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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

stack.[1](i);

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cpeek

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push causes removal instead of addition.

Using peek does not add the element.

Using empty is a check, not a modification.

We use stack.push(i) to add the current index to the stack.

2fill in blank

medium

Complete the code to calculate the width of the rectangle when popping from the stack.

DSA Java

int width = stack.isEmpty() ? i : i - stack.[1]() - 1;

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cpeek

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the element prematurely.

Using push() or empty() are incorrect for accessing the top element.

We use stack.peek() to get the top index without removing it, to calculate width.

3fill in blank

hard

Fix the error in the condition to pop from the stack when current height is less than stack top height.

DSA Java

while (!stack.isEmpty() && heights[i] < heights[stack.[1]()]) {

Drag options to blanks, or click blank then click option'

Apush

Bpop

Cempty

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() in the condition removes elements unexpectedly.

Using push() or empty() are invalid here.

We use stack.peek() to check the top index height without removing it.

4fill in blank

hard

Fill both blanks to calculate the area and update maxArea inside the while loop.

DSA Java

int height = heights[stack.[1]()];
int width = stack.isEmpty() ? i : i - stack.[2]() - 1;
maxArea = Math.max(maxArea, height * width);

Drag options to blanks, or click blank then click option'

Apeek

Bpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of pop() for height causes incorrect height calculation.

Using pop() for width calculation removes elements prematurely.

We pop the top index to get height, then peek the new top to calculate width.

5fill in blank

hard

Fill all three blanks to complete the final while loop that clears the stack after processing all bars.

DSA Java

while (!stack.isEmpty()) {
    int height = heights[stack.[1]()];
    stack.[2]();
    int width = stack.isEmpty() ? n : n - stack.[3]() - 1;
    maxArea = Math.max(maxArea, height * width);
}

Drag options to blanks, or click blank then click option'

Apeek

Bpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() before getting height causes errors.

Using push() instead of pop() or peek() is incorrect.

Peek to get height index, pop to remove it, peek again to calculate width.

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 sort the histogram bars by height

C. To store the maximum area found so far

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 efficiently find boundaries for rectangles.

2. Step 2: Why increasing order is important

   When a lower height bar is found, it signals the end of rectangles for taller bars, allowing area calculation.

3. Final Answer:

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

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 directly
• Ignoring the order of bars in stack

2. Which of the following Java code snippets correctly initializes an empty stack to store integers for the histogram problem?

easy

A. Stack<Integer> stack = new Stack<int>();

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

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

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

Solution

1. Step 1: Identify correct generic syntax in Java

   Java uses generics with angle brackets and wrapper classes like Integer, not primitive int.

2. Step 2: Check each option

   Stack<Integer> stack = new Stack<>(); uses Stack<Integer> stack = new Stack<>(); which is correct syntax for an empty integer stack.

3. Final Answer:

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

4. Quick Check:

   Correct generic syntax = Stack<Integer> stack = new Stack<>(); [OK]

Hint: Use wrapper class Integer with generics and diamond operator [OK]

Common Mistakes:

• Using primitive int in generics
• Missing diamond operator <>
• Not specifying generic type

3. What is the output of the following Java code snippet that calculates the largest rectangle area in histogram heights = {2, 1, 5, 6, 2, 3}?

public class Histogram {
    public static int largestRectangleArea(int[] heights) {
        Stack<Integer> stack = new Stack<>();
        int maxArea = 0, i = 0;
        while (i < heights.length) {
            if (stack.isEmpty() || heights[i] >= heights[stack.peek()]) {
                stack.push(i++);
            } else {
                int top = stack.pop();
                int area = heights[top] * (stack.isEmpty() ? i : i - stack.peek() - 1);
                maxArea = Math.max(maxArea, area);
            }
        }
        while (!stack.isEmpty()) {
            int top = stack.pop();
            int area = heights[top] * (stack.isEmpty() ? i : i - stack.peek() - 1);
            maxArea = Math.max(maxArea, area);
        }
        return maxArea;
    }

    public static void main(String[] args) {
        int[] heights = {2, 1, 5, 6, 2, 3};
        System.out.println(largestRectangleArea(heights));
    }
}

medium

A. 12

B. 10

C. 8

D. 6

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 value

   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 of heights[top] * width during stack pops [OK]

Common Mistakes:

• Miscomputing width when stack is empty
• Not popping all remaining bars after loop
• Confusing indices with heights

4. Identify the bug in the following Java method for Largest Rectangle in Histogram and select the fix:

public static int largestRectangleArea(int[] heights) {
    Stack<Integer> stack = new Stack<>();
    int maxArea = 0, i = 0;
    while (i < heights.length) {
        if (stack.isEmpty() || heights[i] > heights[stack.peek()]) {
            stack.push(i++);
        } else {
            int top = stack.pop();
            int area = heights[top] * (stack.isEmpty() ? i : i - stack.peek() - 1);
            maxArea = Math.max(maxArea, area);
        }
    }
    while (!stack.isEmpty()) {
        int top = stack.pop();
        int area = heights[top] * (stack.isEmpty() ? i : i - stack.peek() - 1);
        maxArea = Math.max(maxArea, area);
    }
    return maxArea;
}

medium

A. Initialize maxArea to Integer.MIN_VALUE instead of 0

B. Replace stack with a queue for correct order

C. Change '>' to '>=' in the if condition to handle equal heights

D. Remove the second while loop to avoid double counting

Solution

1. Step 1: Analyze the if condition in the main loop

   The condition uses '>' which skips pushing bars of equal height, causing incorrect area calculation.

2. Step 2: Fix by changing '>' to '>='

   Using '>=' ensures bars of equal height are pushed, maintaining correct stack order and area calculation.

3. Final Answer:

   Change '>' to '>=' in the if condition to handle equal heights -> Option C

4. Quick Check:

   Use '>=' to handle equal heights correctly [OK]

Hint: Use '>=' to push equal height bars to stack [OK]

Common Mistakes:

• Using '>' causes missing equal height bars
• Removing second loop causes incomplete area calculation
• Using queue breaks LIFO needed for stack

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 as smallest bar.

2. Step 2: Calculate largest area for given heights

   For {6,2,5,4,5,1,6}, largest rectangle area is 15 (bars at indices 0 and 6 with height 6 and width 1, or bars 2,3,4 with height 4 and width 3, but max is 15 from bars 0 to 0 or 6 to 6 or combined area 15 from bars 2,3,4 with height 5 and width 3).

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Stack helps find max area = 15 [OK]

Hint: Stack pops calculate max area using height * width [OK]

Common Mistakes:

• Not popping all bars after iteration
• Incorrect width calculation when stack is empty
• Ignoring bars with same height