Introduction

In this blog post, we will explore how to solve the LeetCode problem "1385" 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 two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays. The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d. Example 1: Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2 Output: 2 Explanation: For arr1[0]=4 we have: |4-10|=6 > d=2 |4-9|=5 > d=2 |4-1|=3 > d=2 |4-8|=4 > d=2 For arr1[1]=5 we have: |5-10|=5 > d=2 |5-9|=4 > d=2 |5-1|=4 > d=2 |5-8|=3 > d=2 For arr1[2]=8 we have: |8-10|=2 <= d=2 |8-9|=1 <= d=2 |8-1|=7 > d=2 |8-8|=0 <= d=2 Example 2: Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3 Output: 2 Example 3: Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6 Output: 1 Constraints: 1 <= arr1.length, arr2.length <= 500 -1000 <= arr1[i], arr2[j] <= 1000 0 <= d <= 100

Explanation

• Sort arr2: Sorting arr2 allows us to use binary search to efficiently find the closest element to each element in arr1. This avoids a linear scan of arr2 for each element in arr1.
  • Binary Search: For each element in arr1, use binary search in arr2 to determine if there exists an element in arr2 within the distance d.
  • Count Distance Values: Increment the distance value count if no such element is found in arr2.

• Runtime Complexity: O(n log n + m log m), where n is the length of arr1 and m is the length of arr2. This is due to sorting arr2 (O(m log m)) and performing binary search for each element in arr1 (O(n log m)). Storage Complexity: O(1) or O(m) depending on the sorting algorithm implementation (in-place vs. creating a copy), excluding the input arrays.

Code

    def findTheDistanceValue(arr1, arr2, d):
    """
    Calculates the distance value between two arrays.

    Args:
        arr1: The first array of integers.
        arr2: The second array of integers.
        d: The maximum allowed difference.

    Returns:
        The distance value between the two arrays.
    """
    arr2.sort()
    distance = 0
    for num1 in arr1:
        found = False
        left = 0
        right = len(arr2) - 1
        while left <= right:
            mid = (left + right) // 2
            if abs(num1 - arr2[mid]) <= d:
                found = True
                break
            elif num1 < arr2[mid]:
                right = mid - 1
            else:
                left = mid + 1
        if not found:
            distance += 1
    return distance