Find K Closest Elements

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given a sorted array of integers in ascending order.
  • We need to find the k closest elements to a given value x in the array.
  • If there are multiple valid answers, we should choose the smallest possible difference between the element and x.

2. Analyze the problem: To solve this problem, we can use a binary search to find the index of the closest element to x. Then, we can use two pointers to expand the window around the found index and include the k closest elements.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize two pointers left and right to 0 and len(arr) - 1 respectively.
  2. Perform a binary search to find the index mid of the element closest to x.
    • While left is less than right:
      • Calculate the middle index mid as (left + right) // 2.
      • If arr[mid] is greater than x, update right to mid.
      • Otherwise, update left to mid + 1.
  3. The window of the k closest elements will be between mid - k and mid + k - 1.
  4. Initialize a new list result and iterate from mid - k to mid + k - 1 (inclusive) and append the corresponding elements from arr to result.
  5. Return result.

4. Explain your approach: The approach involves using a binary search to find the index of the closest element to x. By expanding a window around the found index using two pointers, we can include the k closest elements.

5. Write clean and readable code:

python
def findClosestElements(arr, k, x):
    left = 0
    right = len(arr) - 1

    while left < right:
        mid = (left + right) // 2
        if arr[mid] > x:
            right = mid
        else:
            left = mid + 1

    left = max(0, left - k)
    right = min(len(arr), left + 2 * k)
    result = arr[left:right]

    return result

6. Test your code: Let's test the code with some test cases:

  • Test case 1:

    • arr = [1, 2, 3, 4, 5]
    • k = 4
    • x = 3
    • The expected output is [1, 2, 3, 4] because the closest element to 3 is 3 itself, and we need to include the 4 closest elements.

  • Test case 2:

    • arr = [1, 2, 3, 4, 5]
    • k = 4
    • x = -1
    • The expected output is [1, 2, 3, 4] because the closest element to -1 is 1, and we need to include the 4 closest elements.
python
# Test case 1
arr1 = [1, 2, 3, 4, 5]
k1 = 4
x1 = 3
print(findClosestElements(arr1, k1, x1))  # Expected output: [1, 2, 3, 4]

# Test case 2
arr2 = [1, 2, 3, 4, 5]
k2 = 4
x2 = -1
print(findClosestElements(arr2, k2, x2))  # Expected output: [1, 2, 3, 4]

7. Optimize if necessary: The solution already has a time complexity of O(log n) for the binary search part and O(k) for iterating through the window of closest elements. The space complexity is O(k) to store the result list.

8. Handle error cases: The code assumes that the input array arr is non-empty, the value of k is a positive integer, and the array is sorted in ascending order. If the input does not satisfy these conditions, the behavior of the code may not be as expected. We can add checks at the beginning to handle these cases and return an empty list.

9. Discuss complexity analysis: The time complexity of the solution is O(log n + k) since we perform a binary search and then iterate through the window of closest elements. The space complexity is O(k) to store the result list. The solution has a logarithmic time complexity and linear space complexity.

Next Post Previous Post