Naive Approach

A brute force solution would be to iterate through all possible split points of the strings a and b, and for each split, check if a_prefix + b_suffix or b_prefix + a_suffix is a palindrome. This would involve creating substrings and reversing strings multiple times.

Code (Python)

def is_palindrome(s):
    return s == s[::-1]


def check_palindrome_formation_brute_force(a, b):
    n = len(a)
    for i in range(n + 1):
        a_prefix = a[:i]
        a_suffix = a[i:]
        b_prefix = b[:i]
        b_suffix = b[i:]

        if is_palindrome(a_prefix + b_suffix) or is_palindrome(b_prefix + a_suffix):
            return True

    return False

Time Complexity

O(n^2), where n is the length of the strings. The outer loop iterates n+1 times. Inside the loop, string concatenation takes O(n) time, and is_palindrome takes O(n) time.

Space Complexity

O(n), due to the creation of substrings.

Optimal Approach

The optimal approach involves a more efficient way to check for palindrome formation. The main idea is to check if there exists a split point such that the unmatched parts of the prefix and suffix are palindromes.

1. Implement a helper function check(a, b) that checks if a split exists that forms a palindrome.
2. Start with left = 0 and right = n - 1. Compare a[left] and b[right]. If they are equal, increment left and decrement right. Continue until left >= right or a mismatch is found.
3. If left >= right, the whole string is a palindrome. If a mismatch is found, check if either a[left:right+1] or b[left:right+1] is a palindrome. If either one is, then return True.
4. The main function will call the helper function twice, once with a and b, and another time with b and a.

Code (Python)

def is_palindrome_range(s, left, right):
    while left < right:
        if s[left] != s[right]:
            return False
        left += 1
        right -= 1
    return True


def check(a, b):
    left = 0
    right = len(a) - 1
    while left < right and a[left] == b[right]:
        left += 1
        right -= 1

    if left >= right:
        return True

    return is_palindrome_range(a, left, right) or is_palindrome_range(b, left, right)


def check_palindrome_formation(a, b):
    return check(a, b) or check(b, a)

Time Complexity

O(n), where n is the length of the strings. The check function iterates through the strings at most once. The is_palindrome_range function also takes at most O(n) time.

Space Complexity

O(1). The algorithm uses a constant amount of extra space.

Edge Cases

• Empty strings (though the problem statement specifies lengths are at least 1).
• Strings of length 1. These are always palindromes.
• Strings that are already palindromes.
• Strings where the first half of a matches the reversed second half of b, or vice versa.