin(), arr.end());
10
11 int numElements = arr.size(); // Total number of elements in the array
12 double sum = 0; // Initialize sum to store the sum of the elements
13
14 // Calculate the starting index after trimming 5% from the front
15 int startIndex = static_cast<int>(numElements * 0.05);
16 // Calculate the ending index before trimming 5% from the back
17 int endIndex = numElements - startIndex;
18
19 // Loop through the array excluding the trimmed 5% from both ends
20 for (int i = startIndex; i < endIndex; ++i) {
21 sum += arr[i]; // Add the current element to the sum
22 }
23
24 // Calculate the trimmed mean by dividing sum by the number of elements after trimming
25 double trimmedMean = sum / (numElements * 0.9);
26
27 return trimmedMean; // Return the calculated trimmed mean
28 }
29};
301/**
2 * Calculates the trimmed mean of an array after removing the smallest and
3 * largest 5% of elements.
4 *
5 * @param {number[]} arr - The array of numbers from which to calculate the mean.
6 * @return {number} - The trimmed mean of the array.
7 */
8function trimMean(arr: number[]): number {
9 // Sort the array in ascending order.
10 arr.sort((a, b) => a - b);
11
12 // Calculate the number of elements to remove from each end of the array.
13 let length = arr.length;
14 let removeLength = Math.floor(length * 0.05);
15
16 // Sum the array elements while excluding the top and bottom 5% of elements.
17 let sum = 0;
18 for (let i = removeLength; i < length - removeLength; i++) {
19 sum += arr[i];
20 }
21
22 // Calculate the trimmed mean by dividing the sum by the number of elements included in the calculation.
23 return sum / (length - 2 * removeLength);
24}
25What is the best way of checking if an element exists in a sorted array once in terms of time complexity? Select the best that applies.
Time and Space Complexity
Time Complexity
The time complexity of the given code is mainly determined by the sorting function. The sort() function in Python uses the Timsort algorithm, which has a time complexity of O(n log n) for sorting an array.
Here's the breakdown:
- Sorting the array:
O(n log n) - Slicing the sorted array to remove the 5% of the elements from both ends:
O(n).
Therefore, the overall time complexity is O(n log n + n). Since O(n log n) is the dominating term, the time complexity simplifies to O(n log n).
Space Complexity
As for the space complexity:
- The array
tthat stores the sliced part of the original array introduces additional space. The space taken bytis proportional to the length of the slice, which is 90% of the original array, soO(n). - The
sort()function may requireO(n)space to perform the sorting.
The overall space complexity is O(n) since both the additional array and the sorting space complexity are linear with the input size.
Learn more about how to find time and space complexity quickly using problem constraints.
Which algorithm is best for finding the shortest distance between two points in an unweighted graph?
Recommended Readings
Sorting Summary Comparisons We presented quite a few sorting algorithms and it is essential to know the advantages and disadvantages of each one The basic algorithms are easy to visualize and easy to learn for beginner programmers because of their simplicity As such they will suffice if you don't know any advanced
Top Patterns to Conquer the Technical Coding Interview Should the written word bore you fear not A delightful video alternative awaits iframe width 560 height 315 src https www youtube com embed LW8Io6IPYHw title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture
Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first
Got a question? Ask the Teaching Assistant anything you don't understand.
Still not clear? Ask in the Forum, Discord or Submit the part you don't understand to our editors.