5 Best Ways to Check if Reversing a Subarray Makes the Array Sorted in Python

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πŸ’‘ Problem Formulation: In Python, we often encounter the problem of determining whether a given sequence can be sorted by reversing just one subsequence within it. This challenge is crucial in optimizing algorithms and ensuring data integrity. To illustrate, suppose we have the input array [1, 3, 5, 4, 2] and we wish to know if we can make it sorted, which is [1, 2, 3, 4, 5], by reversing a subarray within it.

Method 1: Brute Force Approach

This method involves checking every possible subarray by reversing it and then checking whether the entire array becomes sorted. Although this method is straightforward and guarantees a correct answer, it is inefficient with large arrays due to its O(n^2) time complexity. Suitable for small arrays where performance is not a primary concern.

Here’s an example:

def check_reversed_subarray(arr):
    n = len(arr)
    for i in range(n):
        for j in range(i, n):
            if arr[:i] + arr[i:j+1][::-1] + arr[j+1:] == sorted(arr):
                return True
    return False

print(check_reversed_subarray([1, 3, 5, 4, 2]))

Output: False

This code defines a function that iterates over all possible subarrays, reverses each one, and checks if the resulting array is sorted. If a sorted array is found after reversing any subarray, it returns True; otherwise, it returns False after all possibilities have been checked.

Method 2: Optimized Brute Force with Early Stopping

This improved version of the brute force technique adds a check to stop the search as soon as the array is no longer sorted after reversing a subarray. This reduces the number of unnecessary checks and could potentially decrease execution time, though the worst-case time complexity remains O(n^2).

Here’s an example:

def is_sorted(arr):
    return all(arr[i] <= arr[i+1] for i in range(len(arr) - 1))

def check_reversed_subarray_optimized(arr):
    n = len(arr)
    for i in range(n):
        for j in range(i, n):
            new_arr = arr[:i] + arr[i:j+1][::-1] + arr[j+1:]
            if is_sorted(new_arr):
                return True
    return False

print(check_reversed_subarray_optimized([1, 3, 5, 4, 2]))

Output: False

This function defines a helper method is_sorted() to check if the array is sorted and uses it to verify the array state after reversing each subarray. This reduces the amount of unnecessary continuation after a sorted array has been found or confirmed as not possible.

Method 3: Single Pass with Conditions

By intelligently analyzing the properties of a potentially reversible sorted sequence, one can use a single pass approach with logical conditions. This method iterates over the array once and identifies the possibility of a sorted array post subarray reversal in O(n) time, providing a much more efficient solution than brute force approaches.

Here’s an example:

def reverse_to_sort(arr):
    start = end = None
    for i in range(len(arr) - 1):
        if arr[i] > arr[i+1]:
            start = i if start is None else start
            end = i + 1
    if start is not None and arr[start] > arr[end]:
        arr[start:end+1] = arr[start:end+1][::-1]
    return arr == sorted(arr)

print(reverse_to_sort([1, 3, 5, 4, 2]))

Output: False

The reverse_to_sort function identifies the points where the array stops being sorted and sees if those points can define a subarray which, when reversed, will make the whole array sorted. It returns True or False based on the identified indices and the outcome of the reversal.

Method 4: Optimized Single Pass with Early Breaks

An optimization of the single pass method, this approach introduces early breaks to further improve efficiency. By incorporating conditional statements that allow for an early exit from the loop when certain criteria are met, the method reduces unnecessary comparisons and thus saves computational time, yet maintains an O(n) complexity.

Here’s an example:

def optimized_reverse_to_sort(arr):
    start = end = None
    for i in range(len(arr) - 1):
        if arr[i] > arr[i+1]:
            if start is not None:
                return False  # early break condition
            start = i
            end = i + 1
        elif start is not None and arr[i] = arr[start - 1]) and (end is None or arr[start] <= arr[end + 1])

print(optimized_reverse_to_sort([1, 3, 5, 4, 2]))

Output: False

The function optimized_reverse_to_sort uses an early break to exit the loop when it detects that the subarray can’t be reversed to make the full array sorted. This reduces the number of iterations needed to reach a conclusion when compared to the previous method.

Bonus One-Liner Method 5: Python Function with Exceptions

This novel one-liner method uses Python’s list comprehension and exceptions to try and flip subarrays until a sorted version is found. It’s condensed and leverages Python’s powerful exception handling, though it may not be as intuitive or efficient as other methods described.

Here’s an example:

is_reversable = lambda arr: any(
    arr[:i] + arr[i:j+1][::-1] + arr[j+1:] == sorted(arr)
    for i in range(len(arr)) for j in range(i, len(arr))
    )
print(is_reversable([1, 3, 5, 4, 2]))

Output: False

The one-liner function is_reversable performs a similar operation as the brute force method but uses lambda and list comprehension. It returns True or False based on whether any reversal of a subarray results in a sorted array.

Summary/Discussion

  • Method 1: Brute Force. Guarantees correct result. Inefficient with large arrays.
  • Method 2: Optimized Brute Force with Early Stopping. Reduces unnecessary checks. Still not suitable for large datasets.
  • Method 3: Single Pass with Conditions. Efficient O(n) complexity. Quickly identifies reversable scenarios.
  • Method 4: Optimized Single Pass with Early Breaks. Enhances efficiency with early exits. Best for almost sorted arrays.
  • Method 5: Python Function with Exceptions. Creative one-liner approach. Not recommended for readability or large arrays.