Largest Rectangle in Histogram Using Stack in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a stack to find the largest rectangle in a histogram?

Use a stack to keep track of bars in increasing height order. When a smaller bar appears, calculate areas with bars popped from the stack as the smallest bar, ensuring all rectangles are considered.

beginner

Why do we push the index of bars onto the stack instead of their heights?

Storing indices allows us to calculate the width of rectangles easily by comparing current index with the index on top of the stack after popping.

intermediate

What happens when we encounter a bar shorter than the bar at the top of the stack?

We pop bars from the stack and calculate the area for each popped bar using its height and the width determined by current index and the new top of the stack.

intermediate

How do we handle remaining bars in the stack after processing all histogram bars?

Pop all remaining bars and calculate areas assuming the right boundary is the end of the histogram, updating the maximum area accordingly.

beginner

What is the time complexity of the largest rectangle in histogram algorithm using a stack?

O(n), because each bar is pushed and popped at most once from the stack.

What data structure is primarily used to solve the largest rectangle in histogram problem efficiently?

A Stack

B Queue

C Linked List

D Hash Map

A stack is used to keep track of bars in increasing height order to calculate areas efficiently.

When do we calculate the area of a rectangle in the stack-based approach?

A When a shorter bar is encountered

B When a taller bar is encountered

C At the end only

D Before pushing any bar

We calculate area when a shorter bar appears, popping taller bars from the stack to find max areas.

What does the width of the rectangle depend on when popping a bar from the stack?

A Difference between current index and popped bar's index

B Height of the popped bar

C Sum of all bar heights

D Difference between current index and new top of stack index minus one

Width is calculated as current index minus the index of the new top of the stack minus one.

What is the maximum number of times a bar can be pushed or popped from the stack?

A Never

B Twice

C Once

D Multiple times

Each bar is pushed and popped at most once, ensuring O(n) time complexity.

What do we do after processing all bars in the histogram?

A Ignore remaining bars in stack

B Pop remaining bars and calculate areas

C Push a zero height bar

D Restart the process

We pop all remaining bars to calculate areas considering the histogram's end as the right boundary.

Explain step-by-step how the stack helps find the largest rectangle in a histogram.

Think about how the stack keeps track of bars and when areas are calculated.

Describe how to calculate the width of the rectangle when popping a bar from the stack.

Consider what the stack top represents after popping.

Practice

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

easy

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

B. To store the heights of all bars directly

C. To sort the histogram bars by height

D. To reverse the order of histogram bars

Solution

1. Step 1: Understand stack usage in the problem

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

2. Step 2: Why store indices, not heights?

   Indices allow calculation of width between bars when popping from the stack, which is essential for area calculation.

3. Final Answer:

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

4. Quick Check:

   Stack stores indices for area calculation [OK]

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

Common Mistakes:

• Storing heights instead of indices
• Using stack to sort bars
• Ignoring width calculation using indices

2. Which of the following C# code snippets correctly initializes a stack to store integers for the histogram problem?

easy

A. int[] stack = new int[10];

B. Stack stack = new Stack();

C. List<int> stack = new List<int>();

D. Stack<int> stack = new Stack<int>();

Solution

1. Step 1: Identify correct stack initialization in C#

   Using generic Stack<int> is the modern and type-safe way to create a stack for integers.

2. Step 2: Why not others?

   The non-generic Stack (like "Stack stack = new Stack();") uses the older style without type safety. A List (like "List<int> stack = new List<int>();") is not a stack. An array (like "int[] stack = new int[10];") is not a stack.

3. Final Answer:

   Stack<int> stack = new Stack<int>(); -> Option D

4. Quick Check:

   Generic Stack<int> is correct stack initialization [OK]

Hint: Use Stack<int> for integer stacks in C# [OK]

Common Mistakes:

• Using non-generic Stack without type safety
• Using List instead of Stack
• Using arrays instead of stack

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

int[] heights = {2, 1, 5, 6, 2, 3};
int largestRectangleArea(int[] heights) {
    Stack<int> stack = new Stack<int>();
    int maxArea = 0, i = 0;
    while (i < heights.Length) {
        if (stack.Count == 0 || heights[i] >= heights[stack.Peek()]) {
            stack.Push(i++);
        } else {
            int top = stack.Pop();
            int width = stack.Count == 0 ? i : i - stack.Peek() - 1;
            maxArea = Math.Max(maxArea, heights[top] * width);
        }
    }
    while (stack.Count > 0) {
        int top = stack.Pop();
        int width = stack.Count == 0 ? i : i - stack.Peek() - 1;
        maxArea = Math.Max(maxArea, heights[top] * width);
    }
    return maxArea;
}

Console.WriteLine(largestRectangleArea(heights));

medium

A. 12

B. 10

C. 15

D. 9

Solution

1. Step 1: Trace the stack operations and area calculations

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

2. Step 2: Confirm final maxArea

   After processing all bars and stack cleanup, maxArea is 10.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Largest rectangle area = 10 [OK]

Hint: Largest area is max height * width using stack [OK]

Common Mistakes:

• Miscomputing width when popping stack
• Forgetting to process remaining stack after loop
• Mixing up indices and heights

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

int largestRectangleArea(int[] heights) {
    Stack<int> stack = new Stack<int>();
    int maxArea = 0, i = 0;
    while (i <= heights.Length) {
        int currentHeight = (i == heights.Length) ? 0 : heights[i];
        if (stack.Count == 0 || currentHeight >= heights[stack.Peek()]) {
            stack.Push(i++);
        } else {
            int top = stack.Pop();
            int width = stack.Count == 0 ? i : i - stack.Peek() - 1;
            maxArea = Math.Max(maxArea, heights[top] * width);
        }
    }
    return maxArea;
}

medium

A. Width calculation misses subtracting 1, causing wrong area

B. Stack should store heights, not indices

C. Loop condition should be i < heights.Length, not <=

D. currentHeight assignment is incorrect

Solution

1. Step 1: Check width calculation formula

   Width should be i - stack.Peek() - 1 when stack is not empty, but code misses '-1'.

2. Step 2: Effect of missing '-1'

   Missing '-1' causes width to be larger than actual, leading to incorrect maxArea.

3. Final Answer:

   Width calculation misses subtracting 1, causing wrong area -> Option A

4. Quick Check:

   Width must subtract 1 for correct area [OK]

Hint: Subtract 1 in width calculation after popping stack [OK]

Common Mistakes:

• Forgetting to subtract 1 in width
• Using stack for heights instead of indices
• Incorrect loop boundary causing index errors

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. 15

B. 10

C. 12

D. 8

Solution

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

   Push indices while heights increase; pop when current height is less, calculating area with popped height and width between indices.

2. Step 2: Calculate largest area for given heights

   For [6,2,5,4,5,1,6], largest rectangle area is 12 (from bars at indices 2,3,4 with min height 4 and width 3, calculated when popping index 3).

3. Final Answer:

   12 -> Option C

4. Quick Check:

   Stack helps find max area = 12 [OK]

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

Common Mistakes:

• Not popping all remaining bars after loop
• Incorrect width calculation
• Confusing heights and indices in stack