Find K Closest Elements

Find K Closest Elements

What is this problem about?

The Find K Closest Elements interview question asks you to find the $k$ integers in a sorted array that are closest to a target value x. An integer $a$ is closer to $x$ than $b$ if $|a - x| < |b - x|$, or if the absolute differences are equal and $a < b$. The output must be sorted in ascending order.

Why is this asked in interviews?

This is a classic problem asked by Meta, Apple, and Microsoft. It tests your ability to leverage Sorted Array properties and optimize search space. While a simple sort by distance takes $O(n \log n)$, the array is already sorted, which hints at a more efficient $O(\log n + k)$ or $O(\log(n-k))$ solution. It evaluations your mastery of Binary Search and Two Pointers patterns.

Algorithmic pattern used

There are two main approaches:

1. Binary Search + Two Pointers ($O(\log n + k)$): Find the closest position to $x$ using binary search. Use two pointers to expand outward, picking the closer of the two available neighbors until you have $k$ elements.
2. Binary Search on Ranges ($O(\log(n-k))$): Binary search for the start of the valid window. The comparison logic is: x - arr[mid] > arr[mid+k] - x. If true, the window must start further to the right.

Example explanation

arr = [1, 2, 3, 4, 5], k = 4, x = 3

1. Initial window of size 4: [1, 2, 3, 4]. Start index 0.
2. Next potential window: [2, 3, 4, 5]. Start index 1.
3. Compare x - arr[0] (3-1=2) vs arr[0+4] - x (5-3=2). Since they are equal, we stay left (rule $a < b$). Result: [1, 2, 3, 4].

Common mistakes candidates make

• Sorting: Using Arrays.sort with a custom comparator, which is $O(n \log n)$ time and $O(n)$ space.
• Distance Ties: Not handling the $|a-x| == |b-x|$ case correctly by picking the smaller value.
• Heap: Using a Max-Heap of size $k$ is $O(n \log k)$, which is better than sorting but still worse than binary search on a sorted array.

Interview preparation tip

When an array is already sorted, always aim for a solution involving Binary Search. If the problem asks for a range of elements, consider using binary search to find the boundaries of that range rather than finding individual elements.