5 Best Ways to Program to Check if Reversing a Sublist Can Form a Second List in Python

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πŸ’‘ Problem Formulation: In Python programming, a common task might involve comparing two lists to determine if one list can be formed from the other by reversing a contiguous sublist. For instance, given the input list [1, 2, 3, 4, 5] and a target list [1, 4, 3, 2, 5], the question is: can the second list be obtained by reversing some part of the first list?

Method 1: Brute Force Comparison

This method involves iterating over the original list to find all possible sublists, reversing them, and checking if the modified list matches the target list. It’s a straightforward function but is not efficient for large lists due to its high time complexity.

Here’s an example:

def can_form_by_reversing_sublist(original, target):
    for i in range(len(original)):
        for j in range(i, len(original)):
            modified = original[:i] + original[i:j+1][::-1] + original[j+1:]
            if modified == target:
                return True
    return False

# Test the function
original_list = [1, 2, 3, 4, 5]
target_list = [1, 4, 3, 2, 5]
print(can_form_by_reversing_sublist(original_list, target_list))

Output: True

This code snippet defines a function can_form_by_reversing_sublist that takes two lists as arguments. It iterates over every possible sublist of the original list, reverses it, and checks if this new list equals the target list. If a match is found, it returns True; otherwise after all possibilities are checked, it returns False.

Method 2: Two-Pointer Approach

Improving on the brute force method, the two-pointer approach reduces the number of comparisons by identifying mismatch positions between the original and target lists and checking if the intervening sublist can be reversed to match.

Here’s an example:

def is_reversible_to_target(original, target):
    if original == target:
        return True
    
    left, right = 0, len(original) - 1
    while left  left and original[right] == target[right]:
        right -= 1

    return original[left:right+1] == target[left:right+1][::-1]

# Test the function
original_list = [1, 2, 3, 4, 5]
target_list = [1, 4, 3, 2, 5]
print(is_reversible_to_target(original_list, target_list))

Output: True

In this code snippet, the function is_reversible_to_target uses two pointers that move inwards from the ends of the list, comparing elements until they find mismatches. Then, it checks if the sublist between those pointers in the original list is the reverse of the corresponding sublist in the target list, returning the result accordingly.

Method 3: Using Python’s Slicing and zip

We can enhance efficiency by utilizing Python’s slicing syntax and the zip function to find and verify the portion that needs to be reversed through element-wise comparison.

Here’s an example:

def can_reach_target_by_reversing(original, target):
    pairs = zip(original, target)
    differences = [(i, orig) for i, (orig, tar) in enumerate(pairs) if orig != tar]

    if not differences: return original == target
    left, right = differences[0][0], differences[-1][0]

    return original[left:right+1] == target[left:right+1][::-1]

# Test the function
original_list = [1, 2, 3, 4, 5]
target_list = [1, 4, 3, 2, 5]
print(can_reach_target_by_reversing(original_list, target_list))

Output: True

The function can_reach_target_by_reversing uses zip to pair up items from the original and target lists, then builds a list of differing elements. If no differences are found, it directly compares the lists. Otherwise, it slices both lists around the differing elements and compares them, thereby verifying the possibility of forming the target list by reversing a sublist.

Method 4: Optimized Sublist Searching

This method seeks to find the smallest window of the sublist to be reversed using a while loop, reducing unnecessary operations by eliminating equal leading and trailing elements and then comparing the rest with the slice reversal.

Here’s an example:

def can_match_by_reversal(original, target):
    if len(original) != len(target):
        return False
    
    i, j = 0, len(original) - 1
    while i < j and original[i] == target[i]:
        i += 1
    while i < j and original[j] == target[j]:
        j -= 1
    
    return original[i:j+1] == target[j:i-1:-1]

# Test the function
original_list = [1, 2, 3, 4, 5]
target_list = [1, 4, 3, 2, 5]
print(can_match_by_reversal(original_list, target_list))

Output: True

The function can_match_by_reversal starts by comparing the lengths of both lists, following that, it carries out a double-ended search to find the bounds of the potential reversal. If the identified slice of the original list matches the reversed slice of the target list within these bounds, this indicates that the reverse operation can form the second list.

Bonus One-Liner Method 5: List Comprehension and any()

Employing list comprehension and the any function, we condense the brute-force approach into a single line, taking advantage of Python’s expressive syntax for concise solutions.

Here’s an example:

original = [1, 2, 3, 4, 5]
target = [1, 4, 3, 2, 5]

print(any(original[:i] + original[i:j][::-1] + original[j:]
          == target for i in range(len(original)) for j in range(i + 1, len(original)+1)))

Output: True

The one-liner uses a list comprehension nested within an any function to test every possible sublist reversal against the target list. If any of the generated configurations match the target list, the entire expression evaluates to True.

Summary/Discussion

  • Method 1: Brute Force Comparison. Simple to understand. Not time-efficient for large lists.
  • Method 2: Two-Pointer Approach. More efficient than brute force. Requires lists with matching non-reversed elements at both ends.
  • Method 3: Using Python’s Slicing and zip. Streamlined and Pythonic. Dependent on identifying all differences between the lists first.
  • Method 4: Optimized Sublist Searching. Targets the smallest possible sublist to reverse. Skips checking already matched elements.
  • Method 5: List Comprehension and any(). Extremely concise. However, not the best in terms of readability and still has a brute force nature.