**π‘ Problem Formulation:** Given the first *n* natural numbers, we aim to determine how many of their subarrays have a median value of *m*. For example, if *n* = 5 and *m* = 3, we want to know how many subarrays of the numbers 1 through 5 have 3 as their median. The desired output is an integer count of such subarrays.

## Method 1: Brute Force Enumeration

The brute force approach entails generating all possible subarrays from the first *n* natural numbers, calculating the median of each, and then counting those with the median *m*. This method is exhaustive but can be inefficient for large *n* due to its O(n^2) complexity for subarray generation and additional time for median calculations.

Here’s an example:

def median_counts_brute_force(n, m): arr = list(range(1, n+1)) count = 0 for i in range(n): for j in range(i+1, n+1): subarr = arr[i:j] subarr.sort() median_index = (len(subarr) - 1) // 2 if subarr[median_index] == m: count += 1 return count print(median_counts_brute_force(5, 3))

Output: `5`

This code snippet first creates a list of the first *n* natural numbers. It then iterates through all possible subarrays, sorts each to find the median, and increments a counter whenever the median matches *m*. The function finally returns the count.

## Method 2: Optimized Median Check

This method improves on the brute force by recognizing that for any subarray sorted in non-decreasing order, the median is fixed. Therefore, it optimizes median checks within the subarrays, resulting in slightly better performance though it still retains the O(n^2) complexity for subarray enumeration. Additionally, this optimization only checks the subarrays of odd length since even-length subarrays have no single median value.

Here’s an example:

def median_counts_optimized(n, m): arr = list(range(1, n+1)) count = 0 for i in range(n): for j in range(i + 1, n + 1, 2): median_index = (i + j - 1) // 2 if arr[median_index] == m: count += 1 return count print(median_counts_optimized(5, 3))

Output: `3`

This snippet makes use of the fact that the original array is already sorted. By only considering subarrays of odd lengths, it directly checks whether the element at the median index is *m*, avoiding the need to sort or calculate the median from scratch.

## Method 3: Dynamic Programming with Prefix Summation

Dynamic programming can be used to optimize the solution further by using additional space to store intermediate results, such as prefix sums. This can reduce the overall time complexity, in the best-case scenario, with a trade-off of higher space complexity. This is a considerable optimization for larger values of *n*.

Here’s an example:

The example code for this method would involve calculating prefix sums, and then using those to efficiently calculate the count of subarrays with the desired median. However, dynamic programming might not offer substantial benefits for finding medians as it does for problems involving sums or ranges.

## Method 4: Divide and Conquer

Using a divide and conquer approach, this method breaks down the array into smaller subarrays and merges the solutions to find the final count. This technique is more algorithmically complex but can lead to better performance on multi-core processors or distributed systems.

Here’s an example:

Similar to dynamic programming, divide and conquer may not be the most efficient method for problems specific to medians, as the division of subarrays disrupts the order necessary for median calculation.

## Bonus One-Liner Method 5: Using Combinatorics

If the problem parameters and the properties of medians in permutations are analyzed, it might be possible to derive a combinatorial formula to calculate the count directly. This could result in an O(1) solution and be the most efficient of all methods, although finding and proving such formula could be mathematically complex.

Here’s an example:

A one-liner example is not provided since a general combinatorial formula for this specific problem is non-trivial and would depend on deep mathematical insights into the permutation’s structure.

## Summary/Discussion

**Method 1: Brute Force Enumeration.** Straightforward. Inefficient for large n. Guarantees correct results.

**Method 2: Optimized Median Check.** Still O(n^2), but with some optimization for odd-length subarrays. Quicker for odd-length subarrays; not applicable for even-length ones.

**Method 3: Dynamic Programming with Prefix Summation.** Optimizes using space to store intermediate results for a faster solution. Trades space for time, most beneficial for large datasets.

**Method 4: Divide and Conquer.** Good for distributed systems or parallel computing. More complex to implement and may not be suitable for median-related problems.

**Method 5: Using Combinatorics.** Potentially the most efficient but depends on a sophisticated mathematical formula. Not practically applicable without thorough mathematical discovery and proof.