What is the time complexity of Heaters?

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

What data structures are used to solve Heaters?

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

What are the key insights for solving Heaters?

Insight 1: Sorting the heaters allows us to use binary search to efficiently find the nearest heaters to each house. Insight 2: For each house, we only need to consider the distance to its closest heater on the left and the distance to its closest heater on the right.

Which companies ask LeetCode 475. Heaters in interviews?

This problem is frequently asked at Anduril, DE Shaw, Intuit, Nutanix, PhonePe.

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

Heaters - LeetCode 475 Solution

Heaters - Complete Solution Guide

Heaters is LeetCode problem 475, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Winter is coming! During the contest, your first job is to design a standard heater with a fixed warm radius to warm all the houses. Every house can be warmed, as long as the house is within the heater's warm radius range. Given the positions of houses and heaters on a horizontal line, return the minimum radius standard of heaters so that those heaters could cover all houses. Notice that all the heaters follow your radius standard, and the warm radius will be the same. Example 1: Input: houses =

Detailed Explanation

The problem asks us to find the minimum radius needed for heaters to cover all houses on a horizontal line. We are given the positions of houses and heaters. The goal is to determine the smallest radius such that every house is within the range of at least one heater. Each heater will have the same radius.

Solution Approach

The solution first sorts the heaters' positions. Then, for each house, it uses binary search to find the closest heater. Once the closest heater (or the index where the house would be inserted) is found, the algorithm calculates the distances to the nearest heaters on the left and right. The minimum of these two distances represents the minimum radius required to cover the current house. Finally, the algorithm takes the maximum of these minimum distances across all houses, which is the required minimum radius.

Step-by-Step Algorithm

Step 1: Sort the `heaters` array in ascending order. This enables the use of binary search.
Step 2: Initialize `radius` to 0. This variable will store the minimum radius needed to cover all houses.
Step 3: Iterate through each `house` in the `houses` array.
Step 4: For each `house`, use binary search (or `bisect_left` in Python) to find the index of the closest heater or the index where the house would be inserted in the sorted `heaters` array.
Step 5: Calculate the distance to the nearest heater on the left (`dist_left`). If the index found in step 4 is 0, the house is to the left of all heaters, so `dist_left` is set to infinity. Otherwise, `dist_left` is the difference between the house and the heater to its immediate left.
Step 6: Calculate the distance to the nearest heater on the right (`dist_right`). If the index found in step 4 is equal to the size of the `heaters` array, the house is to the right of all heaters, so `dist_right` is set to infinity. Otherwise, `dist_right` is the difference between the heater at the found index and the house.
Step 7: Determine the minimum distance required to cover the current house by taking the minimum of `dist_left` and `dist_right`.
Step 8: Update the `radius` by taking the maximum of the current `radius` and the minimum distance calculated in step 7.
Step 9: After iterating through all houses, return the final `radius`.

Key Insights

Insight 1: Sorting the heaters allows us to use binary search to efficiently find the nearest heaters to each house.
Insight 2: For each house, we only need to consider the distance to its closest heater on the left and the distance to its closest heater on the right.
Insight 3: The minimum radius is the maximum of the minimum distances calculated for each house.

Complexity Analysis

Time Complexity: O(n log m)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Anduril, DE Shaw, Intuit, Nutanix, PhonePe.

Heaters | LeetCode 475

Medium

Array Two Pointers Binary Search Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);
        int radius = 0;

        for (int house : houses) {
            int idx = Arrays.binarySearch(heaters, house);
            if (idx < 0) {
                idx = -idx - 1;
            }

            int dist_left = Integer.MAX_VALUE;
            if (idx > 0) {
                dist_left = house - heaters[idx - 1];
            }

            int dist_right = Integer.MAX_VALUE;
            if (idx < heaters.length) {
                dist_right = heaters[idx] - house;
            }

            int min_dist_for_house = Math.min(dist_left, dist_right);
            radius = Math.max(radius, min_dist_for_house);
        }

        return radius;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 475: Heaters - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

AndurilDE ShawIntuitNutanixPhonePe

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:475

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

9 steps

1
Step 1: Sort the `heaters` array in ascending order. This enables the use of binary search.
2
Step 2: Initialize `radius` to 0. This variable will store the minimum radius needed to cover all houses.
3
Step 3: Iterate through each `house` in the `houses` array.
4
Step 4: For each `house`, use binary search (or `bisect_left` in Python) to find the index of the closest heater or the index where the house would be inserted in the sorted `heaters` array.
5
Step 5: Calculate the distance to the nearest heater on the left (`dist_left`). If the index found in step 4 is 0, the house is to the left of all heaters, so `dist_left` is set to infinity. Otherwise, `dist_left` is the difference between the house and the heater to its immediate left.
6
Step 6: Calculate the distance to the nearest heater on the right (`dist_right`). If the index found in step 4 is equal to the size of the `heaters` array, the house is to the right of all heaters, so `dist_right` is set to infinity. Otherwise, `dist_right` is the difference between the heater at the found index and the house.
7
Step 7: Determine the minimum distance required to cover the current house by taking the minimum of `dist_left` and `dist_right`.
8
Step 8: Update the `radius` by taking the maximum of the current `radius` and the minimum distance calculated in step 7.
9
Step 9: After iterating through all houses, return the final `radius`.

💡

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 log m)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the sorting of the heaters and the binary search performed for each house. The space complexity is constant because the algorithm uses a fixed amount of extra space regardless of the input size.

📖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

Two Pointers

Instead of using binary search for each house, we can use two pointers, one for houses and one for heaters. We iterate through both arrays simultaneously, maintaining the closest heater to each house. This approach requires both arrays to be sorted.

Time: O(n log n + m log m), where n is the number of houses and m is the number of heaters.Space: O(1)

Trade-offs: The time complexity is slightly better if m and n are close, because it avoids multiple binary searches. It may be more complex to implement than the binary search approach.

✨

Key Insights

1Insight 1: Sorting the heaters allows us to use binary search to efficiently find the nearest heaters to each house.
2Insight 2: For each house, we only need to consider the distance to its closest heater on the left and the distance to its closest heater on the right.
3Insight 3: The minimum radius is the maximum of the minimum distances calculated for each house.

📚

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: The houses are all to the left of the heaters. The binary search will always return 0, and dist_left will be infinite. The radius is determined by the dist_right.
Edge case 2: The houses are all to the right of the heaters. The binary search will always return the length of heaters, and dist_right will be infinite. The radius is determined by the dist_left.
Edge case 3: Empty houses or heaters array. The problem statement doesn't specify handling empty arrays; assuming it's handled implicitly by the constraints preventing empty inputs.
Edge case 4: Houses and heaters in the same position: the distance should be zero

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the heaters array. Binary search requires the array to be sorted.

✗Common mistake 2: Integer overflow. When calculating the distance between houses and heaters, using `int` can cause overflow if the difference is large. Using `long` (in C++ and potentially Java) can avoid this issue or, using appropriate data type.

✗Common mistake 3: Failing to consider the boundary cases for the binary search results. Specifically, when the house is to the left/right of all heaters, the closest distance may be infinite (or a very large number)

✅

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 cell towers to provide the best coverage to all users in a region. The houses represent users, and the heaters represent cell towers.

🏦Real-world scenario 2: Determining the optimal placement of emergency services (fire stations, hospitals) to minimize the response time to all residents. The houses represent residents, and the heaters represent emergency service locations.

🎮Real-world scenario 3: Finding the optimal placement of Wi-Fi routers to provide coverage in a building. The houses are locations needing coverage and heaters are router positions.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 24, 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Heaters"?

For this medium-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.

🤖 Need help with this problem?