Find K Closest Elements

Given a sorted integer array arr, two integers k and x, return the k closest integers to x in the array. The result should also be sorted in ascending order.

An integer a is closer to x than an integer b if:

  • |a - x| < |b - x|, or
  • |a - x| == |b - x| and a < b

 

Example 1:

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

Example 2:

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

 

Constraints:

  • 1 <= k <= arr.length
  • 1 <= arr.length <= 104
  • arr is sorted in ascending order.
  • -104 <= arr[i], x <= 104
SOLUTION:
import heapq
import bisect

class Solution:
    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
        heap = []
        for num in arr:
            heapq.heappush(heap, (abs(x - num), num))
        kclosest = []
        for i in range(k):
            bisect.insort(kclosest, heapq.heappop(heap)[1])
        return kclosest

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