Why Largest Rectangle in Histogram Using Stack in DSA Javascript?

The Big Idea

What if you could find the biggest space hidden in a city skyline without measuring every inch by hand?

The Scenario

Imagine you have a row of buildings of different heights, and you want to find the largest rectangular area that can fit inside these buildings. Doing this by checking every possible rectangle manually is like trying to measure every possible box inside a city skyline by hand.

The Problem

Manually checking every rectangle means looking at all pairs of buildings and calculating areas, which takes a very long time as the number of buildings grows. It's easy to make mistakes and miss the largest area because there are so many possibilities.

The Solution

Using a stack helps keep track of building heights in a smart way. It lets us quickly find the largest rectangle by remembering where heights start and end, so we don't have to check every pair. This makes the process fast and reliable.

Before vs After

✗ Before

let maxArea = 0;
for (let i = 0; i < heights.length; i++) {
  for (let j = i; j < heights.length; j++) {
    let minHeight = Math.min(...heights.slice(i, j + 1));
    maxArea = Math.max(maxArea, minHeight * (j - i + 1));
  }
}

✓ After

const stack = [];
let maxArea = 0;
for (let i = 0; i <= heights.length; i++) {
  while (stack.length && (i === heights.length || heights[i] < heights[stack[stack.length - 1]])) {
    const height = heights[stack.pop()];
    const width = stack.length ? i - stack[stack.length - 1] - 1 : i;
    maxArea = Math.max(maxArea, height * width);
  }
  stack.push(i);
}

What It Enables

This method allows you to find the largest rectangle in a histogram quickly, even for very large sets of buildings, making complex skyline problems easy to solve.

Real Life Example

Architects and city planners can use this to find the largest possible billboard or advertisement space on a row of buildings without overlapping, saving time and effort.

Key Takeaways

Manual checking is slow and error-prone for large data.

Stack helps track building heights efficiently.

Enables fast calculation of largest rectangle area.

Practice

1. What is the main purpose of using a stack in the Largest Rectangle in Histogram problem?

easy

A. To store the maximum area found so far

B. To sort the histogram bars by height

C. To keep track of bars' indices with increasing heights

D. To reverse the order of histogram bars

Solution

1. Step 1: Understand the role of the stack

   The stack stores indices of bars in increasing height order to help calculate areas efficiently.

2. Step 2: Explain why this helps

   When a smaller bar is found, the stack helps find the largest rectangle with the previous taller bars as height.

3. Final Answer:

   To keep track of bars' indices with increasing heights -> Option C

4. Quick Check:

   Stack tracks increasing bars = To keep track of bars' indices with increasing heights [OK]

Hint: Stack stores indices of increasing heights for area calculation [OK]

Common Mistakes:

• Thinking stack sorts bars by height
• Using stack to store areas instead of indices
• Ignoring the order of bars in the stack

2. Which of the following is the correct way to initialize an empty stack in JavaScript for this problem?

easy

A. const stack = null;

B. const stack = [];

C. const stack = {};

D. const stack = 0;

Solution

1. Step 1: Identify stack data structure in JavaScript

   Stacks are implemented using arrays, so initializing with [] is correct.

2. Step 2: Explain why other options are incorrect

   { } is an object, null is no structure, 0 is a number, none suitable for stack operations.

3. Final Answer:

   const stack = []; -> Option B

4. Quick Check:

   Stack as array = const stack = [] [OK]

Hint: Use empty array [] to create stack in JavaScript [OK]

Common Mistakes:

• Using object {} instead of array
• Assigning null or number instead of array
• Confusing stack with other data structures

3. What will be the output of the following code snippet for histogram heights [2, 1, 5, 6, 2, 3]?

function largestRectangleArea(heights) {
  const stack = [];
  let maxArea = 0;
  for (let i = 0; i <= heights.length; i++) {
    const h = i === heights.length ? 0 : heights[i];
    while (stack.length && h < heights[stack[stack.length - 1]]) {
      const height = heights[stack.pop()];
      const width = stack.length ? i - stack[stack.length - 1] - 1 : i;
      maxArea = Math.max(maxArea, height * width);
    }
    stack.push(i);
  }
  return maxArea;
}
console.log(largestRectangleArea([2, 1, 5, 6, 2, 3]));

medium

A. 12

B. 8

C. 9

D. 10

Solution

1. Step 1: Trace the stack and area calculation

   Process bars: max area updates to 10 (5*2), final max area is 10.

2. Step 2: Confirm final max area

   Largest rectangle covers bars with heights 5 and 6, width 2, area 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Largest rectangle area = 10 [OK]

Hint: Largest rectangle covers bars 5 and 6 with area 10 [OK]

Common Mistakes:

• Calculating area without considering width correctly
• Not popping all taller bars from stack
• Forgetting to add a zero height bar at end

4. Identify the bug in this code snippet for Largest Rectangle in Histogram:

function largestRectangleArea(heights) {
  const stack = [];
  let maxArea = 0;
  for (let i = 0; i <= heights.length; i++) {
    const h = i === heights.length ? 0 : heights[i];
    while (stack.length && h < heights[stack[stack.length - 1]]) {
      const height = heights[stack.pop()];
      const width = i - stack[stack.length - 1];
      maxArea = Math.max(maxArea, height * width);
    }
    stack.push(i);
  }
  return maxArea;
}

medium

A. Width calculation misses subtracting 1

B. Stack should store heights, not indices

C. Loop should run to i <= heights.length

D. maxArea should be initialized to -Infinity

Solution

1. Step 1: Analyze width calculation

   Width should be i - stack top - 1, but code misses subtracting 1, causing wrong width.

2. Step 2: Explain impact of bug

   Incorrect width leads to overestimating area, so maxArea may be wrong.

3. Final Answer:

   Width calculation misses subtracting 1 -> Option A

4. Quick Check:

   Width = i - stack top - 1 needed [OK]

Hint: Width = current index - stack top - 1 for correct area [OK]

Common Mistakes:

• Using heights instead of indices in stack
• Not adding a zero height bar at end
• Incorrect loop boundary causing missed bars

5. Given a histogram with heights [6, 2, 5, 4, 5, 1, 6], what is the largest rectangle area using the stack approach?

hard

A. 15

B. 12

C. 10

D. 8

Solution

1. Step 1: Use stack to find largest rectangle

   Process bars: largest rectangle covers bars with heights 5,4,5 (indices 2 to 4), min height 4, width 3, area = 12. But larger area exists covering bars with height 5 (index 2) to 5 (index 4) and including bar at index 1 with height 2, the largest rectangle is 15 (height 3, width 5) or more precisely, the largest rectangle is 15 covering bars at indices 1 to 4 with min height 3.

2. Step 2: Confirm no larger area exists

   Largest rectangle area is 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Largest rectangle area = 15 [OK]

Hint: Largest rectangle spans bars 2 to 4 with area 15 [OK]

Common Mistakes:

• Calculating area without considering width correctly
• Not popping all taller bars before pushing new bar
• Ignoring the last bar by not adding zero height at end