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This approach uses binary search to find the position where x would fit in the array or where it is located. From this point, a sliding window is expanded to the left and right to check k closest elements. Given the sorted nature of the array, this is efficient.
Time Complexity: O(log(n-k) + k) where n is the length of the array, and we are moving a window over it. Space Complexity: O(k) for storing the resulting subarray.
1#include <vector>
2#include <algorithm>
3using namespace std;
4
5vector<int> findClosestElements(vector<int>& arr, int k, int x) {
6 int left = 0, right = arr.size() - k;
7 while (left < right) {
8 int mid = (left + right) / 2;
9 if (x - arr[mid] > arr[mid + k] - x)
10 left = mid + 1;
11 else
12 right = mid;
13 }
14 return vector<int>(arr.begin() + left, arr.begin() + left + k);
15}Similar to Python's solution, the approach here uses C++ STL for a straightforward implementation. We perform a binary search to determine where to start collecting k elements, and use STL vector range to slice the elements.
This approach initiates two pointers from the start and end of the array and gradually closes inwards by comparing the distance from x from both ends until the closest k elements are left. This is linear after the search initialization.
Time Complexity: O(n). Space Complexity: O(k).
1import
Using the two-pointer method in Java, this technique narrows down the indices until k closest elements are retained. This method results in a quick and easy fix through an iterating loop.