Print the Area of the Largest Rectangle in the Specified Histogram: Tackling the Problem with Three Solutions Using Stack

I'm thrilled to share my latest achievement on Codeforces! Recently, I tackled a challenging problem titled "Print the Area of the Largest Rectangle in the Specified Histogram." (Link to the problem)

Problem Overview

The task was to find and print the area of the largest rectangle that can be formed within a given histogram. This classic problem is a favorite among competitive programmers due to its intriguing combination of data structure manipulation and algorithm design.

My Approach

I managed to solve this problem in three distinct ways, all leveraging the power of the stack data structure. Here’s a breakdown of each solution:

The First Solution:

In this approach, we use a stack to keep track of the indices of the histogram's bars and calculate the nearest smaller bar to the left and right of each bar. Here's a step-by-step breakdown:

1. Initialize Two Arrays: leftsmall and rightsmall to store the indices of the nearest smaller bar to the left and right of each bar, respectively.
2. Left to Right Scan:Traverse the histogram from left to right.For each bar, pop elements from the stack until the stack is empty or the bar at the top of the stack is shorter than the current bar.If the stack is empty, the nearest smaller bar to the left is outside the histogram, so set leftsmall[i] to 0. Otherwise, set leftsmall[i] to the index of the top of the stack plus one.Push the current bar's index onto the stack.
3. Right to Left Scan:Clear the stack and traverse the histogram from right to left.For each bar, pop elements from the stack until the stack is empty or the bar at the top of the stack is shorter than the current bar.If the stack is empty, the nearest smaller bar to the right is outside the histogram, so set rightsmall[i] to n - 1. Otherwise, set rightsmall[i] to the index of the top of the stack minus one.Push the current bar's index onto the stack.
4. Calculate the Maximum Area:Iterate over each bar and calculate the area of the rectangle formed with the current bar as the smallest height.The width of the rectangle is determined by the difference between the indices stored in rightsmall and leftsmall plus one.Keep track of the maximum area found.

Here’s the implementation of this approach

The Second Solution:

This approach also uses a stack but simplifies the process by handling the calculation of the maximum rectangle area during a single pass through the histogram.

1. Single Pass through Histogram:Traverse the histogram from left to right.Use a stack to keep track of the indices of the bars in increasing order of height.If the stack is empty or the current bar is taller than the bar at the index on the top of the stack, push the current index onto the stack.If the current bar is shorter, pop from the stack and calculate the area of the rectangle formed with the height of the bar at the index that was just popped. The width is determined by the difference between the current index and the index of the new top of the stack (or the start if the stack is empty).
2. Process Remaining Bars:After processing all bars, pop any remaining bars from the stack and calculate the area of the rectangles using the same logic.

Here’s the implementation of this approach

The Third Solution:

In this approach, we also use a stack but scan the histogram from right to left, which is a variation of the first solution.

1. Right to Left Scan: Traverse the histogram from right to left.Use a stack to keep track of the indices of the bars in increasing order of height.If the stack is empty or the current bar is taller than the bar at the index on the top of the stack, push the current index onto the stack.If the current bar is shorter, pop from the stack and calculate the area of the rectangle formed with the height of the bar at the index that was just popped. The width is determined by the difference between the current index and the index of the new top of the stack (or the end if the stack is empty).
2. Process Remaining Bars: After processing all bars, pop any remaining bars from the stack and calculate the area of the rectangles using the same logic.

Here’s the implementation of this approach

Additional Solution

While experimenting with different approaches, I discovered another solution. Unfortunately, this method results in a time limit exceed (TLE) error. However, I believe it’s worth sharing for educational purposes and to showcase the exploration process. You can find the code for this solution Here

Learning and Insights

Each solution provided a unique perspective and deepened my understanding of stack operations and their applications in problem-solving. It was particularly rewarding to explore different methodologies and see how they each led to the same correct result.

Conclusion

This experience was a great reminder of the multiple paths one can take to reach a solution in programming. It reinforced the importance of versatility and adaptability in tackling complex problems.

Feel free to share your thoughts or ask any questions in the comments below. I'm always eager to discuss different approaches and learn from others in the community!

#Codeforces #Programming #ProblemSolving #Algorithms #CodingChallenge #ContinuousLearning #Histogram #datastructure #computerScience