Largest Rectangle in Histogram Using Stack in DSA Python - Time & Space Complexity

Time Complexity: Largest Rectangle in Histogram Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the largest rectangle in a histogram changes as the number of bars grows.

How does the algorithm's speed change when the histogram gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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
    

This code finds the largest rectangle area in a histogram by using a stack to track bars and calculate areas efficiently.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop goes through each bar once, and inside it, bars are popped from the stack when they are taller than the current bar.
• How many times: Each bar is pushed and popped at most once, so the total number of stack operations is proportional to the number of bars.

How Execution Grows With Input

As the number of bars (n) increases, the algorithm processes each bar a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 stack operations (push and pop combined)
100	About 200 stack operations
1000	About 2000 stack operations

Pattern observation: The number of operations grows roughly in direct proportion to the number of bars.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the largest rectangle grows linearly with the number of bars in the histogram.

Common Mistake

[X] Wrong: "Because there is a nested while loop inside the for loop, the time complexity must be O(n²)."

[OK] Correct: Each bar is pushed and popped only once, so the inner loop does not run for every iteration of the outer loop, keeping the total operations linear.

Interview Connect

Understanding this stack-based approach shows your ability to use data structures smartly to solve problems efficiently, a skill highly valued in coding interviews.

Self-Check

What if we removed the extra 0 added at the end of the heights list? How would the time complexity change?

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