What is the time complexity of Maximum Building Height?

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

What data structures are used to solve Maximum Building Height?

This problem uses Array, Math, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Building Height?

Insight 1: We can't simply assign the maximum possible height to each building because of the adjacency constraint. We must consider the propagation of height restrictions from left to right and right to left. Insight 2: The restrictions define a series of segments. The tallest building will either be one of the restricted buildings or will lie within a segment between two restricted buildings.

Which companies ask LeetCode 1840. Maximum Building Height in interviews?

This problem is frequently asked at Dataminr.

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

Maximum Building Height - LeetCode 1840 Solution

Maximum Building Height - Complete Solution Guide

Maximum Building Height is LeetCode problem 1840, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You want to build n new buildings in a city. The new buildings will be built in a line and are labeled from 1 to n . However, there are city restrictions on the heights of the new buildings: The height of each building must be a non-negative integer. The height of the first building must be 0 . The height difference between any two adjacent buildings cannot exceed 1 . Additionally, there are city restrictions on the maximum height of specific buildings. These restrictions are given as a 2D integ

Detailed Explanation

The problem asks us to find the maximum possible height of the tallest building given 'n' buildings to be constructed in a line, with the first building's height fixed at 0. We also have a list of restrictions specifying the maximum allowed height for certain buildings. The height difference between adjacent buildings must not exceed 1.

Solution Approach

The solution uses a two-pass approach combined with sorting. First, all restrictions, including the implied ones for the first building and potentially the last building, are collected and sorted by building ID. The first pass propagates height restrictions from left to right, ensuring the height of each building adheres to the adjacency constraint based on its left neighbor. The second pass propagates restrictions from right to left, ensuring height consistency with its right neighbor. Finally, the code iterates through adjacent restriction pairs to calculate the maximum possible height within each segment based on the endpoints and calculates the maximum height by comparing the peak in the segment with current maximum value found.

Step-by-Step Algorithm

Step 1: Create a list of all restrictions, including the restriction for the first building (id=1, height=0). If the last building 'n' doesn't have an explicit restriction, add an implicit restriction (id=n, height=n-1).
Step 2: Sort the restrictions list based on building IDs.
Step 3: Perform a left-to-right pass. For each building, ensure its height is not greater than the height of the previous building plus the distance between them. Update the building's height accordingly (min(current height, previous height + distance)).
Step 4: Perform a right-to-left pass. For each building, ensure its height is not greater than the height of the next building plus the distance between them. Update the building's height accordingly (min(current height, next height + distance)).
Step 5: Iterate through adjacent pairs of restricted buildings. For each pair, calculate the maximum possible height within that segment based on the buildings heights and distance. Formula: peak_height = (height1 + height2 + distance) / 2. Update max_height to the maximum calculated so far.
Step 6: Return the calculated max_height.

Key Insights

Insight 1: We can't simply assign the maximum possible height to each building because of the adjacency constraint. We must consider the propagation of height restrictions from left to right and right to left.
Insight 2: The restrictions define a series of segments. The tallest building will either be one of the restricted buildings or will lie within a segment between two restricted buildings.
Insight 3: Appending the first building (id=1, height=0) and the last building (id=n, height=n-1 if no explicit restriction) simplifies the algorithm and handles the boundary cases.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Sorting.

Companies

Asked at: Dataminr.

Maximum Building Height | LeetCode 1840

Hard

Array Math Sorting

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

class Solution {
    public int maxBuilding(int n, int[][] restrictions) {
        List<int[]> all_restrictions = new ArrayList<>();
        all_restrictions.add(new int[]{1, 0});
        for (int[] restriction : restrictions) {
            all_restrictions.add(restriction);
        }

        all_restrictions.sort(Comparator.comparingInt(a -> a[0]));

        if (all_restrictions.get(all_restrictions.size() - 1)[0] != n) {
            all_restrictions.add(new int[]{n, n - 1});
        }

        // Pass 1: Left to right propagation
        for (int i = 1; i < all_restrictions.size(); i++) {
            int prev_id = all_restrictions.get(i - 1)[0];
            int prev_h = all_restrictions.get(i - 1)[1];
            int curr_id = all_restrictions.get(i)[0];
            int curr_h = all_restrictions.get(i)[1];
            all_restrictions.get(i)[1] = Math.min(curr_h, prev_h + curr_id - prev_id);
        }

        // Pass 2: Right to left propagation
        for (int i = all_restrictions.size() - 2; i >= 0; i--) {
            int next_id = all_restrictions.get(i + 1)[0];
            int next_h = all_restrictions.get(i + 1)[1];
            int curr_id = all_restrictions.get(i)[0];
            int curr_h = all_restrictions.get(i)[1];
            all_restrictions.get(i)[1] = Math.min(curr_h, next_h + next_id - curr_id);
        }

        int max_height = 0;
        for (int i = 0; i < all_restrictions.size() - 1; i++) {
            int id1 = all_restrictions.get(i)[0];
            int h1 = all_restrictions.get(i)[1];
            int id2 = all_restrictions.get(i + 1)[0];
            int h2 = all_restrictions.get(i + 1)[1];

            int dist = id2 - id1;
            int peak_in_segment = (h1 + h2 + dist) / 2;
            max_height = Math.max(max_height, peak_in_segment);
        }

        return max_height;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1840: Maximum Building Height - Detailed explanation and solution approaches.

Topics

Array Math Sorting

Companies

Dataminr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1840

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

6 steps

1
Step 1: Create a list of all restrictions, including the restriction for the first building (id=1, height=0). If the last building 'n' doesn't have an explicit restriction, add an implicit restriction (id=n, height=n-1).
2
Step 2: Sort the restrictions list based on building IDs.
3
Step 3: Perform a left-to-right pass. For each building, ensure its height is not greater than the height of the previous building plus the distance between them. Update the building's height accordingly (min(current height, previous height + distance)).
4
Step 4: Perform a right-to-left pass. For each building, ensure its height is not greater than the height of the next building plus the distance between them. Update the building's height accordingly (min(current height, next height + distance)).
5
Step 5: Iterate through adjacent pairs of restricted buildings. For each pair, calculate the maximum possible height within that segment based on the buildings heights and distance. Formula: peak_height = (height1 + height2 + distance) / 2. Update max_height to the maximum calculated so far.
6
Step 6: Return the calculated max_height.

💡

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 n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dictated by the sorting algorithm which is typically O(n log n). The two linear passes each take O(n) time where n is the number of restrictions. The overall space complexity is O(n) to store the augmented restrictions array.

📖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

Dynamic Programming

This could potentially be solved with a dynamic programming approach by maintaining an array representing the maximum allowed height for each building. However, the large input size of n (up to 10^9) makes it impractical due to memory constraints.

Time: O(n) in the worst case but will cause memory issues for large n.Space: O(n) for a large array.

Trade-offs: Using DP becomes impractical due to large n making our current solution better in terms of space.

✨

Key Insights

1Insight 1: We can't simply assign the maximum possible height to each building because of the adjacency constraint. We must consider the propagation of height restrictions from left to right and right to left.
2Insight 2: The restrictions define a series of segments. The tallest building will either be one of the restricted buildings or will lie within a segment between two restricted buildings.
3Insight 3: Appending the first building (id=1, height=0) and the last building (id=n, height=n-1 if no explicit restriction) simplifies the algorithm and handles the boundary cases.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Empty restrictions array. The algorithm handles this correctly by adding the implied restrictions at id=1 and id=n (if not present), allowing the two-pass algorithm to proceed.
Edge case 2: Restrictions are not sorted. The sorting step ensures the restrictions are processed in the correct order.
Edge case 3: Only one restriction. The algorithm handles this correctly because we add the initial restriction at building 1 and the final restriction at building n so the algorithm can calculate the largest possible height using the two passes.
Edge case 4: Restrictions with the same height, or restrictions implying a smaller maximum height than previously considered. The algorithm correctly handles these cases using the min() function in the left-to-right and right-to-left passes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the restrictions. This leads to incorrect propagation of heights and wrong results.

✗Common mistake 2: Incorrectly calculating the maximum height within a segment. Failing to account for the distance between restricted buildings and the maximum height difference leads to an inaccurate answer.

✗Common mistake 3: Not including the first building restriction (id=1, height=0). This will result in not taking the initial height constraint into account, leading to a wrong answer.

✗Common mistake 4: Not considering that the last building might not be restricted, and incorrectly calculating maximum height in this case.

✅

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: Resource allocation in a linearly constrained environment. For example, bandwidth allocation along a network path where each node has a maximum capacity and adjacent nodes' bandwidth differences are limited.

🏦Real-world scenario 2: Road construction where there are restrictions on the maximum gradient and the height of certain points. The problem is to find the maximum possible elevation of the highest point in the constructed road.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: July 11, 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 14

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

@AlgoMasterJul 7

❓

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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Maximum Building Height"?

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

🤖 Need help with this problem?