Introduction

In this blog post, we will explore how to solve the LeetCode problem "1471" 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 an array of integers arr and an integer k. A value arr[i] is said to be stronger than a value arr[j] if |arr[i] - m| > |arr[j] - m| where m is the centre of the array. If |arr[i] - m| == |arr[j] - m|, then arr[i] is said to be stronger than arr[j] if arr[i] > arr[j]. Return a list of the strongest k values in the array. return the answer in any arbitrary order. The centre is the middle value in an ordered integer list. More formally, if the length of the list is n, the centre is the element in position ((n - 1) / 2) in the sorted list (0-indexed). For arr = [6, -3, 7, 2, 11], n = 5 and the centre is obtained by sorting the array arr = [-3, 2, 6, 7, 11] and the centre is arr[m] where m = ((5 - 1) / 2) = 2. The centre is 6. For arr = [-7, 22, 17, 3], n = 4 and the centre is obtained by sorting the array arr = [-7, 3, 17, 22] and the centre is arr[m] where m = ((4 - 1) / 2) = 1. The centre is 3. Example 1: Input: arr = [1,2,3,4,5], k = 2 Output: [5,1] Explanation: Centre is 3, the elements of the array sorted by the strongest are [5,1,4,2,3]. The strongest 2 elements are [5, 1]. [1, 5] is also accepted answer. Please note that although |5 - 3| == |1 - 3| but 5 is stronger than 1 because 5 > 1. Example 2: Input: arr = [1,1,3,5,5], k = 2 Output: [5,5] Explanation: Centre is 3, the elements of the array sorted by the strongest are [5,5,1,1,3]. The strongest 2 elements are [5, 5]. Example 3: Input: arr = [6,7,11,7,6,8], k = 5 Output: [11,8,6,6,7] Explanation: Centre is 7, the elements of the array sorted by the strongest are [11,8,6,6,7,7]. Any permutation of [11,8,6,6,7] is accepted. Constraints: 1 <= arr.length <= 105 -105 <= arr[i] <= 105 1 <= k <= arr.length

Explanation

Here's a solution to the problem, addressing efficiency and optimality:

• Sort and Find Median: Sort the input array arr to find the median m. The median serves as the center point for strength calculation.
• Two-Pointer Approach: Use two pointers, one starting from the beginning and the other from the end of the sorted array. Compare the strengths of the elements pointed to by these pointers relative to the median. Add the stronger element to the result and move the corresponding pointer.

• Build Result: Repeat the two-pointer comparison until k strongest elements are collected.

• Runtime Complexity: O(n log n) due to the initial sorting.

• Storage Complexity: O(n) in the worst case, where n is the size of the input array. This is because sorting might take O(n) space

Code

    def k_strongest(arr, k):
    """
    Finds the k strongest values in an array based on distance from the median.

    Args:
        arr: A list of integers.
        k: The number of strongest values to return.

    Returns:
        A list of the k strongest values.
    """
    n = len(arr)
    sorted_arr = sorted(arr)
    m = sorted_arr[(n - 1) // 2]  # Calculate the median

    strongest = []
    left = 0
    right = n - 1

    while len(strongest) < k:
        if abs(sorted_arr[right] - m) > abs(sorted_arr[left] - m):
            strongest.append(sorted_arr[right])
            right -= 1
        elif abs(sorted_arr[right] - m) < abs(sorted_arr[left] - m):
            strongest.append(sorted_arr[left])
            left += 1
        else:
            if sorted_arr[right] > sorted_arr[left]:
                strongest.append(sorted_arr[right])
                right -= 1
            else:
                strongest.append(sorted_arr[left])
                left += 1
    return strongest