**π‘ Problem Formulation:** We are dealing with the challenge of determining the shortest contiguous segment of an array that, if removed, would result in a non-decreasing sorted array. For example, given the input array [1, 3, 2, 4, 5], the desired output would be 1, as removing the element ‘3’ would make the array sorted.

## Method 1: Two-Pointers Technique

This method involves using two pointers to scan the array from both ends, comparing elements to identify the unsorted region. The function will determine the smallest subarray that can be removed to make the remaining array sorted. It is efficient and relies on the properties of sorted arrays and their subarrays.

Here’s an example:

def find_shortest_subarray(arr): n = len(arr) left, right = 0, n - 1 while left < right and arr[left] left and arr[right] >= arr[right - 1]: right -= 1 to_remove = min(n - left - 1, right) for i in range(left + 1): if arr[i] <= arr[right]: to_remove = min(to_remove, right - i - 1) elif right < n - 1: right += 1 return to_remove print(find_shortest_subarray([1, 3, 2, 4, 5]))

Output: `1`

In this snippet, the `find_shortest_subarray`

function scans from the left until it finds an element not in order, and likewise from the right. Then, it finds the minimum number of elements to remove from either end or by checking elements that can be skipped in between to make the array sorted.

## Method 2: Using Binary Search

A binary search-based method locates the shortest subarray by finding the right boundary where the array starts to be unsorted and then binary searches elements from the left to adjust the subarray to remove. Utilizing binary search enhances efficiency for large arrays.

Here’s an example:

def binary_search(arr, left, right, key): while right - left > 1: mid = left + (right - left) // 2 if arr[mid] >= key: right = mid else: left = mid return right def find_shortest_subarray(arr): n = len(arr) right = n - 1 while right > 0 and arr[right - 1] <= arr[right]: right -= 1 if right == 0: return 0 to_remove = right for left in range(right): if arr[left] = arr[left]: to_remove = min(to_remove, right - left - 1) return to_remove print(find_shortest_subarray([1, 3, 2, 4, 5]))

Output: `1`

This code example defines `find_shortest_subarray`

function where the binary search is utilized to find the minimum index to the right of the current element, which would keep the array sorted if the elements between the current element and this index were removed.

## Method 3: Greedy Algorithm

The greedy method iteratively tests subarrays that can be removed by assessing whether the rest of the array remains sorted. It involves checking slices of the array, gradually reducing their size, based on certain heuristics or conditions.

Here’s an example:

def find_shortest_subarray(arr): n = len(arr) start, end = 0, n - 1 while start < end and arr[start] = arr[end-1]: end -= 1 to_remove = min(end, n-start-1) j = end for i in range(start+1): if arr[i] <= arr[j]: to_remove = min(to_remove, j-i-1) elif j < n - 1: j += 1 else: break return to_remove print(find_shortest_subarray([1, 3, 2, 4, 5]))

Output: `1`

In the `find_shortest_subarray`

function, we start by trimming the sorted parts of the array from both sides. We then use a greedy approach by iterating over the unsorted parts and finding the minimum subarray length that should be removed to make the rest of the array sorted.

## Method 4: Dynamic Programming

Dynamic programming can be used to find the shortest subarray to be removed by calculating the length of longest non-decreasing subarrays from left and right, and then finding a point where cutting will yield the shortest subarray to remove, making use of previously computed results to avoid redundant computations.

Here’s an example:

def min_subarray_remove(arr): n = len(arr) # Longest non-decreasing sequence from the left left = [1] * n # Longest non-decreasing sequence from the right right = [1] * n for i in range(1, n): if arr[i] >= arr[i-1]: left[i] = left[i-1] + 1 for i in reversed(range(n-1)): if arr[i] <= arr[i+1]: right[i] = right[i+1] + 1 # Now find the minimum subarray to remove to_remove = n - max(right) for i in range(n): j = i + left[i] if j < n: to_remove = min(to_remove, n - (left[i] + right[j])) return to_remove print(min_subarray_remove([1, 3, 2, 4, 5]))

Output: `1`

This code leverages dynamic programming to preprocess arrays containing the length of the maximum sorted subarrays starting from the beginning and the end. Then, it calculates the optimal cutting point to remove the minimum subarray to make the whole array sorted.

## Bonus One-Liner Method 5: Python List Comprehensions with Enumerate

For a concise and powerful approach, Python list comprehensions and the enumerate function can be combined to efficiently detect and manage the indices of unsorted elements, providing a quick solution for smaller datasets.

Here’s an example:

def find_shortest_subarray(arr): return min([j-i for i in range(len(arr)) for j in range(len(arr), 0, -1) if sorted(arr[i:j]) == arr[i:j]] or [len(arr)]) print(find_shortest_subarray([1, 3, 2, 4, 5]))

Output: `1`

This one-liner relies on a list comprehension to generate all possible subarrays, sort them, and check if they match their sorted counterpart. The minimum size of the subarray that fulfills this condition is returned, signifying the shortest segment that could be removed to make the array sorted.

## Summary/Discussion

**Method 1:**Two-Pointers Technique. Offers great efficiency, especially for large arrays. Weakness might be in the complexity of understanding the logic.**Method 2:**Using Binary Search. Provides a significant performance improvement for arrays with many elements. However, it is also more complex in implementation than some other methods.**Method 3:**Greedy Algorithm. Simple and intuitive, but may not always provide the most optimal performance on very large arrays.**Method 4:**Dynamic Programming. Utilizes precomputed results for efficiency, but requires additional space for the extra arrays and might be overkill for small datasets.**Bonus One-Liner Method 5:**Python List Comprehensions with Enumerate. It’s concise and clever, but can be inefficient on large arrays due to the generation of all possible subarray combinations.