What is the time complexity of Max Increase to Keep City Skyline?

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

What data structures are used to solve Max Increase to Keep City Skyline?

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

What are the key insights for solving Max Increase to Keep City Skyline?

Insight 1: The increase in height of each building at grid[r][c] is limited by the minimum of the maximum height of its row (max_row_heights[r]) and the maximum height of its column (max_col_heights[c]). Insight 2: We need to first precompute the maximum height for each row and each column.

Which companies ask LeetCode 807. Max Increase to Keep City Skyline in interviews?

This problem is frequently asked at Rivian.

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Max Increase to Keep City Skyline - LeetCode 807 Solution

Max Increase to Keep City Skyline - Complete Solution Guide

Max Increase to Keep City Skyline is LeetCode problem 807, 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

There is a city composed of n x n blocks, where each block contains a single building shaped like a vertical square prism. You are given a 0-indexed n x n integer matrix grid where grid[r][c] represents the height of the building located in the block at row r and column c . A city's skyline is the outer contour formed by all the building when viewing the side of the city from a distance. The skyline from each cardinal direction north, east, south, and west may be different. We are allowed to inc

Detailed Explanation

The problem asks us to find the maximum total increase in the height of buildings in a city represented by an n x n grid, such that the skyline of the city remains unchanged when viewed from any cardinal direction (north, east, south, west). We are given a grid where grid[r][c] represents the height of the building at row r and column c. The skyline is determined by the maximum height in each row and each column. We can increase the height of any building, including zero-height buildings, without changing these maximum row and column heights. The goal is to find the largest possible sum of height increases.

Solution Approach

The solution involves first calculating the maximum heights for each row and column. Then, for each building in the grid, it determines the maximum height the building can be increased to without changing the skyline. This is the minimum of the maximum row height and maximum column height the building belongs to. Finally, it sums the differences between this maximum allowed height and the original height of each building to find the total increase.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `max_row_heights` and `max_col_heights`, to store the maximum heights of each row and each column, respectively. Both arrays have size 'n', where 'n' is the dimension of the grid.
Step 2: Iterate through each row of the grid. For each row 'r', find the maximum height among all buildings in that row. Store this maximum height in `max_row_heights[r]`.
Step 3: Iterate through each column of the grid. For each column 'c', find the maximum height among all buildings in that column. Store this maximum height in `max_col_heights[c]`.
Step 4: Initialize a variable `total_increase` to 0. This variable will store the total increase in building heights.
Step 5: Iterate through each building in the grid using nested loops (row 'r' and column 'c').
Step 6: For each building at `grid[r][c]`, determine the maximum allowed height, which is the minimum of `max_row_heights[r]` and `max_col_heights[c]`.
Step 7: Calculate the increase in height for the current building by subtracting the original height `grid[r][c]` from the maximum allowed height. Add this increase to `total_increase`.
Step 8: After iterating through all buildings, return `total_increase`.

Key Insights

Insight 1: The increase in height of each building at grid[r][c] is limited by the minimum of the maximum height of its row (max_row_heights[r]) and the maximum height of its column (max_col_heights[c]).
Insight 2: We need to first precompute the maximum height for each row and each column.
Insight 3: The total increase is the sum of the difference between the allowed height and the original height for each building.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Matrix.

Companies

Asked at: Rivian.

Max Increase to Keep City Skyline | LeetCode 807

Medium

Array Greedy Matrix

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

Solution Code

Available in:

class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int n = grid.length;

        int[] max_row_heights = new int[n];
        int[] max_col_heights = new int[n];

        for (int r = 0; r < n; r++) {
            int max_height = 0;
            for (int c = 0; c < n; c++) {
                max_height = Math.max(max_height, grid[r][c]);
            }
            max_row_heights[r] = max_height;
        }

        for (int c = 0; c < n; c++) {
            int max_height = 0;
            for (int r = 0; r < n; r++) {
                max_height = Math.max(max_height, grid[r][c]);
            }
            max_col_heights[c] = max_height;
        }

        int total_increase = 0;

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < n; c++) {
                int max_allowed_height = Math.min(max_row_heights[r], max_col_heights[c]);
                total_increase += max_allowed_height - grid[r][c];
            }
        }

        return total_increase;
    }
}

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Problem Info

Description

Problem 807: Max Increase to Keep City Skyline - Detailed explanation and solution approaches.

Topics

Array Greedy Matrix

Companies

Rivian

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:807

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 two arrays, `max_row_heights` and `max_col_heights`, to store the maximum heights of each row and each column, respectively. Both arrays have size 'n', where 'n' is the dimension of the grid.
2
Step 2: Iterate through each row of the grid. For each row 'r', find the maximum height among all buildings in that row. Store this maximum height in `max_row_heights[r]`.
3
Step 3: Iterate through each column of the grid. For each column 'c', find the maximum height among all buildings in that column. Store this maximum height in `max_col_heights[c]`.
4
Step 4: Initialize a variable `total_increase` to 0. This variable will store the total increase in building heights.
5
Step 5: Iterate through each building in the grid using nested loops (row 'r' and column 'c').
6
Step 6: For each building at `grid[r][c]`, determine the maximum allowed height, which is the minimum of `max_row_heights[r]` and `max_col_heights[c]`.
7
Step 7: Calculate the increase in height for the current building by subtracting the original height `grid[r][c]` from the maximum allowed height. Add this increase to `total_increase`.
8
Step 8: After iterating through all buildings, return `total_increase`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops iterate through all elements of the n x n grid, hence O(n^2) time complexity. The extra space is due to the two arrays of length n for row and column maximums, leading to O(n) space complexity.

📖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

In-place Calculation

Instead of storing the row and column maximums in separate arrays, one could modify the input grid to store these maximums along the first row and first column. However, this approach modifies the input grid, which might not be desirable in all cases.

Time: O(n^2)Space: O(1) (excluding the input grid)

Trade-offs: Saves space but modifies the input. Best if modifying the input grid is acceptable.

✨

Key Insights

1Insight 1: The increase in height of each building at grid[r][c] is limited by the minimum of the maximum height of its row (max_row_heights[r]) and the maximum height of its column (max_col_heights[c]).
2Insight 2: We need to first precompute the maximum height for each row and each column.
3Insight 3: The total increase is the sum of the difference between the allowed height and the original height for each building.

📚

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: A grid filled with zeros. The algorithm correctly calculates max_row_heights and max_col_heights as zeros and hence total_increase will be 0.
Edge case 2: A grid with only one row or column. The algorithm still calculates the correct max heights and the increases appropriately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the minimum of the maximum row height and maximum column height. If you only consider one of them, you might violate the skyline constraint.

✗Common mistake 2: Forgetting to initialize the `total_increase` variable to 0. This will lead to incorrect results.

✅

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: City planning: Optimizing building heights in a city while preserving the aesthetic appeal and view from different directions.

🏦Real-world scenario 2: Resource allocation: Optimizing the distribution of resources across different departments/regions while maintaining certain performance targets (akin to the skyline).

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 27, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Max Increase to Keep City Skyline"?

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

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