What is the time complexity of Two Furthest Houses With Different Colors?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Two Furthest Houses With Different Colors?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Two Furthest Houses With Different Colors?

Insight 1: The problem can be solved by iterating through all pairs of houses and checking if their colors differ. We only need to consider pairs where the second house's index is greater than the first. Insight 2: A brute-force approach is sufficient because the constraints limit the input array size to a maximum of 100 elements. More efficient algorithms (like using a two-pointer approach) aren't strictly necessary but could be explored as an optimization.

Which companies ask LeetCode 2078. Two Furthest Houses With Different Colors in interviews?

This problem is frequently asked at Visa.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Two Furthest Houses With Different Colors - LeetCode 2078 Solution

Two Furthest Houses With Different Colors - Complete Solution Guide

Two Furthest Houses With Different Colors is LeetCode problem 2078, 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

There are n houses evenly lined up on the street, and each house is beautifully painted. You are given a 0-indexed integer array colors of length n , where colors[i] represents the color of the i th house. Return the maximum distance between two houses with different colors . The distance between the i th and j th houses is abs(i - j) , where abs(x) is the absolute value of x . Example 1: Input: colors = [ 1 ,1,1, 6 ,1,1,1] Output: 3 Explanation: In the above image, color 1 is blue, and color 6

Detailed Explanation

The problem asks you to find the maximum distance between any two houses with different colors. You are given an array `colors` where each element represents the color of a house. The distance between two houses is the absolute difference of their indices. The goal is to find the largest distance between any pair of houses with differing colors.

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible pairs of houses. For each pair, it checks if the colors are different. If they are, it updates the `max_dist` variable with the larger of the current `max_dist` and the distance between the houses. Finally, it returns the `max_dist`.

Step-by-Step Algorithm

Step 1: Initialize `max_dist` to 0. This variable will store the maximum distance found so far.
Step 2: Iterate through the `colors` array using nested loops. The outer loop iterates from the beginning of the array, and the inner loop iterates from the current index of the outer loop to the end of the array.
Step 3: For each pair of houses (indices `i` and `j`), compare the colors `colors[i]` and `colors[j]`. If the colors are different, calculate the distance `j - i` (since j > i) and update `max_dist` if this distance is greater.
Step 4: After iterating through all pairs, return the value of `max_dist`.

Key Insights

Insight 1: The problem can be solved by iterating through all pairs of houses and checking if their colors differ. We only need to consider pairs where the second house's index is greater than the first.
Insight 2: A brute-force approach is sufficient because the constraints limit the input array size to a maximum of 100 elements. More efficient algorithms (like using a two-pointer approach) aren't strictly necessary but could be explored as an optimization.
Insight 3: The problem statement guarantees that at least two houses have different colors, eliminating the need for special handling of cases with all houses having the same color.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Visa.

Two Furthest Houses With Different Colors | LeetCode 2078

Easy

Array Greedy

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxDistance(int[] colors) {
        int n = colors.length;
        int maxDist = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (colors[i] != colors[j]) {
                    maxDist = Math.max(maxDist, Math.abs(i - j));
                }
            }
        }

        return maxDist;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2078: Two Furthest Houses With Different Colors - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

Visa

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2078

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

4 steps

1
Step 1: Initialize `max_dist` to 0. This variable will store the maximum distance found so far.
2
Step 2: Iterate through the `colors` array using nested loops. The outer loop iterates from the beginning of the array, and the inner loop iterates from the current index of the outer loop to the end of the array.
3
Step 3: For each pair of houses (indices `i` and `j`), compare the colors `colors[i]` and `colors[j]`. If the colors are different, calculate the distance `j - i` (since j > i) and update `max_dist` if this distance is greater.
4
Step 4: After iterating through all pairs, return the value of `max_dist`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through all pairs of houses, resulting in a time complexity of O(n^2). The space complexity is O(1) because we only use a constant amount of extra space to store the `max_dist` variable and loop indices.

📖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

Optimized Brute Force (using only one loop)

Instead of nested loops, iterate once and keep track of the index of the furthest house with a different color from the current house.

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

Trade-offs: This approach reduces time complexity to linear time. It is more efficient for larger inputs, while the original nested loops are easier to understand.

✨

Key Insights

1Insight 1: The problem can be solved by iterating through all pairs of houses and checking if their colors differ. We only need to consider pairs where the second house's index is greater than the first.
2Insight 2: A brute-force approach is sufficient because the constraints limit the input array size to a maximum of 100 elements. More efficient algorithms (like using a two-pointer approach) aren't strictly necessary but could be explored as an optimization.
3Insight 3: The problem statement guarantees that at least two houses have different colors, eliminating the need for special handling of cases with all houses having the same color.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: All houses have the same color except for one. The algorithm will still correctly identify the maximum distance to the outlier.
Edge case 2: Only two houses exist with different colors. The algorithm correctly calculates the distance between them.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for different colors before calculating and updating the distance. This could lead to incorrect results.

✗Common mistake 2: Incorrectly calculating the distance. Using `abs(i-j)` is crucial for handling cases where `i > j`.

✅

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: Optimizing the placement of facilities (e.g., fire stations) to ensure quick response times to areas with different characteristics. The algorithm can help in determining the maximum distance covered by facilities.

🏦Real-world scenario 2: In sensor networks, maximizing the coverage area by selecting sensor locations with different sensor types to ensure comprehensive data acquisition. This problem is analogous to maximizing distance between differing house colors.

💡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.

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: September 7, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Two Furthest Houses With Different Colors"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?