Why Largest Rectangle in Histogram Using Stack in DSA Rust?

The Big Idea

Discover how a simple stack can turn a huge guessing game into a quick, exact answer!

The Scenario

Imagine you have a row of buildings with different heights, and you want to find the largest flat area you can cover by choosing some consecutive buildings. Doing this by checking every possible group of buildings one by one is like trying to find a needle in a haystack by looking at every straw.

The Problem

Manually checking all groups of buildings is very slow and tiring. It takes a lot of time because you have to look at many combinations, and it's easy to make mistakes counting or missing some groups. This method does not work well when there are many buildings.

The Solution

Using a stack helps keep track of building heights in a smart way. It lets you quickly find the largest rectangle by remembering where heights start and end without checking every group. This method is fast and avoids repeated work, making the problem easy to solve even for many buildings.

Before vs After

✗ Before

fn largest_rectangle_manual(heights: &[i32]) -> i32 {
    let mut max_area = 0;
    for start in 0..heights.len() {
        let mut min_height = i32::MAX;
        for end in start..heights.len() {
            min_height = min_height.min(heights[end]);
            let area = min_height * (end - start + 1) as i32;
            if area > max_area {
                max_area = area;
            }
        }
    }
    max_area
}

✓ After

fn largest_rectangle_stack(heights: &[i32]) -> i32 {
    let mut stack = Vec::new();
    let mut max_area = 0;
    for (index, &height) in heights.iter().enumerate() {
        while let Some(&top) = stack.last() {
            if heights[top] > height {
                let height_top = heights[stack.pop().unwrap()];
                let width = if let Some(&new_top) = stack.last() {
                    index - new_top - 1
                } else {
                    index
                };
                max_area = max_area.max(height_top * width as i32);
            } else {
                break;
            }
        }
        stack.push(index);
    }
    let length = heights.len();
    while let Some(top) = stack.pop() {
        let height_top = heights[top];
        let width = if let Some(&new_top) = stack.last() {
            length - new_top - 1
        } else {
            length
        };
        max_area = max_area.max(height_top * width as i32);
    }
    max_area
}

What It Enables

This method makes it possible to quickly find the largest rectangle in a histogram, even when there are thousands of buildings, without wasting time checking every possibility.

Real Life Example

Think of a city planner who wants to find the biggest flat area to build a park between tall buildings. Using this method, they can quickly decide where to place the park without measuring every possible space.

Key Takeaways

Manually checking all rectangles is slow and error-prone.

Using a stack tracks building heights efficiently.

This approach finds the largest rectangle quickly and correctly.

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 reverse the order of histogram bars

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

Solution

1. Step 1: Understand stack usage in the problem

   The stack stores indices of bars in increasing height order to find boundaries for rectangles.

2. Step 2: Explain why this helps

   This allows quick calculation of largest rectangle areas when a smaller height bar is found.

3. Final Answer:

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

4. Quick Check:

   Stack tracks indices with increasing heights = C [OK]

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

Common Mistakes:

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

2. Which of the following Rust code snippets correctly initializes an empty stack to store indices for the histogram bars?

easy

A. let mut stack: Vec<usize> = Vec::new();

B. let stack = Vec::with_capacity(0);

C. let stack: Vec<i32> = Vec::new();

D. let mut stack: Vec<i32> = Vec::new();

Solution

1. Step 1: Identify correct stack type for indices

   Indices are usize type, so stack must be Vec<usize>.

2. Step 2: Check mutability and initialization

   Stack must be mutable and initialized empty with Vec::new().

3. Final Answer:

   let mut stack: Vec<usize> = Vec::new(); -> Option A

4. Quick Check:

   Mutable Vec of usize initialized empty = B [OK]

Hint: Use mutable Vec<usize> for indices stack [OK]

Common Mistakes:

• Using wrong type like i32 for indices
• Not making stack mutable
• Using Vec::with_capacity(0) without mut

3. What is the output of the following Rust code that calculates the largest rectangle area in the histogram heights = [2, 1, 5, 6, 2, 3]?

fn largest_rectangle_area(heights: Vec<i32>) -> i32 {
    let mut stack: Vec<usize> = Vec::new();
    let mut max_area = 0;
    let mut i = 0;
    while i <= heights.len() {
        let h = if i == heights.len() { 0 } else { heights[i] };
        if stack.is_empty() || h >= heights[*stack.last().unwrap()] {
            stack.push(i);
            i += 1;
        } else {
            let top = stack.pop().unwrap();
            let width = if stack.is_empty() { i } else { i - stack.last().unwrap() - 1 };
            max_area = max_area.max(heights[top] * width as i32);
        }
    }
    max_area
}

fn main() {
    println!("{}", largest_rectangle_area(vec![2, 1, 5, 6, 2, 3]));
}

medium

A. 12

B. 10

C. 9

D. 8

Solution

1. Step 1: Trace stack and area calculation

   For heights [2,1,5,6,2,3], largest rectangle area is 10 (from bars 5 and 6 with width 2).

2. Step 2: Confirm max_area value

   Code correctly calculates max_area as 10 after processing all bars.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Largest rectangle area = 10 [OK]

Hint: Largest rectangle area for example histogram is 10 [OK]

Common Mistakes:

• Confusing max area with sum of heights
• Miscomputing width when popping stack
• Forgetting to add sentinel zero height

4. Identify the bug in this Rust code snippet for Largest Rectangle in Histogram and select the fix:

fn largest_rectangle_area(heights: Vec<i32>) -> i32 {
    let mut stack: Vec<usize> = Vec::new();
    let mut max_area = 0;
    let mut i = 0;
    while i < heights.len() {
        if stack.is_empty() || heights[i] >= heights[*stack.last().unwrap()] {
            stack.push(i);
            i += 1;
        } else {
            let top = stack.pop().unwrap();
            let width = if stack.is_empty() { i } else { i - stack.last().unwrap() - 1 };
            max_area = max_area.max(heights[top] * width as i32);
        }
    }
    max_area
}

medium

A. Subtract 1 from width calculation: i - stack.last().unwrap() - 1

B. Change while condition to i <= heights.len() and use 0 height sentinel

C. Make stack mutable by adding 'mut' keyword

D. Initialize max_area with i32::MIN instead of 0

Solution

1. Step 1: Identify width calculation error

   Width should be i - stack.last().unwrap() - 1 to exclude the popped bar's index.

2. Step 2: Explain fix impact

   Without subtracting 1, width is overcounted, causing wrong area calculation.

3. Final Answer:

   Subtract 1 from width calculation: i - stack.last().unwrap() - 1 -> Option A

4. Quick Check:

   Correct width calculation subtracts 1 [OK]

Hint: Width = i - previous index - 1 when popping stack [OK]

Common Mistakes:

• Missing -1 in width calculation
• Not handling sentinel zero height
• Incorrect while loop condition

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

hard

A. 8

B. 10

C. 12

D. 15

Solution

1. Step 1: Use stack to track indices with increasing heights

   Push indices while heights increase; pop when current height is less to calculate areas.

2. Step 2: Calculate largest rectangle area

   Largest rectangle area is 12 (from bars at indices 2,3,4 with min height 4 and width 3).

3. Final Answer:

   12 -> Option C

4. Quick Check:

   Stack helps find max area 12 [OK]

Hint: Stack finds max area 12 by tracking increasing heights [OK]

Common Mistakes:

• Calculating area without considering width correctly
• Ignoring bars after popping stack
• Not using sentinel zero height to flush stack