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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:

Clarifying Questions

  1. What are the constraints on the input values?
    • The array arr is sorted in ascending order.
    • The length of the array (n) and the integers k and x can vary within certain constraints typically found in competitive programming (e.g., 1 <= k <= n and 1 <= n <= 10^4 or 10^5).
  2. Could there be duplicate values in the array?
    • Yes, the array may have duplicate values.
  3. If k is greater than the number of elements in the array, what should be returned?
    • Given the constraints, we assume 1 <= k <= n. Hence, this situation should not occur.

Strategy

To find the k closest integers to x, we can use a binary search combined with a two-pointer technique:

  1. Binary Search: Identify the position to potentially insert x in the array.
  2. Two-Pointer Technique: Extend outwards from the position to find the k closest elements.

Here’s the step-by-step solution:

  1. Binary Search: Use bisect_left to find the index where x would be inserted to maintain the sorted order.
  2. Two-Pointer Technique: Initialize two pointers, left and right, around the found index.
  3. Expand the window: Expand outward from the initial window’s left and right to include the closest elements until you get k elements.
  4. Sort the result: Finally, sort the extracted elements in ascending order.

Code

Here’s the Python implementation of the above strategy:

from bisect import bisect_left

def findClosestElements(arr, k, x):
    # Find the index where `x` would be inserted to keep the sorted order
    position = bisect_left(arr, x)
    
    # Initialize two pointers
    left = position - 1
    right = position
    
    # While we need to find k elements
    while k > 0:
        # If left pointer is out of bounds, move right pointer
        if left < 0:
            right += 1
        # If right pointer is out of bounds, move left pointer
        elif right >= len(arr):
            left -= 1
        # Compare the distance of two pointers to x
        elif abs(arr[left] - x) <= abs(arr[right] - x):
            left -= 1
        else:
            right += 1
        # Decrease k since we have selected one element
        k -= 1
    
    # Return the sub-array from left+1 to right (exclusive)
    return arr[left + 1:right]

# Example usage
arr = [1, 2, 3, 4, 5]
k = 4
x = 3
print(findClosestElements(arr, k, x))  # Output: [1, 2, 3, 4]

Time Complexity

The time complexity of this algorithm is:

Overall, the time complexity is O(log n + k). Given the constraints, this approach is efficient and should perform well for large inputs.

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