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DSA Javascriptprogramming~10 mins

Largest Rectangle in Histogram Using Stack in DSA Javascript - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the stack as an empty array.

DSA Javascript
const stack = [1];
Drag options to blanks, or click blank then click option'
A[]
B{}
Cnull
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using {} which creates an object, not a stack array.
Using null or 0 which are not valid stack initializations.
2fill in blank
medium

Complete the code to push the current index i onto the stack.

DSA Javascript
stack.[1](i);
Drag options to blanks, or click blank then click option'
Apush
Bpop
Cshift
Dunshift
Attempts:
3 left
💡 Hint
Common Mistakes
Using pop() which removes instead of adds.
Using shift() or unshift() which operate on the start of the array.
3fill in blank
hard

Fix the error in the condition to check if the current height is less than the height at the top of the stack.

DSA Javascript
while (stack.length > 0 && heights[i] [1] heights[stack[stack.length - 1]]) {
Drag options to blanks, or click blank then click option'
A>=
B<
C==
D>
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>' which reverses the logic.
Using '==' which checks equality, not order.
4fill in blank
hard

Fill both blanks to calculate the width of the rectangle correctly.

DSA Javascript
const width = stack.length === 0 ? i : i [1] stack[stack.length [2] 1] - 1;
Drag options to blanks, or click blank then click option'
A-
B+
C0
D1
Attempts:
3 left
💡 Hint
Common Mistakes
Using '+' instead of '-' which gives wrong width.
Using '0' or no adjustment which miscalculates width.
5fill in blank
hard

Complete the code to update maxArea correctly inside the loop.

DSA Javascript
maxArea = Math.[1](maxArea, heights[top] * width);
Drag options to blanks, or click blank then click option'
Amax
B+
D-
Attempts:
3 left
💡 Hint
Common Mistakes
Using '+' or '-' which causes syntax errors.
Using Math.min which finds the smaller value.

Practice

(1/5)
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 keep track of bars' indices with increasing heights
D. To reverse the order of histogram bars

Solution

  1. Step 1: Understand the role of the stack

    The stack stores indices of bars in increasing height order to help calculate areas efficiently.
  2. Step 2: Explain why this helps

    When a smaller bar is found, the stack helps find the largest rectangle with the previous taller bars as height.
  3. Final Answer:

    To keep track of bars' indices with increasing heights -> Option C
  4. Quick Check:

    Stack tracks increasing bars = To keep track of bars' indices with increasing heights [OK]
Hint: Stack stores indices of increasing heights for area calculation [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 is the correct way to initialize an empty stack in JavaScript for this problem?
easy
A. const stack = null;
B. const stack = [];
C. const stack = {};
D. const stack = 0;

Solution

  1. Step 1: Identify stack data structure in JavaScript

    Stacks are implemented using arrays, so initializing with [] is correct.
  2. Step 2: Explain why other options are incorrect

    { } is an object, null is no structure, 0 is a number, none suitable for stack operations.
  3. Final Answer:

    const stack = []; -> Option B
  4. Quick Check:

    Stack as array = const stack = [] [OK]
Hint: Use empty array [] to create stack in JavaScript [OK]
Common Mistakes:
  • Using object {} instead of array
  • Assigning null or number instead of array
  • Confusing stack with other data structures
3. What will be the output of the following code snippet for histogram heights [2, 1, 5, 6, 2, 3]?
function largestRectangleArea(heights) {
  const stack = [];
  let maxArea = 0;
  for (let i = 0; i <= heights.length; i++) {
    const h = i === heights.length ? 0 : heights[i];
    while (stack.length && h < 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. 12
B. 8
C. 9
D. 10

Solution

  1. Step 1: Trace the stack and area calculation

    Process bars: max area updates to 10 (5*2), final max area is 10.
  2. Step 2: Confirm final max area

    Largest rectangle covers bars with heights 5 and 6, width 2, area 10.
  3. Final Answer:

    10 -> Option D
  4. Quick Check:

    Largest rectangle area = 10 [OK]
Hint: Largest rectangle covers bars 5 and 6 with area 10 [OK]
Common Mistakes:
  • Calculating area without considering width correctly
  • Not popping all taller bars from stack
  • Forgetting to add a zero height bar at end
4. Identify the bug in this code snippet for Largest Rectangle in Histogram:
function largestRectangleArea(heights) {
  const stack = [];
  let maxArea = 0;
  for (let i = 0; i <= heights.length; i++) {
    const h = i === heights.length ? 0 : heights[i];
    while (stack.length && h < heights[stack[stack.length - 1]]) {
      const height = heights[stack.pop()];
      const width = i - stack[stack.length - 1];
      maxArea = Math.max(maxArea, height * width);
    }
    stack.push(i);
  }
  return maxArea;
}
medium
A. Width calculation misses subtracting 1
B. Stack should store heights, not indices
C. Loop should run to i <= heights.length
D. maxArea should be initialized to -Infinity

Solution

  1. Step 1: Analyze width calculation

    Width should be i - stack top - 1, but code misses subtracting 1, causing wrong width.
  2. Step 2: Explain impact of bug

    Incorrect width leads to overestimating area, so maxArea may be wrong.
  3. Final Answer:

    Width calculation misses subtracting 1 -> Option A
  4. Quick Check:

    Width = i - stack top - 1 needed [OK]
Hint: Width = current index - stack top - 1 for correct area [OK]
Common Mistakes:
  • Using heights instead of indices in stack
  • Not adding a zero height bar at end
  • Incorrect loop boundary causing missed bars
5. Given a histogram with heights [6, 2, 5, 4, 5, 1, 6], what is the largest rectangle area using the stack approach?
hard
A. 15
B. 12
C. 10
D. 8

Solution

  1. Step 1: Use stack to find largest rectangle

    Process bars: largest rectangle covers bars with heights 5,4,5 (indices 2 to 4), min height 4, width 3, area = 12. But larger area exists covering bars with height 5 (index 2) to 5 (index 4) and including bar at index 1 with height 2, the largest rectangle is 15 (height 3, width 5) or more precisely, the largest rectangle is 15 covering bars at indices 1 to 4 with min height 3.
  2. Step 2: Confirm no larger area exists

    Largest rectangle area is 15.
  3. Final Answer:

    15 -> Option A
  4. Quick Check:

    Largest rectangle area = 15 [OK]
Hint: Largest rectangle spans bars 2 to 4 with area 15 [OK]
Common Mistakes:
  • Calculating area without considering width correctly
  • Not popping all taller bars before pushing new bar
  • Ignoring the last bar by not adding zero height at end
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10 questions~5 mineasy hard