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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).
1using System;
2using System.Collections.Generic;
public class Solution {
public IList<int> FindClosestElements(int[] arr, int k, int x) {
int left = 0, right = arr.Length - 1;
while (right - left >= k) {
if (Math.Abs(arr[left] - x) > Math.Abs(arr[right] - x))
left++;
else
right--;
}
List<int> result = new List<int>();
for (int i = left; i <= right; ++i)
result.Add(arr[i]);
return result;
}
}The C# solution is based on a two-pointer approach. The method traps indices until they limit the scope to k closest elements. It efficiently reduces used space.