Why Largest Rectangle in Histogram Using Stack in DSA Java?

The Big Idea

What if you could find the biggest flat space among many buildings without checking every possible group?

The Scenario

Imagine you have a row of buildings with different heights, and you want to find the largest flat area you can cover by choosing some consecutive buildings. Doing this by checking every possible group one by one is like measuring every possible patch of land manually, which takes a lot of time.

The Problem

Manually checking all groups of buildings means trying every start and end point, which grows very fast as the number of buildings increases. This is slow and tiring, and easy to make mistakes when counting or comparing areas.

The Solution

Using a stack helps keep track of building heights in a smart way. It lets you quickly find the biggest rectangle by remembering where heights go up and down, so you don't have to check every group. This saves time and avoids errors.

Before vs After

✗ Before

int maxArea = 0;
for (int start = 0; start < heights.length; start++) {
  int minHeight = heights[start];
  for (int end = start; end < heights.length; end++) {
    minHeight = Math.min(minHeight, heights[end]);
    maxArea = Math.max(maxArea, minHeight * (end - start + 1));
  }
}

✓ After

Stack<Integer> stack = new Stack<>();
int maxArea = 0;
for (int i = 0; i <= heights.length; i++) {
  int currentHeight = (i == heights.length) ? 0 : heights[i];
  while (!stack.isEmpty() && currentHeight < heights[stack.peek()]) {
    int height = heights[stack.pop()];
    int width = stack.isEmpty() ? i : i - stack.peek() - 1;
    maxArea = Math.max(maxArea, height * width);
  }
  stack.push(i);
}

What It Enables

This method lets you find the largest rectangle in a histogram quickly and efficiently, even for very tall and many buildings.

Real Life Example

Architects or city planners can use this to find the largest possible flat area for building or placing solar panels on uneven rooftops.

Key Takeaways

Manual checking of all groups is slow and error-prone.

Stack helps track building heights to find largest rectangle fast.

This approach saves time and works well for large inputs.

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