Why Largest Rectangle in Histogram Using Stack in DSA Go?

The Big Idea

Discover how a simple stack can save you hours of tedious work finding the biggest rectangle!

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 of buildings one by one is like measuring every possible rectangle manually, which takes a lot of time.

The Problem

Manually checking all groups means trying every start and end point, calculating the smallest building height in that group, and then the area. This is very slow and tiring, especially if there are many buildings. It's easy to make mistakes and miss the biggest area.

The Solution

Using a stack helps us keep track of building heights in a smart way. We can quickly find where a building's height stops being the smallest and calculate the largest rectangle possible without checking every group. This saves a lot of time and effort.

Before vs After

✗ Before

maxArea := 0
for i := 0; i < len(heights); i++ {
  for j := i; j < len(heights); j++ {
    minHeight := heights[i]
    for k := i; k <= j; k++ {
      if heights[k] < minHeight {
        minHeight = heights[k]
      }
    }
    area := minHeight * (j - i + 1)
    if area > maxArea {
      maxArea = area
    }
  }
}

✓ After

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

Think of a city planner who wants to find the biggest flat area in a row of buildings to build a park or a billboard. Using this stack method, they can quickly find the best spot without checking every possible area manually.

Key Takeaways

Manually checking all rectangles is slow and error-prone.

Using a stack helps track building heights smartly to find the largest rectangle fast.

This approach saves time and works well even for large histograms.

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