Largest Rectangle in Histogram Using Stack in DSA Typescript - 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 lower height bar appears, calculate areas with bars popped from the stack as the smallest height, ensuring all rectangles are considered.

beginner

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

Storing indices allows us to calculate the width of rectangles easily by subtracting indices when popping bars, which is essential for area calculation.

intermediate

What triggers the calculation of rectangle areas in the stack-based approach?

When the current bar's height is less than the height of the bar at the top of the stack, we pop bars and calculate areas because the current bar limits the rectangle width for those taller bars.

intermediate

How do you calculate the width of the rectangle when popping a bar from the stack?

Width = current index - index of new top of stack - 1. If the stack is empty after popping, width = current index.

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 pop bars from the stack during the algorithm?

A When the stack is empty

B When the current bar is taller than the top of the stack

C When the current bar is shorter than the top of the stack

D At the end of the array only

We pop bars when the current bar is shorter, indicating the end of rectangles with those heights.

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 Number of bars in the stack

C Height of the popped bar

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 and popped in this algorithm?

A Multiple times

B Once

C Twice

D Never popped

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

What happens after processing all bars in the histogram array?

A The stack is cleared by popping remaining bars to calculate areas

B The algorithm ends immediately

C Bars are pushed again

D No further action is needed

Remaining bars in the stack are popped to calculate areas for rectangles extending to the end.

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 to calculate areas.

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

Width depends on indices around the popped bar.

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 histogram bars

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

C. To store the maximum area found so far

D. To sort the histogram bars by height

Solution

1. Step 1: Understand the stack's role

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

2. Step 2: Why increasing heights?

   This helps identify when a bar is lower than the previous, signaling to calculate area for taller bars.

3. Final Answer:

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

4. Quick Check:

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

Hint: Stack holds indices of increasing bars to find rectangle widths [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 TypeScript code snippets correctly initializes an empty stack for indices in the Largest Rectangle problem?

easy

A. const stack = new Stack<number>();

B. const stack: string[] = [];

C. const stack: number[] = [];

D. const stack = []; stack.push('index');

Solution

1. Step 1: Identify correct stack type

   We need a stack of numbers (indices), so number[] is correct.

2. Step 2: Check other options

   const stack: string[] = []; uses string[], wrong type. const stack = new Stack<number>(); uses a Stack class which is not standard in TypeScript. const stack = []; stack.push('index'); pushes a string 'index', wrong type.

3. Final Answer:

   const stack: number[] = []; -> Option C

4. Quick Check:

   Stack of numbers = const stack: number[] = []; [OK]

Hint: Use number[] for stack of indices in TypeScript [OK]

Common Mistakes:

• Using string[] instead of number[]
• Assuming Stack class exists by default
• Pushing wrong data types into stack

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

function largestRectangleArea(heights: number[]): number {
  const stack: number[] = [];
  let maxArea = 0;
  heights.push(0);
  for (let i = 0; i < heights.length; i++) {
    while (stack.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);
  }
  return maxArea;
}

console.log(largestRectangleArea([2, 1, 5, 6, 2, 3]));

medium

A. 10

B. 12

C. 8

D. 6

Solution

1. Step 1: Add sentinel and iterate

   Heights become [2,1,5,6,2,3,0]. We use stack to track indices of increasing bars.

2. Step 2: Calculate max area during pops

   Pop bars when current height is less, calculate area: max area found is 10 (from bars 5 and 6 with width 2).

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max rectangle area = 10 [OK]

Hint: Add 0 height at end to flush stack and find max area [OK]

Common Mistakes:

• Not adding sentinel 0 at end
• Wrong width calculation after popping
• Ignoring last bars in stack

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

function largestRectangleArea(heights: number[]): number {
  const stack: number[] = [];
  let maxArea = 0;
  for (let i = 0; i <= heights.length; i++) {
    while (stack.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);
  }
  return maxArea;
}

medium

A. maxArea is not updated inside the loop

B. Stack is not initialized properly

C. Width calculation is incorrect

D. Loop goes out of bounds accessing heights[i] when i == heights.length

Solution

1. Step 1: Check loop boundary

   The loop runs i from 0 to heights.length inclusive, so heights[i] is accessed when i == heights.length, causing out-of-bounds error.

2. Step 2: Fix by adding sentinel or changing loop

   Adding a 0 height at end or changing loop to i < heights.length fixes this.

3. Final Answer:

   Loop goes out of bounds accessing heights[i] when i == heights.length -> Option D

4. Quick Check:

   Out-of-bounds access = Loop goes out of bounds accessing heights[i] when i == heights.length [OK]

Hint: Avoid accessing heights[i] when i equals length without sentinel [OK]

Common Mistakes:

• Ignoring loop boundary causing runtime error
• Assuming stack initialization is wrong
• Misreading width calculation

5. Given a histogram with heights [6, 2, 5, 4, 5, 1, 6], what is the largest rectangle area using the stack approach? Explain why the rectangle with area 12 is the largest.

hard

A. 12

B. 10

C. 15

D. 8

Solution

1. Step 1: Use stack to find max area

   Push indices while heights increase. When a lower bar appears, pop and calculate area. For this histogram, max area is 12 from bars at indices 2,3,4 (heights 5,4,5) width 3.

2. Step 2: Why area 12 is largest

   The rectangle covers bars with heights >=4 over width 3, area = 4*3=12. No other rectangle covers more area.

3. Final Answer:

   12 -> Option A

4. Quick Check:

   Max rectangle area = 12 [OK]

Hint: Pop stack on lower bar, calculate area with width between indices [OK]

Common Mistakes:

• Calculating area only for single bars
• Ignoring width when popping multiple bars
• Missing the largest rectangle spanning multiple bars