# Leetcode – Find the Distance Value Between Two Arrays Solution

Leetcode is a platform that provides a collection of coding challenges to help programmers prepare for technical interviews. Among these challenges, the problem “Find the Distance Value Between Two Arrays” stands out as a particularly interesting one. This problem focuses on array manipulation and efficient searching techniques. In this article, we’ll go through a deep dive to understand, solve, and optimize this problem.

1. Understanding the Problem Statement
2. Breaking Down the Problem into Sub-Problems
3. The Brute-Force Approach
4. Optimizing with Sorting
5. Optimizing with Binary Search Trees
6. Time and Space Complexity Analysis
7. Testing and Edge Cases
8. Conclusion

## 1. Understanding the Problem Statement

The problem asks you to find the “distance value” between two integer arrays, arr1 and arr2, given an integer d. A value x from arr1 is said to have a distance value d from arr2 if no value in arr2 lies in the range [x - d, x + d].In other words, for a value x in arr1 to contribute to the distance value, each value y in arr2 must satisfy either y < x - d or y > x + d.

### Constraints and Assumptions

• The length of both arr1 and arr2 will be between 1 and 100.
• The elements of both arrays will be integers between 1 and 10^4.
• d will be a non-negative integer.

## 2. Breaking Down the Problem into Sub-Problems

The core of this problem is to efficiently compare each element in arr1 against all elements in arr2.

## 3. The Brute-Force Approach

A straightforward way to solve this problem is to iterate through each element x in arr1 and then iterate through each element y in arr2 to check the distance condition.

Here’s a Python code snippet to achieve this:

def findTheDistanceValue(arr1, arr2, d):
count = 0
for x in arr1:
if all(abs(x - y) > d for y in arr2):
count += 1
return count

## 4. Optimizing with Sorting

We can improve the algorithm by first sorting arr2. Once arr2 is sorted, we can use binary search for each element in arr1 to find whether a violating element exists in arr2, thereby reducing the time complexity.

import bisect

def findTheDistanceValue(arr1, arr2, d):
count = 0
arr2.sort()
for x in arr1:
index = bisect.bisect_left(arr2, x)
if (index == 0 or arr2[index-1] < x - d) and (index == len(arr2) or arr2[index] > x + d):
count += 1
return count

## 5. Optimizing with Binary Search Trees

We could take advantage of data structures like Binary Search Trees (BSTs) or balanced trees like AVL trees to maintain a sorted order of arr2 and perform efficient lookups.

## 6. Time and Space Complexity Analysis

• Brute-force Approach: Time Complexity is O(n * m) and Space Complexity is O(1).
• Sorting and Binary Search: Time Complexity is O(n log m + m log m) and Space Complexity is O(1).
• BST: Time Complexity is O(n log m) for balanced trees and Space Complexity is O(m).

## 7. Testing and Edge Cases

Make sure to test your solution against various cases, such as:

• Small arrays.
• Large values of d.
• Negative integers in arrays.

## 8. Conclusion

The problem “Find the Distance Value Between Two Arrays” offers a perfect opportunity to understand array manipulation techniques, sorting algorithms, and searching algorithms. Multiple approaches exist to solve it, each with its pros and cons in terms of time and space complexity.