Largest Rectangle in Histogram Using Stack in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the stack for indices.

DSA Go

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

Drag options to blanks, or click blank then click option'

Ai

Bheights[i]

CmaxArea

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Appending the height value instead of the index.

Appending the maxArea variable instead of index.

We append the current index 'i' to the stack to keep track of histogram bars.

2fill in blank

medium

Complete the code to calculate the width when the stack is not empty.

DSA Go

width := i
if len(stack) > 0 {
    width = i - [1] - 1
}

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Astack[len(stack)-1]

Btop

CmaxArea

Dheights[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using the popped top index instead of the new top of the stack.

Using height values instead of indices.

Width is calculated using the index at the top of the stack after popping.

3fill in blank

hard

Fix the error in the final loop to process remaining bars in the stack.

DSA Go

n := len(heights)
for len(stack) > 0 {
    top := stack[len(stack)-1]
    stack = stack[:len(stack)-1]
    width := n
    if len(stack) > 0 {
        width = n - [1] - 1
    }
    area := heights[top] * width
    if area > maxArea {
        maxArea = area
    }
}

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Atop

Bstack[len(stack)-1]

CmaxArea

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using the popped index 'top' instead of the new stack top.

Using the total length 'n' directly without adjustment.

Width is calculated using the new top of the stack after popping, not the popped top.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA Go

func [1](heights []int) int {
    maxArea := 0
    stack := make([]int, 0)
    for i := 0; i < len(heights); i++ {
        for len(stack) > 0 && heights[stack[len(stack)-1]] > heights[i] {
            top := stack[len(stack)-1]
            stack = stack[:len(stack)-1]
            width := i
            if len(stack) > 0 {
                width = i - stack[len(stack)-1] - 1
            }
            area := heights[top] * width
            if area > maxArea {
                maxArea = area
            }
        }
        stack = append(stack, i)
    }
    n := len(heights)
    for len(stack) > 0 {
        top := stack[len(stack)-1]
        stack = stack[:len(stack)-1]
        width := n
        if len(stack) > 0 {
            width = n - stack[len(stack)-1] - 1
        }
        area := heights[top] * width
        if area > maxArea {
            maxArea = area
        }
    }
    return [2]
}

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AlargestRectangleArea

BmaxArea

Cheights

Darea

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the wrong variable like 'area' or 'heights'.

Using a generic function name that doesn't match the task.

The function is named 'largestRectangleArea' and returns the maximum area found.

5fill in blank

hard

Fill all three blanks to complete the code that calculates area inside the loop.

DSA Go

for len(stack) > 0 && heights[stack[len(stack)-1]] > heights[i] {
    top := stack[len(stack)-1]
    stack = stack[:len(stack)-1]
    width := [1]
    if len(stack) > 0 {
        width = i - [2] - 1
    }
    area := heights[[3]] * width
    if area > maxArea {
        maxArea = area
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bstack[len(stack)-1]

Ctop

Dlen(stack)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'len(stack)' instead of the top index for width calculation.

Using wrong index for height in area calculation.

Width starts as current index 'i', adjusted by the new top of stack; area uses height at 'top' index.

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