What is the time complexity of Beautiful Towers I?

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

What data structures are used to solve Beautiful Towers I?

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

What are the key insights for solving Beautiful Towers I?

Insight 1: The core idea is to iterate through each element of the input array `heights` and consider it as a potential peak of the mountain. Insight 2: For each potential peak, we need to construct the left and right slopes of the mountain by ensuring non-decreasing and non-increasing properties, respectively. This involves reducing the height of towers where necessary.

Which companies ask LeetCode 2865. Beautiful Towers I in interviews?

This problem is frequently asked at Salesforce.

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Beautiful Towers I - LeetCode 2865 Solution

Beautiful Towers I - Complete Solution Guide

Beautiful Towers I is LeetCode problem 2865, 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

You are given an array heights of n integers representing the number of bricks in n consecutive towers. Your task is to remove some bricks to form a mountain-shaped tower arrangement. In this arrangement, the tower heights are non-decreasing, reaching a maximum peak value with one or multiple consecutive towers and then non-increasing. Return the maximum possible sum of heights of a mountain-shaped tower arrangement. Example 1: Input: heights = [5,3,4,1,1] Output: 13 Explanation: We remove some

Detailed Explanation

The problem asks us to find the maximum possible sum of heights of a 'mountain-shaped' tower arrangement that can be formed from a given array `heights`. A mountain-shaped arrangement is defined as a sequence of heights that are non-decreasing until a peak (maximum value) and then non-increasing after the peak. We need to remove bricks from the original towers (represented by the `heights` array) to achieve this mountain shape. The goal is to find the mountain shape that maximizes the sum of the tower heights.

Solution Approach

The solution uses a brute-force approach. It iterates through each element of the `heights` array, considering each element as a potential peak of the mountain. For each potential peak, it constructs the left and right slopes by minimizing the heights of the towers to satisfy the mountain shape property (non-decreasing to the left and non-increasing to the right). It calculates the sum of heights for the constructed mountain and updates the maximum sum found so far. This process is repeated for all possible peaks, and the maximum sum found across all peaks is the final result.

Step-by-Step Algorithm

Step 1: Initialize `max_sum` to 0. This variable will store the maximum sum of heights found so far.
Step 2: Iterate through the `heights` array using a loop (from index `i = 0` to `n-1`), considering each element `heights[i]` as a potential peak.
Step 3: For each potential peak `heights[i]`, initialize `current_sum` to `heights[i]`. This will accumulate the sum of heights for the current mountain shape.
Step 4: Construct the left slope (non-decreasing). Start from `heights[i-1]` and move towards the beginning of the array. In each step, take the minimum between the previous tower's height (`last_h`, initially `heights[i]`) and the current tower's height (`heights[j]`). Update `last_h` with this minimum and add it to `current_sum`.
Step 5: Construct the right slope (non-increasing). Start from `heights[i+1]` and move towards the end of the array. Similar to the left slope, take the minimum between the previous tower's height (`last_h`, initially `heights[i]`) and the current tower's height (`heights[j]`). Update `last_h` with this minimum and add it to `current_sum`.
Step 6: After constructing both slopes, update `max_sum` with the maximum between `max_sum` and `current_sum`.
Step 7: After iterating through all potential peaks, return `max_sum`.

Key Insights

Insight 1: The core idea is to iterate through each element of the input array `heights` and consider it as a potential peak of the mountain.
Insight 2: For each potential peak, we need to construct the left and right slopes of the mountain by ensuring non-decreasing and non-increasing properties, respectively. This involves reducing the height of towers where necessary.
Insight 3: Since the constraint n <= 1000, a brute-force approach iterating through each possible peak and calculating the sum around it is feasible within the time limit.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Salesforce.

Beautiful Towers I | LeetCode 2865

Medium

Array Stack Monotonic Stack

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

Solution Code

Available in:

class Solution {
    public long maximumSumOfHeights(int[] heights) {
        int n = heights.length;
        long maxSum = 0;

        for (int i = 0; i < n; i++) {
            long currentSum = heights[i];
            
            // Calculate sum for the left slope (non-decreasing)
            int lastH = heights[i];
            for (int j = i - 1; j >= 0; j--) {
                lastH = Math.min(lastH, heights[j]);
                currentSum += lastH;
            }
            
            // Calculate sum for the right slope (non-increasing)
            lastH = heights[i];
            for (int j = i + 1; j < n; j++) {
                lastH = Math.min(lastH, heights[j]);
                currentSum += lastH;
            }
            
            maxSum = Math.max(maxSum, currentSum);
            
        }
        return maxSum;
    }
}

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Problem Info

Description

Problem 2865: Beautiful Towers I - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

Salesforce

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2865

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

7 steps

1
Step 1: Initialize `max_sum` to 0. This variable will store the maximum sum of heights found so far.
2
Step 2: Iterate through the `heights` array using a loop (from index `i = 0` to `n-1`), considering each element `heights[i]` as a potential peak.
3
Step 3: For each potential peak `heights[i]`, initialize `current_sum` to `heights[i]`. This will accumulate the sum of heights for the current mountain shape.
4
Step 4: Construct the left slope (non-decreasing). Start from `heights[i-1]` and move towards the beginning of the array. In each step, take the minimum between the previous tower's height (`last_h`, initially `heights[i]`) and the current tower's height (`heights[j]`). Update `last_h` with this minimum and add it to `current_sum`.
5
Step 5: Construct the right slope (non-increasing). Start from `heights[i+1]` and move towards the end of the array. Similar to the left slope, take the minimum between the previous tower's height (`last_h`, initially `heights[i]`) and the current tower's height (`heights[j]`). Update `last_h` with this minimum and add it to `current_sum`.
6
Step 6: After constructing both slopes, update `max_sum` with the maximum between `max_sum` and `current_sum`.
7
Step 7: After iterating through all potential peaks, return `max_sum`.

💡

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 outer loop iterates `n` times (for each potential peak). The inner loops (for constructing left and right slopes) each iterate up to `n` times in the worst case. Therefore, the overall time complexity is O(n * (n + n)) which simplifies to O(n^2). The algorithm uses a constant amount of extra space, regardless of the input size, resulting in a space complexity of O(1).

📖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

Monotonic Stack

This approach calculates the left and right contributions for each element using a monotonic stack. It stores the indices in stack such that the values at those indices are in non-decreasing/non-increasing order. This can help optimize the process of finding valid mountain configurations.

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

Trade-offs: The monotonic stack approach offers a better time complexity (O(n)) compared to the brute-force (O(n^2)). However, it requires additional space for the stack. It's beneficial for larger input sizes where the O(n^2) complexity becomes a bottleneck.

✨

Key Insights

1Insight 1: The core idea is to iterate through each element of the input array `heights` and consider it as a potential peak of the mountain.
2Insight 2: For each potential peak, we need to construct the left and right slopes of the mountain by ensuring non-decreasing and non-increasing properties, respectively. This involves reducing the height of towers where necessary.
3Insight 3: Since the constraint n <= 1000, a brute-force approach iterating through each possible peak and calculating the sum around it is feasible within the time limit.

📚

Key Terms Explained

For Students

Array

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

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: `heights` array with only one element. The algorithm correctly handles this because the left and right slope loops will not execute, and the `max_sum` will be initialized with the single element's value.
Edge case 2: `heights` array is already a mountain shape. The algorithm correctly identifies the mountain shape and calculates the sum without unnecessary modifications.
Edge case 3: All elements in `heights` array are equal. The algorithm still works correctly, setting each element in the final 'mountain' to the same value, which is equal to original height.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the sum of heights for a given peak. Forgetting to include the peak's own height or incorrectly applying the min function in the left/right slope calculations can lead to inaccurate sums.

✗Common mistake 2: Not updating the `max_sum` variable properly, or failing to consider all potential peaks. This will result in a suboptimal solution.

✗Common mistake 3: Confusing non-decreasing and non-increasing sequences when constructing the mountain's slopes.

✅

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 with constraints. Imagine allocating resources along a timeline, where the amount of resources allocated needs to increase until a peak (maximum resource usage) and then decrease. The problem helps to find the optimal resource allocation to maximize the total resource utilization while adhering to the increasing/decreasing constraints.

🏦Real-world scenario 2: Signal processing where a signal needs to be smoothed but with the requirement that values can only decrease compared to the original signal. The algorithm finds the optimal smoothed signal (mountain shape) that maximizes the overall signal strength.

💡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.

👨‍💻 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: November 13, 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

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

@InterviewProJun 30

❓

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 is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

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 "Beautiful Towers I"?

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

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