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General Rule of Thumb
Use right = mid - 1 when you are certain the current midpoint cannot be part of the solution.
Use right = mid when the current midpoint is a valid candidate and you want to keep it in the search space.
In this problem, the latter applies because mid is always a candidate for the starting index. This ensures we find the smallest valid left.

when a[mid] == a[mid+k-1]  which is right boundary , then we need to move towards left or right 
solution : to move towards left
by setting high = mid;

why right is set to mid , usually right = mid -1 correct??
ChatGPT said:
Good question! Let’s carefully analyze why right = mid (not mid - 1) is used in this particular binary search implementation.

Key Difference: This is a Range Search Problem
Standard Binary Search
In a typical binary search for an exact element:

When the mid is not the desired value, we exclude it from the search space:
If the target is less than arr[mid], set right = mid - 1.
If the target is greater than arr[mid], set left = mid + 1.
Range Search: Why right = mid
Here, we are looking for a starting index left for a range of k elements, not for a specific element. When we update right = mid:

Inclusion: We keep mid in the search space because it is still a valid candidate for the starting index.
Narrowing the Range:
If mid could be the correct starting point, shrinking the search range by setting right = mid ensures we do not exclude potential valid candidates prematurely.
Moving right to mid - 1 might accidentally skip valid solutions.

Why "Smallest" Index?
The array is sorted. Among multiple subarrays that could satisfy the conditions:

Smaller values of left are preferred when there's a tie.
This ensures that the resulting subarray remains sorted naturally in ascending order.

import java.util.*;

public class KClosestElements {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int left = 0;
        int right = arr.length - k;

        // Binary search for the left bound
        while (left < right) {
            int mid = left + (right - left) / 2;

            // Compare distances to determine the closer side
            if (Math.abs(arr[mid] - x) > Math.abs(arr[mid + k] - x)) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        // Collect the result from the final window
        List<Integer> result = new ArrayList<>();
        for (int i = left; i < left + k; i++) {
            result.add(arr[i]);
        }

        return result;
    }

    // Driver Method
    public static void main(String[] args) {
        KClosestElements obj = new KClosestElements();

        int[] arr1 = {1, 2, 3, 4, 5};
        int k1 = 4, x1 = 3;
        System.out.println(obj.findClosestElements(arr1, k1, x1)); // Output: [1, 2, 3, 4]

        int[] arr2 = {1, 1, 2, 3, 4, 5};
        int k2 = 4, x2 = -1;
        System.out.println(obj.findClosestElements(arr2, k2, x2)); // Output: [1, 1, 2, 3]
    }
}