Find K Closest Elements

class Solution {
    public List<Integer> findKClosestElements(int[] numbers, int count, int target) {
        int start = 0;
        int end = numbers.length - count;

        while (start < end) {
            int center = (start + end) / 2;
            if (target - numbers[center] > numbers[center + count] - target) {
                start = center + 1;
            } else {
                end = center;
            }
        }

        List<Integer> closestElements = new ArrayList<Integer>();
        for (int index = start; index < start + count; index++) {
            closestElements.add(numbers[index]);
        }

        return closestElements;
    }
}

In this Java-based solution, you define a method to find the k closest elements to a given target from a sorted array of integers. The approach used here is efficiently based on a binary search to minimize the time complexity, typically optimizing it to O(log(N-k) + k), where N is the total number of elements in the input array.

The function findKClosestElements takes three parameters:

• numbers: The sorted array of integers.
• count: The number k representing how many closest elements to find.
• target: The target number to which the closest elements are sought.

The implementation initiates by determining the plausible search space using two pointers, start and end, where end is initialized as numbers.length - count to ensure the group of k elements stays within array bounds.

• Employ a while loop running as long as start is less than end. The midpoint, center, between start and end is calculated.

• Compare the absolute differences between the target and the elements at the positions center and center + count to decide the search direction. This helps in narrowing down the region where the closest elements would exist:

  • If the element at center is farther away from the target than the element at position center + count, move the start to center + 1.
  • Otherwise, adjust end to center.

Once the closest interval is localized, populate a list, closestElements, with count numbers from the array starting from the start position.

This method efficiently locates and returns the k closest integers to the target in the list, leveraging binary search for an optimal performance in scenarios where the input list is considerably large.

class Solution:
    def getClosestKElements(self, elements: List[int], count: int, target: int) -> List[int]:
        start = 0
        end = len(elements) - count

        while start < end:
            middle = (start + end) // 2
            if target - elements[middle] > elements[middle + count] - target:
                start = middle + 1
            else:
                end = middle

        return elements[start:start + count]

The Python solution detailed below addresses the task of fetching the k closest elements to a given target from a sorted array. This functionality is encapsulated within a class method, getClosestKElements, which utilizes a binary search approach to efficiently find the desired segment of the array.

• Define a class Solution which contains the method getClosestKElements. This method takes three parameters:

  • elements: A list of sorted integers.
  • count: The number of closest integers to the target you need to find (k).
  • target: The target number you want to find the closest elements to.

• Initialize two pointers, start and end. Set start to 0 and end to the difference between the number of elements and the count. This positions end such that the slice from start to start + count doesn't exceed the list boundaries.

• Implement a binary search:

  • Calculate the middle index of the current window as (start + end) // 2.
  • Compare the difference between target and the element at middle, to the difference between the element at middle + count and target.
  • Adjust the start or end pointer based on which side of the middle point is closer to the target. Specifically, if the left side is farther from the target than the right side, shift the start to middle + 1; otherwise, update end to middle.

• To complete the method, return the subarray starting from start and extending count elements.

This solution ensures that the calculation is both space and time-efficient, using binary search to minimize the iterations required to find the closest elements rather than scanning all possibilities. By aligning to the middle in each iteration, the algorithm effectively narrows down the range, concentrating on the part of the list most likely to contain the closest elements to the target.