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

For example:

arr = [1,2,3,4,5], k = 4, x = 3 should return [1,2,3,4]

arr = [1,1,2,3,4,5], k = 4, x = -1 should return [1,1,2,3]

Write an efficient algorithm in Python to solve this problem, considering various edge cases and analyze its time and space complexity. Make sure to also provide a naive solution and contrast it with the optimal approach.

## Naive Solution

The most straightforward approach is to calculate the absolute difference between each element in the array and `x`, then sort the array based on these differences. Finally, we select the first `k` elements. This approach works but isn't the most efficient due to the sorting step.

```python
def find_closest_elements_naive(arr, k, x):
    # Sort the array based on the absolute difference from x
    arr.sort(key=lambda num: abs(num - x))
    
    # Return the first k elements
    return sorted(arr[:k])

Optimal Solution (Binary Search and Sliding Window)

Since the array is sorted, we can use binary search to find the element closest to x. Once we find this element (or the closest position), we can use a sliding window approach to expand outwards and find the k closest elements.

def find_closest_elements(arr, k, x):
    # Find the index of the element closest to x using binary search
    left = 0
    right = len(arr) - k

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

    return arr[left:left + k]

Explanation

1. Binary Search: We don't need to find the exact element x. Instead, we adjust the binary search to find the starting point of a window of size k that contains the k closest elements to x.
2. Sliding Window: Once we've found the optimal starting point, we simply return the subarray of size k starting from that point.

Example

Let's say arr = [1, 2, 3, 4, 5], k = 4, and x = 3. The function would perform binary search to find the best starting index for the window of size 4, which in this case would be 0. The function then returns arr[0:4], which is [1, 2, 3, 4].

Big(O) Run-time Analysis

• Naive Solution: The time complexity is dominated by the sorting step, which is O(n log n), where n is the length of the array.
• Optimal Solution: The binary search takes O(log(n-k)) time, and creating the subarray takes O(k) time. Therefore, the overall time complexity is O(log(n-k) + k). Because k is typically small, this reduces to nearly O(log n).

Big(O) Space Usage Analysis

• Naive Solution: The space complexity is O(n) because sort() sorts in place (but may take O(n) space in worst case) and the result array can have up to n elements.
• Optimal Solution: The space complexity is O(1) because we are only using a constant amount of extra space for variables, regardless of the input size. The output array is not considered extra space since it is part of the problem requirements.

Edge Cases

1. Empty Array: If the array is empty, we should return an empty list.
2. k >= len(arr): If k is greater than or equal to the length of the array, we should return the entire array.
3. x Smaller than All Elements: If x is smaller than all elements in the array, the first k elements will be the closest.
4. x Larger than All Elements: If x is larger than all elements in the array, the last k elements will be the closest.
5. Duplicate Elements: If there are duplicate elements in the array, the algorithm should still work correctly as it compares absolute differences.

def find_closest_elements_edge_cases(arr, k, x):
    if not arr:  # Empty array
        return []

    n = len(arr)

    if k >= n:  # k is greater than or equal to the length of the array
        return arr

    if x <= arr[0]:  # x is smaller than all elements
        return arr[:k]

    if x >= arr[-1]:  # x is larger than all elements
        return arr[n - k:]

    # Optimal solution implementation
    left = 0
    right = n - k

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

    return arr[left:left + k]