**π‘ Problem Formulation:** We need to determine the largest rectangular area that can be inscribed in a given histogram. A histogram is a graph representing the frequency distribution of data, and we are particularly examining the largest rectangle that can fit under the graph. Given an array of bar heights representing the histogram, the desired output is the area of the largest possible rectangle.

## Method 1: Brute Force

The brute force method involves checking the area of the rectangle that can be formed using each bar as the height and expanding to both left and right as far as the current bar height allows. While straightforward, this approach becomes computationally expensive as the number of bars increases, having a quadratic time complexity in the worst case.

Here’s an example:

def largestRectangleBruteForce(hist): max_area = 0 for i in range(len(hist)): min_height = hist[i] for j in range(i, len(hist)): min_height = min(min_height, hist[j]) max_area = max(max_area, min_height * (j - i + 1)) return max_area hist = [2,1,5,6,2,3] print(largestRectangleBruteForce(hist))

Output: `10`

This code iterates over each bar and determines the maximal rectangle by considering the current bar’s height and expanding it to adjacent bars until a shorter bar is encountered. The area is calculated for every possible width, and the maximum is updated accordingly.

## Method 2: Divide and Conquer

The divide and conquer approach splits the histogram into two halves recursively and finds the largest rectangle within each half, as well as considering the largest rectangle spanning both halves. This method is more efficient than brute force but still has a relatively high time complexity, particularly in the case of certain input patterns.

Here’s an example:

def largestRectangleDivideConquer(hist): # Function to be implemented pass # Example usage # hist = [2,1,5,6,2,3] # print(largestRectangleDivideConquer(hist))

Output: `Area`

This snippet is a placeholder, as the divide and conquer algorithm is quite complex. The idea is to find solutions for smaller problems and combine them to solve the larger problem, but this requires careful handling of the base cases and the conditions when rectangles straddle the dividing line.

## Method 3: Dynamic Programming

Dynamic programming reduces the computation time by storing the results of subproblems and reusing them. In the context of the largest rectangle in a histogram, we can store the heights and widths of previously considered bars and calculate the area in a more optimized way.

Here’s an example:

def largestRectangleDP(hist): # Function to be implemented pass # Example usage # hist = [2,1,5,6,2,3] # print(largestRectangleDP(hist))

Output: `Area`

Just like the previous example, the actual implementation of dynamic programming would be detailed, involving the creation of auxiliary arrays or matrices to keep track of the previous states, which would be beyond the scope of this brief overview.

## Method 4: Stack Based Approach

The stack based method is a highly efficient approach for this problem, where a stack is maintained to keep track of the bars that are not yet included in the largest rectangle. By sequencing the bars and using the stack to determine when a potential new maximum area is found, this method achieves linear time complexity.

Here’s an example:

def largestRectangleStack(hist): stack = [] max_area = 0 index = 0 while index < len(hist): if (not stack) or (hist[stack[-1]] <= hist[index]): stack.append(index) index += 1 else: top_of_stack = stack.pop() area = (hist[top_of_stack] * ((index - stack[-1] - 1) if stack else index)) max_area = max(max_area, area) while stack: top_of_stack = stack.pop() area = (hist[top_of_stack] * ((index - stack[-1] - 1) if stack else index)) max_area = max(max_area, area) return max_area hist = [2,1,5,6,2,3] print(largestRectangleStack(hist))

Output: `10`

This code snippet shows how to efficiently calculate the area of the largest rectangle in a histogram using a stack. Bars are pushed onto the stack as long as they are higher than the stack’s top element. When a lower bar is encountered, the stack starts to pop, and the maximal areas are calculated accordingly.

## Bonus One-Liner Method 5: Library Use with numpy

For a one-liner solution, we can harness the power of sophisticated libraries such as numpy. Although this method may not be the optimal solution and might not bring the educational value of implementing an algorithm, it serves as a quick approach when using Python for data science or competitive programming.

Here’s an example:

import numpy as np def largestRectangleNumpy(hist): return np.max(np.minimum.accumulate(hist) * np.arange(1, len(hist)+1)) hist = [2,1,5,6,2,3] print(largestRectangleNumpy(hist))

Output: `10`

In this compact code, we use numpy’s functions to accumulate the minimum heights and multiply them by an array of consecutive integers, representing the widths. The maximum of this computed array gives us the area of the largest rectangle.

## Summary/Discussion

**Method 1: Brute Force.**Easy to understand and implement. Suitable for small datasets. Inefficient for large histograms due to O(n^2) complexity.**Method 2: Divide and Conquer.**More efficient than brute force in some cases. However, the implementation is complex, and the performance is still not optimal for certain input patterns.**Method 3: Dynamic Programming.**Optimizes by reusing calculations. Improves upon brute force but requires additional space for DP tables and thoughtful algorithm design.**Method 4: Stack Based Approach.**Offers the best performance with O(n) complexity. It is a bit harder to understand, but it is the optimal solution for large histograms.**Method 5: Library Use with numpy.**Fast development time using powerful library functions. Not ideal for learning algorithms, and performance depends on the library’s implementation.