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.

## Table of Contents

- Understanding the Problem Statement
- Breaking Down the Problem into Sub-Problems
- The Brute-Force Approach
- Optimizing with Sorting
- Optimizing with Binary Search Trees
- Time and Space Complexity Analysis
- Testing and Edge Cases
- 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.