Introduction

In this blog post, we will explore how to solve the LeetCode problem "658" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 = [1,1,2,3,4,5], k = 4, x = -1 Output: [1,1,2,3] Constraints: 1 <= k <= arr.length 1 <= arr.length <= 104 arr is sorted in ascending order. -104 <= arr[i], x <= 104

Explanation

Here's a breakdown of the solution:

• Binary Search for Initial Point: Use binary search to find the element in the array that's closest to x. This helps narrow down the search space quickly.
• Expand Around the Initial Point: Once the closest element (or a suitable starting point) is found, expand outwards (left and right) from that point, comparing the distances of elements to x.

• Maintain a k-sized Window: Keep track of the k closest elements found so far, ensuring that the window always contains the closest elements in ascending order.

• Runtime Complexity: O(log n + k), where n is the length of the input array. O(log n) for binary search, and O(k) for expanding around the initial point. Storage Complexity: O(k) to store the result.

Code

    def find_closest_elements(arr: list[int], k: int, x: int) -> list[int]:
    """
    Finds the k closest integers to x in the sorted array arr.

    Args:
        arr: A sorted integer array.
        k: The number of closest integers to find.
        x: The target integer.

    Returns:
        A list of the k closest integers to x, sorted in ascending order.
    """

    n = len(arr)
    left = 0
    right = n - k

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

    return arr[left:left + k]