Largest Rectangle in Histogram Using Stack in DSA Java - Build from Scratch

Largest Rectangle in Histogram Using Stack

📖 Scenario: Imagine you have a row of buildings of different heights. You want to find the largest rectangular area that can fit under the skyline formed by these buildings.

🎯 Goal: Build a Java program that uses a stack to find the largest rectangle area in a histogram represented by an array of building heights.

📋 What You'll Learn

Create an integer array called heights with the exact values {2, 1, 5, 6, 2, 3}.

Create an empty stack of integers called stack to store indices.

Use a for loop with variable i to iterate over heights and apply the stack logic to find the largest rectangle.

Print the largest rectangle area as an integer.

💡 Why This Matters

🌍 Real World

Finding the largest rectangle in a histogram is useful in image processing, stock market analysis, and land area calculations.

💼 Career

This problem is a common interview question to test understanding of stacks and array manipulation.

1

Create the histogram data

Create an integer array called heights with these exact values: {2, 1, 5, 6, 2, 3}.

DSA Java

// Create the histogram heights array
// Your code here

Hint

Use int[] heights = {2, 1, 5, 6, 2, 3}; to create the array.

2

Set up the stack and max area variable

Create a Stack<Integer> called stack and an integer variable called maxArea initialized to 0.

DSA Java

int[] heights = {2, 1, 5, 6, 2, 3};
// Create stack and maxArea variable
// Your code here

Hint

Use Stack<Integer> stack = new Stack<>(); and int maxArea = 0;.

3

Implement the stack logic to find largest rectangle

Use a for loop with variable i from 0 to heights.length inclusive. Inside the loop, use a while loop to pop from stack when current height is less than height at top of stack. Calculate area using popped height and update maxArea. Push i to stack when appropriate.

DSA Java

import java.util.Stack;

int[] heights = {2, 1, 5, 6, 2, 3};
Stack stack = new Stack();
int maxArea = 0;

// Implement the main logic to find largest rectangle
// Your code here

Hint

Remember to handle the case when i == heights.length by setting currentHeight to 0 to clear the stack at the end.

4

Print the largest rectangle area

Print the value of maxArea using System.out.println(maxArea);.

DSA Java

import java.util.Stack;

int[] heights = {2, 1, 5, 6, 2, 3};
Stack stack = new Stack();
int maxArea = 0;

for (int i = 0; i  maxArea) {
            maxArea = area;
        }
    }
    stack.push(i);
}

// Print the largest rectangle area
// Your code here

Hint

Use System.out.println(maxArea); to display the result.

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