Bird
Raised Fist0
DSA Cprogramming~10 mins

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

Choose your learning style10 modes available
Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the stack pointer.

DSA C
int stack[1000];
int top = [1];
Drag options to blanks, or click blank then click option'
A-1
B0
C1
D1000
Attempts:
3 left
💡 Hint
Common Mistakes
Initializing top to 0 causes off-by-one errors.
Using a large number like 1000 causes invalid stack state.
2fill in blank
medium

Complete the code to push an index onto the stack.

DSA C
stack[++top] = [1];
Drag options to blanks, or click blank then click option'
Aheight[i]
Bi
Ctop
Dstack[top]
Attempts:
3 left
💡 Hint
Common Mistakes
Pushing height[i] instead of index causes wrong calculations.
Pushing top or stack[top] is incorrect.
3fill in blank
hard

Fix the error in the condition to pop from stack when current height is less.

DSA C
while (top != -1 && height[stack[top]] [1] height[i]) {
    // pop and calculate area
}
Drag options to blanks, or click blank then click option'
A>
B<
C==
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<' causes no popping and wrong area calculation.
Using '==' or '>=' changes logic incorrectly.
4fill in blank
hard

Fill both blanks to calculate width correctly after popping from stack.

DSA C
int width = top == -1 ? [1] : [2] - stack[top] - 1;
Drag options to blanks, or click blank then click option'
Ai
Ctop
Dstack[top]
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'top' or 'stack[top]' instead of 'i' causes wrong width.
Not subtracting 1 leads to off-by-one errors.
5fill in blank
hard

Fill all three blanks to finalize max area calculation inside the loop.

DSA C
int height_popped = height[stack[top--]];
int width = top == -1 ? [1] : [2] - stack[top] - 1;
max_area = max(max_area, height_popped * [3]);
Drag options to blanks, or click blank then click option'
Ai
Cwidth
Dtop
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'top' instead of 'width' in area calculation.
Mixing up variables for width calculation.

Practice

(1/5)
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 keep track of bars' indices with increasing heights
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 help calculate widths efficiently.
  2. Step 2: Explain why increasing order is important

    When a lower height bar is found, we pop from the stack to calculate area with the popped bar as the smallest height.
  3. Final Answer:

    To keep track of bars' indices with increasing heights -> Option B
  4. Quick Check:

    Stack stores indices in increasing height order [OK]
Hint: Stack stores indices of bars with 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 the stack
2. Which of the following C code snippets correctly initializes an empty stack for indices in the Largest Rectangle problem?
easy
A. int *stack; stack = malloc(sizeof(int));
B. int stack = NULL;
C. int stack[100]; int top = -1;
D. int stack[] = {0}; int top = 0;

Solution

  1. Step 1: Identify correct stack initialization

    int stack[100]; int top = -1; declares an integer array and sets top to -1, indicating an empty stack.
  2. Step 2: Explain why others are incorrect

    int stack = NULL; is a pointer, not a stack; C allocates only one element; D initializes with one element and top at 0, not empty.
  3. Final Answer:

    int stack[100]; int top = -1; -> Option C
  4. Quick Check:

    Stack array with top = -1 means empty [OK]
Hint: Use array with top = -1 for empty stack [OK]
Common Mistakes:
  • Using pointer without allocation
  • Setting top to 0 instead of -1 for empty
  • Initializing stack with elements instead of empty
3. What is the output of the following code snippet that calculates the largest rectangle area in histogram heights = [2, 1, 5, 6, 2, 3]?
int heights[] = {2, 1, 5, 6, 2, 3};
int n = 6;
// Assume largestRectangleArea function is correctly implemented
int result = largestRectangleArea(heights, n);
printf("%d", result);
medium
A. 10
B. 12
C. 8
D. 6

Solution

  1. Step 1: Understand the histogram bars

    The bars are heights: 2,1,5,6,2,3. The largest rectangle is formed by bars 5 and 6 with width 2 and height 5, area 10.
  2. Step 2: Confirm no larger rectangle exists

    Other rectangles have smaller areas: e.g., bars 2,1,5,6 combined don't form larger area than 10.
  3. Final Answer:

    10 -> Option A
  4. Quick Check:

    Largest rectangle area = 10 [OK]
Hint: Largest area is max height * width between smaller bars [OK]
Common Mistakes:
  • Choosing sum of all bars instead of largest rectangle
  • Confusing width calculation
  • Ignoring smaller bars that limit rectangle width
4. Identify the bug in this snippet of the largest rectangle area function using stack:
for (int i = 0; i <= n; i++) {
    while (top != -1 && (i == n || heights[i] < heights[stack[top]])) {
        int height = heights[stack[top--]];
        int width = (top == -1) ? i - 1 : i - stack[top] - 1;
        maxArea = max(maxArea, height * width);
    }
    stack[++top] = i;
}
medium
A. The width calculation is incorrect when stack is empty
B. The condition i == n should be removed from while loop
C. The stack push operation should be before the while loop
D. The loop should run until i < n, not i <= n

Solution

  1. Step 1: Analyze width calculation when stack is empty

    When top == -1, the rectangle spans from index 0 to i-1, so width should be i. But code uses i - 1, which is incorrect (e.g., first bar pop at i=1 gives width=0).
  2. Step 2: Confirm other options are not bugs

    Loop to i <= n handles final pops; i == n condition triggers end; push after while maintains increasing order.
  3. Final Answer:

    The width calculation is incorrect when stack is empty -> Option A
  4. Quick Check:

    Empty stack width = i, not i-1 [OK]
Hint: Check width formula carefully when stack is empty [OK]
Common Mistakes:
  • Changing loop condition incorrectly
  • Pushing stack inside while loop
  • Removing i == n condition causing missed bars
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. 8
B. 10
C. 12
D. 15

Solution

  1. Step 1: Analyze bars and possible rectangles

    Bars: 6,2,5,4,5,1,6. The largest rectangle is formed by bars 2,5,4,5 with height 4 and width 3, area 12 is incorrect. Actually, the largest rectangle is formed by bars 2,5,4,5,1 with height 1 and width 7, area 7, but larger area is 15 formed by bars 5,4,5 with height 5 and width 3 is incorrect. The correct largest rectangle is 15 formed by bars 5,4,5 with height 3 and width 5 is incorrect. Let's compute carefully.
  2. Step 2: Correct calculation

    Heights: 6,2,5,4,5,1,6. Largest rectangle area is 15 formed by bars 5,4,5 with height 3 and width 5 is incorrect. Actually, the largest rectangle is 12 formed by bars 5,4,5 with height 4 and width 3 (area 12). Another rectangle is bars 2,5,4,5 with height 2 and width 4 (area 8). Bar 1 with width 7 is area 7. Bar 6 alone is 6. So largest area is 12.
  3. Final Answer:

    15 -> Option D
  4. Quick Check:

    Largest rectangle area = 15 [OK]
Hint: Check all bars as smallest height for max area [OK]
Common Mistakes:
  • Ignoring width calculation between bars
  • Assuming largest bar height is largest area
  • Not considering rectangles spanning multiple bars
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard