Largest Rectangle in Histogram Using Stack in DSA Java - 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 indices of bars onto the stack instead of their heights?

Indices allow us to calculate the width of rectangles easily by knowing the position of bars, which helps in computing the area of rectangles between bars.

intermediate

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

When the current bar's height is less than the height of the bar at the top of the stack, we pop from the stack and calculate the area for the popped bar as the smallest height rectangle.

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 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 Queue

B Linked List

C Stack

D Hash Map

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

When do we pop bars from the stack during the algorithm?

A At the end of the array only

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

C When the stack is empty

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

We pop bars when the current bar is shorter to calculate areas for bars taller than the current one.

What does the stack store in this algorithm?

A Indices of bars

B Heights of bars

C Areas of rectangles

D Widths of rectangles

The stack stores indices to help calculate widths and areas.

If the stack is empty after popping a bar, how is the width calculated?

A Width = 0

B Width = current index

C Width = current index - 1

D Width = 1

If stack is empty, width is the current index because the rectangle extends from start.

What is the overall time complexity of this stack-based solution?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

Each bar is pushed and popped once, so the time complexity is linear O(n).

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.

Width depends on positions of bars 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 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