Largest Rectangle in Histogram Using Stack in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a stack to find the largest rectangle in a histogram?

Use a stack to keep track of bars in increasing height order. When a smaller bar appears, calculate areas with bars popped from the stack as the smallest bar until the stack is in order again.

beginner

Why do we push the index of bars onto the stack instead of their heights?

Storing indices helps calculate the width of rectangles easily by using the current index and the index of the previous bar in the stack.

intermediate

What happens when we encounter a bar shorter than the bar at the top of the stack?

We pop bars from the stack and calculate the area for each popped bar using its height and the width determined by the current index and the new top of the stack.

intermediate

How do we handle remaining bars in the stack after processing all histogram bars?

Pop all remaining bars and calculate their areas using the total length of the histogram as the right boundary.

beginner

What is the time complexity of the largest rectangle in histogram algorithm using a stack?

O(n), because each bar is pushed and popped at most once.

What data structure is used to solve the largest rectangle in histogram problem efficiently?

A Hash Map

B Queue

C Linked List

D Stack

A stack is used to keep track of bars in increasing height order for efficient area calculation.

When do we calculate the area of a rectangle in the stack-based approach?

A When a taller bar is found

B When a shorter bar is found

C At the end only

D Before pushing any bar

We calculate area when a shorter bar appears, popping taller bars from the stack.

What does the width of the rectangle depend on when popping from the stack?

A Difference between current index and new top of stack index minus one

B Difference between current index and popped index

C Height of the popped bar

D Sum of all bar heights

Width is calculated as current index minus the index of the new top of the stack minus one.

What is the maximum number of times a bar can be pushed or popped in this algorithm?

A Multiple times

B Never popped

C Once

D Twice

Each bar is pushed once and popped once, so at most twice involved in stack operations.

What is the overall time complexity of the stack-based largest rectangle in histogram algorithm?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

The algorithm runs in O(n) time because each bar is pushed and popped at most once.

Explain step-by-step how the stack helps find the largest rectangle in a histogram.

Think about how the stack keeps track of bars and when you calculate areas.

Describe how to calculate the width of the rectangle when popping a bar from the stack.

Use the indices stored in the stack to find the width.

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