Largest Rectangle in Histogram Using Stack in DSA Python - Build from Scratch

Largest Rectangle in Histogram Using Stack

📖 Scenario: Imagine you have a row of buildings of different heights. You want to find the largest rectangular area that can fit inside the skyline formed by these buildings.

🎯 Goal: Build a program that finds the largest rectangle area in a histogram using a stack to keep track of building heights.

📋 What You'll Learn

Create a list called heights with exact building heights

Create an empty list called stack to store indices

Use a for loop with variable i to iterate over range(len(heights) + 1)

Print the largest rectangle area as an integer

💡 Why This Matters

🌍 Real World

Finding the largest rectangle in a histogram helps in image processing, stock market analysis, and land area calculations.

💼 Career

This algorithm is often asked in coding interviews and is useful for roles involving algorithm design and optimization.

1

Create the histogram heights list

Create a list called heights with these exact values: 2, 1, 5, 6, 2, 3

DSA Python

# Create the list heights with values 2, 1, 5, 6, 2, 3
# Your code here

Hint

Use square brackets and separate numbers with commas.

2

Initialize the stack and max_area variable

Create an empty list called stack and a variable called max_area set to 0

DSA Python

heights = [2, 1, 5, 6, 2, 3]
# Create an empty list stack and set max_area to 0
# Your code here

Hint

Use [] for an empty list and assign 0 to max_area.

3

Use a for loop to find the largest rectangle

Use a for loop with variable i to iterate over range(len(heights) + 1). Inside the loop, use a while loop to pop from stack while stack is not empty and either i == len(heights) or heights[i] <= heights[stack[-1]]. Calculate the area for each popped index and update max_area if the area is larger. Append i to stack if the while loop ends.

DSA Python

heights = [2, 1, 5, 6, 2, 3]
stack = []
max_area = 0
# Use a for loop with i in range(len(heights) + 1) and implement the stack logic
# Your code here

Hint

Use a for loop with range(len(heights) + 1) and a while loop inside to pop from stack and calculate areas.

4

Print the largest rectangle area

Print the value of max_area to display the largest rectangle area

DSA Python

heights = [2, 1, 5, 6, 2, 3]
stack = []
max_area = 0
for i in range(len(heights) + 1):
    while stack and (i == len(heights) or heights[i]  max_area:
            max_area = area
    stack.append(i)
# Print the max_area
# Your code here

Hint

Use print(max_area) to show the result.

Practice

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

easy

A. To reverse the order of bars for easier calculation

B. To store the heights of all bars in sorted order

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

D. To count the total number of bars in the histogram

Solution

1. Step 1: Understand the role of the stack

   The stack is used to store indices of bars in increasing height order to efficiently find the largest rectangle.

2. Step 2: Explain why increasing order is important

   When a bar with smaller height appears, we pop from the stack to calculate areas with bars taller than the current one.

3. Final Answer:

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

4. Quick Check:

   Stack stores indices in increasing height order [OK]

Hint: Stack stores indices of increasing height bars [OK]

Common Mistakes:

• Thinking stack stores heights instead of indices
• Assuming stack sorts bars by height
• Using stack to count bars instead of track indices

2. Which of the following Python code snippets correctly initializes an empty stack for the Largest Rectangle in Histogram algorithm?

easy

A. stack = ()

B. stack = {}

C. stack = None

D. stack = []

Solution

1. Step 1: Identify correct data structure for stack

   In Python, a list is commonly used as a stack because it supports append() and pop() operations.

2. Step 2: Check each option

   stack = [] uses a list ([]), which is correct. stack = {} uses a dictionary ({}), not suitable. stack = None assigns None, not a stack. stack = () uses a tuple (()), which is immutable.

3. Final Answer:

   stack = [] -> Option D

4. Quick Check:

   List [] is correct stack initialization [OK]

Hint: Use list [] for stack in Python [OK]

Common Mistakes:

• Using dictionary {} instead of list
• Using tuple () which is immutable
• Assigning None instead of a list

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

def largest_rectangle_area(heights):
    stack = []
    max_area = 0
    for i, h in enumerate(heights + [0]):
        while stack and heights[stack[-1]] > h:
            height = heights[stack.pop()]
            width = i if not stack else i - stack[-1] - 1
            max_area = max(max_area, height * width)
        stack.append(i)
    return max_area

print(largest_rectangle_area([2, 1, 5, 6, 2, 3]))

medium

A. 10

B. 12

C. 8

D. 6

Solution

1. Step 1: Trace the stack and area calculation

   Iterate through heights with an extra 0 at the end to pop all bars. The largest rectangle area found is 10 (from bars with heights 5 and 6).

2. Step 2: Confirm max area calculation

   The max area is calculated as max(10, others), so final answer is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Largest rectangle area = 10 [OK]

Hint: Add zero height at end to pop all bars [OK]

Common Mistakes:

• Forgetting to add zero at end of heights
• Incorrect width calculation
• Not popping all bars from stack

4. Find the error in this code snippet for Largest Rectangle in Histogram:

def largest_rectangle_area(heights):
    stack = []
    max_area = 0
    for i, h in enumerate(heights + [0]):
        while stack and heights[stack[-1]] >= h:
            height = heights[stack.pop()]
            width = i - stack[-1] - 1
            max_area = max(max_area, height * width)
        stack.append(i)
    return max_area

print(largest_rectangle_area([2, 1, 5, 6, 2, 3]))

medium

A. Width calculation should be i - stack[-1] - 1

B. Missing adding zero height at end of heights

C. Comparison should be > instead of >=

D. Stack should store heights, not indices

Solution

1. Step 1: Identify width calculation mistake

   Width is calculated as i - stack[-1], but it should be i - stack[-1] - 1 to exclude the popped bar's index.

2. Step 2: Confirm other parts

   Adding zero at end is optional but common; comparison >= is acceptable; stack stores indices correctly.

3. Final Answer:

   Width calculation should be i - stack[-1] - 1 -> Option A

4. Quick Check:

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

Hint: Width = current index - previous stack top - 1 [OK]

Common Mistakes:

• Incorrect width calculation causing wrong area
• Forgetting to add zero at end (optional)
• Confusing stack contents (indices vs heights)

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

hard

A. 10

B. 15

C. 12

D. 8

Solution

1. Step 1: Use stack to find largest rectangle

   Traverse bars with an extra 0 at end. The largest rectangle is formed by heights 5, 4, 5 spanning width 3, area = 12, but a larger rectangle is formed by heights 6 spanning width 2, area = 12, and also by heights 6 spanning width 1, area = 6. However, the maximum area is 15 formed by heights 5, 4, 5, 1, 6 combined correctly.

2. Step 2: Confirm no larger area exists

   After careful calculation, the largest rectangle area is 15 (from bars at indices 2 to 4 with height 5).

3. Final Answer:

   15 -> Option B

4. Quick Check:

   Max area = 15 from bars 5,4,5 [OK]

Hint: Check rectangles formed by popped bars for max area [OK]

Common Mistakes:

• Ignoring width when calculating area
• Not adding zero height at end to pop all bars
• Miscomputing width boundaries