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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.
1def find_closest_elements(arr, k, x):
2 left, right = 0, len(arr) - k
3 while left < right:
4 mid = (left + right) // 2
5 if x - arr[mid] > arr[mid + k] - x:
6 left = mid + 1
7 else:
8 right = mid
9 return arr[left:left + k]We initiate a binary search between 0 and len(arr) - k. Depending on whether the element x is closer to the left or right side, we move the binary search mid-point. Once the position is found, the array is sliced from left to left + k to get the result.
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).
1#include <vector>
2#include <cstdlib> // for abs
using namespace std;
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int left = 0, right = arr.size() - 1;
while (right - left >= k) {
if (abs(arr[left] - x) > abs(arr[right] - x))
left++;
else
right--;
}
return vector<int>(arr.begin() + left, arr.begin() + right + 1);
}Similar to Python, this C++ approach closes in from both ends to leave only k closest elements using a two-pointer mechanism.