Longest Palindrome After Substring Concatenation I

You are given two strings, s and t.

You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.

Return the length of the longest palindrome that can be formed this way.

Example 1:

Input: s = "a", t = "a" Output: 2 Explanation: Concatenating "a" from s and "a" from t results in "aa", which is a palindrome of length 2.

Example 2:

Input: s = "abc", t = "def" Output: 1 Explanation: Since all characters are different, the longest palindrome is any single character, so the answer is 1.

Example 3:

Input: s = "b", t = "aaaa" Output: 4 Explanation: Selecting "aaaa" from t is the longest palindrome, so the answer is 4.

Example 4:

Input: s = "abcde", t = "ecdba" Output: 5 Explanation: Concatenating "abc" from s and "ba" from t results in "abcba", which is a palindrome of length 5.

Constraints:

1 <= s.length, t.length <= 30 s and t consist of lowercase English letters.

Naive Approach

A brute-force solution would involve generating all possible substrings from both strings s and t, concatenating each pair, and then checking if the resulting string is a palindrome. We keep track of the longest palindrome found so far.

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

def longest_palindrome_substring_naive(s, t):
    max_len = 0
    for i in range(len(s) + 1):
        for j in range(i, len(s) + 1):
            sub_s = s[i:j]
            for k in range(len(t) + 1):
                for l in range(k, len(t) + 1):
                    sub_t = t[k:l]
                    combined = sub_s + sub_t
                    if is_palindrome(combined):
                        max_len = max(max_len, len(combined))
    return max_len

# Example Usage
s = "abcde"
t = "ecdba"
print(longest_palindrome_substring_naive(s, t))  # Output: 5

s = "a"
t = "a"
print(longest_palindrome_substring_naive(s, t)) # Output: 2

s = "abc"
t = "def"
print(longest_palindrome_substring_naive(s, t)) # Output: 1

s = "b"
t = "aaaa"
print(longest_palindrome_substring_naive(s, t)) # Output: 4

Optimal Approach (Dynamic Programming)

We can use dynamic programming to optimize this. The main idea is to create a DP table that stores the lengths of the longest palindromic substrings formed by combining prefixes of s and t. Instead of generating substrings and checking, we build up the palindromes iteratively.

def longest_palindrome_substring_dp(s, t):
    n, m = len(s), len(t)
    max_len = 0

    # Check palindromes formed only by s
    for i in range(n):
        # Odd length palindromes centered at i
        l, r = i, i
        while l >= 0 and r < n and s[l] == s[r]:
            max_len = max(max_len, r - l + 1)
            l -= 1
            r += 1

        # Even length palindromes centered between i and i+1
        l, r = i, i + 1
        while l >= 0 and r < n and s[l] == s[r]:
            max_len = max(max_len, r - l + 1)
            l -= 1
            r += 1

    # Check palindromes formed only by t
    for i in range(m):
        # Odd length palindromes centered at i
        l, r = i, i
        while l >= 0 and r < m and t[l] == t[r]:
            max_len = max(max_len, r - l + 1)
            l -= 1
            r += 1

        # Even length palindromes centered between i and i+1
        l, r = i, i + 1
        while l >= 0 and r < m and t[l] == t[r]:
            max_len = max(max_len, r - l + 1)
            l -= 1
            r += 1

    # Check palindromes formed by combining s and t
    for i in range(n + 1):
        for j in range(m + 1):
            # Consider s[:i] + t[j:]
            combined = s[:i] + t[m-j:]
            if is_palindrome(combined):
                max_len = max(max_len, len(combined))

    return max_len


# Example Usage
s = "abcde"
t = "ecdba"
print(longest_palindrome_substring_dp(s, t))  # Output: 5

s = "a"
t = "a"
print(longest_palindrome_substring_dp(s, t)) # Output: 2

s = "abc"
t = "def"
print(longest_palindrome_substring_dp(s, t)) # Output: 1

s = "b"
t = "aaaa"
print(longest_palindrome_substring_dp(s, t)) # Output: 4

Big(O) Runtime Analysis

• Naive Approach: The naive approach iterates through all possible substrings of s and t, resulting in O(n^2 * m^2) combinations. For each combination, it checks if the combined string is a palindrome, which takes O(n+m) time. Therefore, the overall time complexity is O(n^2 * m^2 * (n+m)).
• Optimal (DP) Approach: The dynamic programming approach first finds palindromes only within s and t, which takes O(n^2) and O(m^2) respectively. Then it checks some combinations of substrings of s and t. The number of combinations could be up to O(nm). The palindrome check takes O(n+m). Therefore, the worst case complexity becomes O(n^2 + m^2 + nm*(n+m)), which simplifies to O((n+m)^3) in the worst case. This can be further optimized if we consider only relevant combinations.

Big(O) Space Usage Analysis

• Naive Approach: The naive approach does not use any significant extra space other than the space needed to store the substrings, which is O(n+m) in the worst case. Therefore, the space complexity is O(1).
• Optimal (DP) Approach: The dynamic programming approach doesn't explicitly create a DP table as in standard DP problems. The extra space needed is mainly for forming the combined strings of substrings, which is O(n+m) in the worst case. Therefore the space complexity is O(n+m).

Edge Cases

• Empty Strings: If both s and t are empty, the longest palindrome is an empty string, so the length is 0.
• One String is Empty: If one string is empty, we just need to find the longest palindromic substring within the non-empty string.
• No Common Characters: If there are no common characters between the two strings, the longest palindrome will be a single character (length 1) if either string is non-empty, or 0 if both are empty.
• Large Strings: The constraint mentions the strings can be up to length 30. For very large strings, further optimizations may be needed to avoid exceeding time limits, possibly involving more advanced palindrome search algorithms or data structures.