How do you solve Find K Closest Elements on LeetCode?

Find K Closest Elements (LeetCode #658) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The array is sorted, which allows for efficient searching and selection.

What is the brute force solution for Find K Closest Elements?

The brute force approach for Find K Closest Elements is Brute Force, with O(n log n) time complexity and O(n) space complexity. The brute force approach involves calculating the distance of each element in the array from x, sorting these distances, and then selecting the k closest elements. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find K Closest Elements?

Always consider the properties of the input data (sorted, unique, etc.). Think about edge cases, such as when k equals the length of the array. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Find K Closest Elements?

Not leveraging the sorted property of the array. Confusing the distance calculation when elements are equidistant from x.

#658

Find K Closest Elements

Medium

ArrayTwo PointersBinary SearchSliding WindowSortingHeap (Priority Queue)Two PointersBinary SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n log n)	O(n)
Space	O(n)	O(1)

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Intuition

Time O(n)Space O(1)

Using a two-pointer approach allows us to efficiently find the k closest elements without sorting the entire array. This method leverages the fact that the array is already sorted, making it faster and more space-efficient.

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Algorithm

3 steps

1Step 1: Use binary search to find the closest index to x.
2Step 2: Initialize two pointers, one to the left and one to the right of the found index.
3Step 3: Expand the pointers outward until k elements are collected, choosing the closer element at each step.

solution.py8 lines

1def findClosestElements(arr, k, x):
2    left, right = 0, len(arr) - 1
3    while right - left >= k:
4        if abs(arr[left] - x) <= abs(arr[right] - x):
5            right -= 1
6        else:
7            left += 1
8    return arr[left:left + k]

ℹ

Complexity note: The time complexity is O(n) in the worst case for the two-pointer approach, and the space complexity is O(1) since we are using pointers instead of additional data structures.

1The array is sorted, which allows for efficient searching and selection.
2Using two pointers can significantly reduce the time complexity compared to sorting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.