Find K Closest Elements

Given a sorted array, two integerskandx, find thekclosest elements toxin the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

Example 1:

Input: [1,2,3,4,5], k=4, x=3
Output: [1,2,3,4]

Example 2:

Input: [1,2,3,4,5], k=4, x=-1
Output: [1,2,3,4]

Note:

1. The value k is positive and will always be smaller than the length of the sorted array.

2. Length of the given array is positive and will not exceed 10^4

3. Absolute value of elements in the array and x will not exceed 10^4

Solution

Using sort.

public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
     Collections.sort(arr, (a,b) -> a == b ? a - b : Math.abs(a-x) - Math.abs(b-x));
     arr = arr.subList(0, k);
     Collections.sort(arr);
     return arr;
}

Using binary search.

public class Solution {
    public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
        int n = arr.size();
        if (x <= arr.get(0)) {
            return arr.subList(0, k);
        } else if (arr.get(n - 1) <= x) {
        return arr.subList(n - k, n);
        } else {
            int index = Collections.binarySearch(arr, x);
            if (index < 0) {
                index = -index - 1;
            }
            int low = Math.max(0, index - k - 1), high = Math.min(arr.size() - 1, index + k - 1);
            while (high - low > k - 1) {
                if (low < 0 || (x - arr.get(low)) <= (arr.get(high) - x)) {
                    high--;
                } else if (high > arr.size() - 1 || (x - arr.get(low)) > (arr.get(high) - x)) {
                    low++;
                } else {
                    System.out.println("unhandled case: " + low + " " + high);
                }
            }
            return arr.subList(low, high + 1);
         }
     }
}