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

Largest Rectangle in Histogram Using Stack

📖 Scenario: Imagine you have a row of buildings with different heights. You want to find the largest rectangular area that can fit inside these buildings when viewed from the side.

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

📋 What You'll Learn

Create a slice called heights with exact values: 2, 1, 5, 6, 2, 3

Create an integer variable maxArea to store the largest rectangle area

Use a stack (slice of integers) to help find the largest rectangle

Implement the logic to calculate the largest rectangle area using the stack

Print the final maxArea value

💡 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 teaches stack usage and efficient area calculation, common in technical interviews and real-world algorithm design.

1

Create the histogram heights slice

Create a slice of integers called heights with these exact values: 2, 1, 5, 6, 2, 3.

DSA Go

// Create the heights slice with the given values
// Your code here

Hint

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

2

Initialize maxArea and stack

Create an integer variable called maxArea and set it to 0. Also create an empty slice of integers called stack to use as a stack.

DSA Go

package main

func main() {
    heights := []int{2, 1, 5, 6, 2, 3}
    // Create maxArea variable and stack slice
    // Your code here
}

Hint

Use maxArea := 0 and stack := []int{}.

3

Implement the stack logic to find largest rectangle

Write a for loop with variable i from 0 to len(heights) inclusive. Inside the loop, use a for loop to pop from stack while stack is not empty and the current height is less than the height at the top of stack. Calculate the area for each popped height and update maxArea if the area is larger. After the inner loop, append i to stack. Use heights with an extra 0 height at the end for the loop.

DSA Go

package main

func main() {
    heights := []int{2, 1, 5, 6, 2, 3}
    maxArea := 0
    stack := []int{}

    // Implement the stack logic to find the largest rectangle
    // Your code here
}

Hint

Use a loop over extendedHeights with an extra 0 at the end. Pop from stack while current height is less than top of stack height. Calculate area and update maxArea. Append i to stack.

4

Print the largest rectangle area

Write a fmt.Println(maxArea) statement to print the largest rectangle area found.

DSA Go

package main

import "fmt"

func main() {
    heights := []int{2, 1, 5, 6, 2, 3}
    maxArea := 0
    stack := []int{}

    extendedHeights := append(heights, 0)
    for i := 0; i  0 && extendedHeights[i]  maxArea {
                maxArea = area
            }
        }
        stack = append(stack, i)
    }

    // Print the maxArea
    // Your code here
}

Hint

Use fmt.Println(maxArea) to print the result.

Practice

1. What is the main purpose of using a stack in the Largest Rectangle in Histogram problem?

easy

A. To sort the histogram bars by height

B. To store the maximum area found so far

C. To keep track of bars' indices with increasing heights

D. To reverse the order of histogram bars

Solution

1. Step 1: Understand stack usage

   The stack stores indices of bars in increasing height order to find boundaries efficiently.

2. Step 2: Purpose in algorithm

   This helps quickly calculate the largest rectangle when a smaller height bar is found.

3. Final Answer:

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

4. Quick Check:

   Stack tracks increasing bars = B [OK]

Hint: Stack stores indices of increasing heights [OK]

Common Mistakes:

• Thinking stack sorts bars by height
• Using stack to store areas instead of indices
• Ignoring the order of bars in stack

2. Which Go syntax correctly declares a stack to hold indices for the histogram bars?

easy

A. var stack []int

B. stack := make(map[int]int)

C. stack := []string{}

D. var stack int

Solution

1. Step 1: Identify stack type

   The stack holds indices, which are integers, so it should be a slice of int.

2. Step 2: Check syntax correctness

   "var stack []int" declares an empty slice of int, suitable for stack operations.

3. Final Answer:

   var stack []int -> Option A

4. Quick Check:

   Stack as []int slice = C [OK]

Hint: Use slice of int for stack indices [OK]

Common Mistakes:

• Using map instead of slice for stack
• Declaring stack as int instead of slice
• Using string slice for integer indices

3. What is the output of the following Go code snippet for histogram heights [2, 1, 5, 6, 2, 3]?

func largestRectangleArea(heights []int) int {
    stack := []int{}
    maxArea := 0
    for i := 0; i <= len(heights); i++ {
        var h int
        if i == len(heights) {
            h = 0
        } else {
            h = heights[i]
        }
        for len(stack) > 0 && h < heights[stack[len(stack)-1]] {
            height := heights[stack[len(stack)-1]]
            stack = stack[:len(stack)-1]
            var width int
            if len(stack) == 0 {
                width = i
            } else {
                width = i - stack[len(stack)-1] - 1
            }
            area := height * width
            if area > maxArea {
                maxArea = area
            }
        }
        stack = append(stack, i)
    }
    return maxArea
}

fmt.Println(largestRectangleArea([]int{2, 1, 5, 6, 2, 3}))

medium

A. 6

B. 12

C. 8

D. 10

Solution

1. Step 1: Trace stack and area calculation

   Process bars: max area found is 10 (from bars 5 and 6 with width 2).

2. Step 2: Confirm max area

   No larger rectangle found; 10 is the largest rectangle area.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Largest rectangle area = 10 [OK]

Hint: Largest area is max of all popped rectangles [OK]

Common Mistakes:

• Forgetting to add a zero height bar at end
• Incorrect width calculation after popping stack
• Not updating maxArea correctly

4. Identify the bug in this Go code snippet for Largest Rectangle in Histogram:

for i := 0; i < len(heights); i++ {
    for len(stack) > 0 && heights[i] < heights[stack[len(stack)-1]] {
        height := heights[stack[len(stack)-1]]
        stack = stack[:len(stack)-1]
        width := i
        if len(stack) > 0 {
            width = i - stack[len(stack)-1] - 1
        }
        area := height * width
        if area > maxArea {
            maxArea = area
        }
    }
    stack = append(stack, i)
}
return maxArea

medium

A. Loop should run until i <= len(heights) with a sentinel zero height

B. Width calculation is incorrect when stack is empty

C. Stack should store heights, not indices

D. maxArea should be reset inside the loop

Solution

1. Step 1: Check loop boundary

   The loop misses processing the last bars fully without a sentinel zero height at the end.

2. Step 2: Fix loop to include sentinel

   Loop should run until i <= len(heights) and treat i == len(heights) as height 0 to pop all remaining bars.

3. Final Answer:

   Loop should run until i <= len(heights) with a sentinel zero height -> Option A

4. Quick Check:

   Loop boundary with sentinel zero fixes bug [OK]

Hint: Add sentinel zero height to process all bars [OK]

Common Mistakes:

• Not adding sentinel zero height
• Incorrect width calculation when stack empty
• Resetting maxArea inside loop incorrectly

5. Given a histogram with heights [6, 2, 5, 4, 5, 1, 6], what is the largest rectangle area using the stack method?

hard

A. 12

B. 15

C. 8

D. 10

Solution

1. Step 1: Analyze bars and stack operations

   Largest rectangle is formed by bars with min height 5 spanning width 3 (indices 2 to 4: heights 5,4,5), area = 5*3 = 15.

2. Step 2: Confirm no larger area

   Other rectangles like 4*3=12 or 2*5=10 are smaller; 15 is maximum.

3. Final Answer:

   15 -> Option B

4. Quick Check:

   Max rectangle area = 15 [OK]

Hint: Check width between popped bars for max area [OK]

Common Mistakes:

• Calculating width incorrectly
• Missing wider rectangle spanning multiple bars
• Not popping all bars at the end