I'm trying to solve the following leetcode problem:
Given a sorted array, two integers k and x, find the k closest elements to x in 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]
My incorrect solution for now is the following:
class Solution {
public List<Integer> findClosestElements(int[] arr, int k, int x) {
PriorityQueue<Integer> pq = new PriorityQueue<>(arr.length, (a,b) -> a == b ? a - b : Math.abs(a-x) - Math.abs(b-x));
for(int i=0; i<arr.length; i++) {
pq.add(arr[i]);
}
ArrayList ints = new ArrayList<>();
for(int i=0;i<k;i++) {
ints.add(pq.poll());
}
return ints;
}
}
The problem is with the comparator I'm passing to the constructor. The idea is that I want my comparator to sort the integers with respect to the minimum distance between any integer i and the input x
and then poll k
elements from the queue. How can I impelement a comparator function that sorts the elements that way?
I would take advantage of the default Integer.compare
method. Basically what you want is to first check the compare of the absolute difference, and if its a tie do a normal compare.
static int compare(int x, int a, int b) {
int comp = Integer.compare(Math.abs(a - x), Math.abs(b - x));
if (comp == 0) {
return Integer.compare(a, b);
}
return comp;
}
This makes it pretty clean to write the actual priority queue implementation
static List<Integer> findClosestElements(int[] arr, int k, int x) {
PriorityQueue<Integer> queue = new PriorityQueue<>(
arr.length, (a,b) -> compare(x, a, b));
Arrays.stream(arr).forEach(queue::add);
return queue.stream().limit(k).sorted().collect(Collectors.toList());
}