What is the time complexity of Largest Rectangle in Histogram?

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

What data structures are used to solve Largest Rectangle in Histogram?

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

What are the key insights for solving Largest Rectangle in Histogram?

Insight 1: The largest rectangle's height must be equal to one of the heights in the `heights` array. Therefore, we can iterate through each height and treat it as the potential height of the largest rectangle. Insight 2: For each height, we need to find the leftmost and rightmost boundaries that extend to bars with heights greater than or equal to the current height. The width of the rectangle would be the distance between these boundaries minus one, and the area is then the height multiplied by this width.

Which companies ask LeetCode 84. Largest Rectangle in Histogram in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Capital One and 20 more companies.

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Largest Rectangle in Histogram - LeetCode 84 Solution

Q: How to solve Largest Rectangle in Histogram?

The provided solution uses a monotonic stack to efficiently compute the largest rectangular area. The stack stores the indices of the bars in increasing order of their heights. The algorithm iterates through the `heights` array (with a 0 appended at the end as a sentinel value). If the current bar's height is less than the height of the bar at the top of the stack, it means we've found the right boundary for the rectangle whose height is the top of the stack. We then pop the top of the stack, calculate the area of the rectangle using the popped height and the width (calculated from the current index and the index at the new top of the stack), and update the `max_area` if necessary. The stack stores indices to allow width calculations using index differences.

Largest Rectangle in Histogram - Complete Solution Guide

Largest Rectangle in Histogram is LeetCode problem 84, 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

Given an array of integers heights representing the histogram's bar height where the width of each bar is 1 , return the area of the largest rectangle in the histogram . Example 1: Input: heights = [2,1,5,6,2,3] Output: 10 Explanation: The above is a histogram where width of each bar is 1. The largest rectangle is shown in the red area, which has an area = 10 units. Example 2: Input: heights = [2,4] Output: 4 Constraints: 1 <= heights.length <= 10 5 0 <= heights[i] <= 10 4

Detailed Explanation

The problem asks us to find the largest rectangular area that can be formed within a histogram represented by an array of integers, `heights`. Each integer in the array represents the height of a bar, and the width of each bar is assumed to be 1. The goal is to find the maximum area of a rectangle that can be inscribed within these bars. The input is an array of non-negative integers, and the output is a single integer representing the largest possible rectangular area.

Solution Approach

The provided solution uses a monotonic stack to efficiently compute the largest rectangular area. The stack stores the indices of the bars in increasing order of their heights. The algorithm iterates through the `heights` array (with a 0 appended at the end as a sentinel value). If the current bar's height is less than the height of the bar at the top of the stack, it means we've found the right boundary for the rectangle whose height is the top of the stack. We then pop the top of the stack, calculate the area of the rectangle using the popped height and the width (calculated from the current index and the index at the new top of the stack), and update the `max_area` if necessary. The stack stores indices to allow width calculations using index differences.

Step-by-Step Algorithm

Step 1: Initialize an empty stack and push -1 onto it. This -1 serves as a sentinel value to handle the case when the stack becomes empty.
Step 2: Append 0 to the `heights` array. This ensures that all elements in the stack are processed at the end.
Step 3: Iterate through the `heights` array using index `i` and value `h`.
Step 4: While the stack is not empty (excluding only the sentinel -1) and the current height `h` is less than or equal to the height at the index on the top of the stack, pop the top element from the stack.
Step 5: Calculate the area of the rectangle using the popped height. The width is calculated as `i - stack[-1] - 1`, where `i` is the current index and `stack[-1]` is the index of the next smaller element to the left of the popped element.
Step 6: Update `max_area` with the maximum of the current `max_area` and the calculated area.
Step 7: Push the current index `i` onto the stack.
Step 8: After iterating through the entire `heights` array, remove the appended 0. The `max_area` now contains the largest rectangular area.
Step 9: Return the `max_area`.

Key Insights

Insight 1: The largest rectangle's height must be equal to one of the heights in the `heights` array. Therefore, we can iterate through each height and treat it as the potential height of the largest rectangle.
Insight 2: For each height, we need to find the leftmost and rightmost boundaries that extend to bars with heights greater than or equal to the current height. The width of the rectangle would be the distance between these boundaries minus one, and the area is then the height multiplied by this width.
Insight 3: A monotonic stack helps efficiently track the potential boundaries and find the previous and next smaller elements for each bar in the histogram.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Capital One, Cisco, DE Shaw, DoorDash, Flipkart, MAQ Software, Meta, Microsoft, Myntra, PhonePe, Roblox, Sprinklr, Uber, Walmart Labs, Yahoo, Zepto, Zoho, Zomato, Zynga, oyo, tcs.

Largest Rectangle in Histogram | LeetCode 84

Hard

Array Stack Monotonic Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int largestRectangleArea(int[] heights) {
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);
        int max_area = 0;
        int[] newHeights = new int[heights.length + 1];
        System.arraycopy(heights, 0, newHeights, 0, heights.length);
        newHeights[heights.length] = 0;

        for (int i = 0; i < newHeights.length; i++) {
            int h = newHeights[i];
            while (stack.peek() != -1 && newHeights[stack.peek()] >= h) {
                int height = newHeights[stack.pop()];
                int width = i - stack.peek() - 1;
                max_area = Math.max(max_area, height * width);
            }
            stack.push(i);
        }

        return max_area;
    }
}

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Problem Info

Description

Problem 84: Largest Rectangle in Histogram - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

AdobeAmazonAppleBloombergCapital OneCiscoDE ShawDoorDash

Stats

Difficulty:Hard

Languages:4

Companies:25

Problem #:84

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: Initialize an empty stack and push -1 onto it. This -1 serves as a sentinel value to handle the case when the stack becomes empty.
2
Step 2: Append 0 to the `heights` array. This ensures that all elements in the stack are processed at the end.
3
Step 3: Iterate through the `heights` array using index `i` and value `h`.
4
Step 4: While the stack is not empty (excluding only the sentinel -1) and the current height `h` is less than or equal to the height at the index on the top of the stack, pop the top element from the stack.
5
Step 5: Calculate the area of the rectangle using the popped height. The width is calculated as `i - stack[-1] - 1`, where `i` is the current index and `stack[-1]` is the index of the next smaller element to the left of the popped element.
6
Step 6: Update `max_area` with the maximum of the current `max_area` and the calculated area.
7
Step 7: Push the current index `i` onto the stack.
8
Step 8: After iterating through the entire `heights` array, remove the appended 0. The `max_area` now contains the largest rectangular area.
9
Step 9: Return the `max_area`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because we iterate through the input array once, and the while loop inside the for loop processes each element at most once. The space complexity is linear because, in the worst case, the stack can store all the indices of the input 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

Brute Force

For each bar, consider it as the minimum height of a rectangle and expand left and right to find the maximum width possible.

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

Trade-offs: Simple to understand but inefficient for large inputs due to the nested loops. The stack based solution is much faster.

Divide and Conquer

Find the minimum height in the array. The maximum area will be either the minimum height * length, or the maximum area in the left part or the maximum area in the right part.

Time: O(n log n) in the average case, O(n^2) in the worst case.Space: O(log n) for the recursion stack in the average case, O(n) in the worst case.

Trade-offs: A more complex approach, can be more efficient than brute force in some cases, but not as efficient as the monotonic stack solution.

✨

Key Insights

1Insight 1: The largest rectangle's height must be equal to one of the heights in the `heights` array. Therefore, we can iterate through each height and treat it as the potential height of the largest rectangle.
2Insight 2: For each height, we need to find the leftmost and rightmost boundaries that extend to bars with heights greater than or equal to the current height. The width of the rectangle would be the distance between these boundaries minus one, and the area is then the height multiplied by this width.
3Insight 3: A monotonic stack helps efficiently track the potential boundaries and find the previous and next smaller elements for each bar in the histogram.

📚

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: Empty histogram (heights array). The initial `max_area` of 0 handles this case correctly.
Edge case 2: Monotonically increasing or decreasing histograms. The monotonic stack algorithm efficiently processes these cases. The sentinel value 0 appended to the heights ensures all remaining bars are processed and areas calculated.
Edge case 3: All bars have the same height. The algorithm still works correctly because it will push all indices onto the stack, then when the sentinel is encountered, calculate the correct maximum area.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the width. The width should be calculated as the difference between the current index and the previous smaller element's index, minus 1.

✗Common mistake 2: Not handling edge cases properly, such as an empty histogram or a monotonically increasing/decreasing sequence of heights.

✗Common mistake 3: Forgetting to include a sentinel value at the end of the heights array. This is crucial for ensuring that all remaining bars on the stack are processed.

✅

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 storage allocation. Imagine each histogram bar represents the size of available contiguous memory blocks. The largest rectangle area represents the largest contiguous chunk you can allocate.

🏦Real-world scenario 2: Analyzing stock prices. The histogram bars could represent the daily price ranges of a stock. Finding the largest rectangle can help identify periods of sustained high or low prices.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

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 "Largest Rectangle in Histogram"?

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

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