What is the time complexity of Find the Distance Value Between Two Arrays?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Find the Distance Value Between Two Arrays?

This problem uses Array, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find the Distance Value Between Two Arrays?

Insight 1: The problem involves nested loops to compare each element in `arr1` with each element in `arr2`. This inherently leads to a time complexity related to the product of the lengths of the two arrays. Insight 2: A brute-force approach using nested loops is straightforward and easy to implement. While not necessarily optimal, it's a good starting point for understanding the problem.

Which companies ask LeetCode 1385. Find the Distance Value Between Two Arrays in interviews?

This problem is frequently asked at Zepto.

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Find the Distance Value Between Two Arrays - LeetCode 1385 Solution

Find the Distance Value Between Two Arrays - Complete Solution Guide

Find the Distance Value Between Two Arrays is LeetCode problem 1385, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks you to calculate the 'distance value' between two integer arrays, `arr1` and `arr2`, given a distance threshold `d`. The distance value is the count of elements in `arr1` where the absolute difference between each element in `arr1` and every element in `arr2` is strictly greater than `d`. In simpler terms, for each element in `arr1`, we check if there's any element in `arr2` that's within a distance `d` of it. If there isn't any such element in `arr2`, we increment the distance value. The inputs are two integer arrays and an integer `d`, and the output is a single integer representing the distance value.

Solution Approach

The provided code utilizes a brute-force approach. It iterates through each element in `arr1`. For each element in `arr1`, it iterates through all elements in `arr2`, calculating the absolute difference. If any difference is less than or equal to `d`, a flag is set to indicate that a close element exists in `arr2`. If the flag remains unset after checking all elements in `arr2`, it means no element in `arr2` is within the distance `d`, and the distance value is incremented.

Step-by-Step Algorithm

Step 1: Initialize a `distance` variable to 0. This variable will store the final distance value.
Step 2: Iterate through each element `i` in `arr1`.
Step 3: For each element `i` in `arr1`, initialize a boolean flag (e.g., `found` or `flag`) to `false`. This flag will track whether a close element is found in `arr2`.
Step 4: Iterate through each element `j` in `arr2`.
Step 5: Calculate the absolute difference `abs(i - j)`.
Step 6: If `abs(i - j) <= d`, set the flag to `true` and break the inner loop (since we've found a close element).
Step 7: After the inner loop, if the flag is still `false`, it means no element in `arr2` was within distance `d` of `i`. Increment `distance`.
Step 8: After iterating through all elements in `arr1`, return the final `distance` value.

Key Insights

Insight 1: The problem involves nested loops to compare each element in `arr1` with each element in `arr2`. This inherently leads to a time complexity related to the product of the lengths of the two arrays.
Insight 2: A brute-force approach using nested loops is straightforward and easy to implement. While not necessarily optimal, it's a good starting point for understanding the problem.
Insight 3: Optimizations might be possible if the arrays were sorted. Binary search could then reduce the inner loop's complexity from O(m) to O(log m), where m is the length of `arr2`.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Zepto.

Find the Distance Value Between Two Arrays | LeetCode 1385

Easy

Array Two Pointers Binary Search Sorting

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
        int distance = 0;
        for (int i = 0; i < arr1.length; i++) {
            boolean found = false;
            for (int j = 0; j < arr2.length; j++) {
                if (Math.abs(arr1[i] - arr2[j]) <= d) {
                    found = true;
                    break;
                }
            }
            if (!found) {
                distance++;
            }
        }
        return distance;
    }
}

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Problem Info

Description

Problem 1385: Find the Distance Value Between Two Arrays - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

Zepto

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1385

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a `distance` variable to 0. This variable will store the final distance value.
2
Step 2: Iterate through each element `i` in `arr1`.
3
Step 3: For each element `i` in `arr1`, initialize a boolean flag (e.g., `found` or `flag`) to `false`. This flag will track whether a close element is found in `arr2`.
4
Step 4: Iterate through each element `j` in `arr2`.
5
Step 5: Calculate the absolute difference `abs(i - j)`.
6
Step 6: If `abs(i - j) <= d`, set the flag to `true` and break the inner loop (since we've found a close element).
7
Step 7: After the inner loop, if the flag is still `false`, it means no element in `arr2` was within distance `d` of `i`. Increment `distance`.
8
Step 8: After iterating through all elements in `arr1`, return the final `distance` value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates `n` times, and for each iteration, the inner loop iterates `m` times. Therefore, the total number of operations is proportional to `n*m`. The space used is constant because the number of variables remains fixed irrespective of the input array lengths.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorted Array with Binary Search

Sort `arr2`. For each element in `arr1`, use binary search on `arr2` to find the closest element. This avoids the nested loop.

Time: O(m log m + n log m)Space: O(log m) for binary search recursion stack (or O(1) with iterative binary search), plus O(1) for extra variables. Sorting adds O(m log m) if you use an efficient algorithm.

Trade-offs: This approach is more efficient than the brute-force solution when m (length of `arr2`) is significantly large. The initial sorting cost is amortized over multiple queries of `arr1` elements.

✨

Key Insights

1Insight 1: The problem involves nested loops to compare each element in `arr1` with each element in `arr2`. This inherently leads to a time complexity related to the product of the lengths of the two arrays.
2Insight 2: A brute-force approach using nested loops is straightforward and easy to implement. While not necessarily optimal, it's a good starting point for understanding the problem.
3Insight 3: Optimizations might be possible if the arrays were sorted. Binary search could then reduce the inner loop's complexity from O(m) to O(log m), where m is the length of `arr2`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty `arr1` or `arr2`. The code correctly handles this; the outer or inner loop won't execute, resulting in a distance value of 0.
Edge case 2: `d` equals 0. The code accurately handles this; it will only increment `distance` if there's an element in `arr1` that is exactly equal to no element in `arr2`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the absolute difference calculation. Forgetting to use `abs()` will lead to incorrect results.

✗Common mistake 2: Forgetting to break the inner loop when a close element is found. This will unnecessarily continue checking other elements in `arr2`, leading to incorrect `distance` value.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Identifying outliers. Imagine you have sensor readings (`arr1`) and a set of expected values (`arr2`). This algorithm helps find readings that differ significantly from the expected values, exceeding a tolerance level (`d`).

🏦Real-world scenario 2: Database querying. Given a set of query points (`arr1`) and a dataset of known points (`arr2`), you might want to find points from `arr1` that are sufficiently far from any point in `arr2`. This could be useful in spatial databases or geographic information systems.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find the Distance Value Between Two Arrays"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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