Problem Description

You are given two strings s and t.

You need to form a palindrome by:

1. Selecting a substring from s (which can be empty)
2. Selecting a substring from t (which can be empty)
3. Concatenating them in that specific order (substring from s first, then substring from t)

The goal is to find the length of the longest palindrome that can be created using this method.

For example, if you select substring "ab" from s and substring "ba" from t, concatenating them gives "abba", which is a palindrome of length 4.

Note that:

• The substrings can be empty (meaning you could use just a substring from s or just from t)
• The order matters - the substring from s must come before the substring from t in the final concatenation
• A palindrome reads the same forwards and backwards

The task is to determine the maximum possible length of such a palindrome.

Intuition

To form a palindrome by concatenating substrings from s and t, we need to think about what makes a valid palindrome structure.

A palindrome reads the same forwards and backwards. When we concatenate a substring from s with a substring from t, there are several possible scenarios:

1. The entire palindrome comes from just s - This is simply finding the longest palindromic substring in s
2. The entire palindrome comes from just t - This is finding the longest palindromic substring in t
3. The palindrome spans both strings - This is the interesting case

For case 3, if we have a palindrome that spans both strings, the part from s must be the mirror image of the part from t. This means if we take some prefix from s and some suffix from t, they should form mirror images of each other.

The key insight is to reverse string t. Why? Because if we're looking for matching parts that form a palindrome, the end of t should match the beginning of s in reverse. By reversing t, we can now look for common matching sequences between s and reversed t.

Additionally, there might be extra palindromic parts within either string. For instance, we might have:

• A matching sequence between s and reversed t (forming the outer palindrome)
• Plus an additional palindrome extending from where the matching ends in either s or t

This leads us to precompute for each position in both strings: what's the longest palindrome that starts from that position? This helps us quickly check if we can extend our palindrome further after the matching part.

The dynamic programming approach then finds the longest matching sequences between s and reversed t, and for each match, checks if we can extend it with additional palindromic substrings from either string. The formula f[i][j] tracks the length of matching characters ending at position i in s and position j in reversed t, essentially building up our palindrome piece by piece.

Learn more about Two Pointers and Dynamic Programming patterns.

Solution Approach

The solution implements the intuition through three main components:

1. Preprocessing Palindrome Extensions

First, we preprocess arrays g1 and g2 where:

• g1[i] = length of the longest palindrome starting at index i in string s
• g2[i] = length of the longest palindrome starting at index i in reversed string t

The expand function implements the classic palindrome expansion technique:

• For each position, it tries to expand around that center (both odd and even length palindromes)
• Updates g[l] with the maximum palindrome length starting at position l

2. String Reversal

We reverse string t with t = t[::-1]. This transformation allows us to find matching sequences that would form palindromes when concatenated in the original order.

3. Dynamic Programming for Matching Sequences

We create a 2D DP table f[i][j] where:

• f[i][j] represents the length of matching characters ending at position i-1 in s and position j-1 in reversed t

The DP transition is:

if s[i-1] == t[j-1]:
    f[i][j] = f[i-1][j-1] + 1

This builds up matching sequences between s and reversed t.

4. Answer Calculation

For each matching position where s[i-1] == t[j-1], we have found a palindrome of length f[i][j] * 2 (the matching part from s plus its mirror from t).

We can potentially extend this palindrome:

• Add g1[i] if i < m (additional palindrome continuing in s)
• Add g2[j] if j < n (additional palindrome continuing in reversed t)

The formulas:

ans = max(ans, f[i][j] * 2 + (0 if i >= m else g1[i]))
ans = max(ans, f[i][j] * 2 + (0 if j >= n else g2[j]))

The initial answer is set to the maximum single palindrome found in either string: ans = max(*g1, *g2).

Time Complexity: O(m*n) where m and n are the lengths of strings s and t Space Complexity: O(m*n) for the DP table plus O(m+n) for the preprocessing arrays

Example Walkthrough

Let's walk through the solution with s = "abc" and t = "bac".

Step 1: Preprocess palindrome extensions

For s = "abc":

• g1[0] = 1 (longest palindrome starting at index 0 is "a")
• g1[1] = 1 (longest palindrome starting at index 1 is "b")
• g1[2] = 1 (longest palindrome starting at index 2 is "c")

For t = "bac", we first reverse it to get "cab", then:

• g2[0] = 1 (longest palindrome starting at index 0 in "cab" is "c")
• g2[1] = 1 (longest palindrome starting at index 1 in "cab" is "a")
• g2[2] = 1 (longest palindrome starting at index 2 in "cab" is "b")

Step 2: Build DP table for matching sequences

We compare s = "abc" with reversed t = "cab":

    ''  c   a   b
''  0   0   0   0
a   0   0   1   0
b   0   0   0   2
c   0   1   0   0

Key matches:

• s[0]='a' matches t[1]='a': f[1][2] = 1
• s[1]='b' matches t[2]='b': f[2][3] = f[1][2] + 1 = 2
• s[2]='c' matches t[0]='c': f[3][1] = 1

Step 3: Calculate maximum palindrome length

For each match, we calculate possible palindrome lengths:

1. At f[1][2] = 1 (matching 'a'):

   • Base palindrome: 1 * 2 = 2 ("aa")
   • Can extend with g1[1] = 1: total = 2 + 1 = 3
   • Can extend with g2[2] = 1: total = 2 + 1 = 3

2. At f[2][3] = 2 (matching "ab"):

   • Base palindrome: 2 * 2 = 4 ("abba")
   • Can extend with g1[2] = 1: total = 4 + 1 = 5
   • No extension from g2 (j=3 is out of bounds)

3. At f[3][1] = 1 (matching 'c'):

   • Base palindrome: 1 * 2 = 2 ("cc")
   • No extension from g1 (i=3 is out of bounds)
   • Can extend with g2[1] = 1: total = 2 + 1 = 3

The maximum palindrome length is 5, achieved by:

• Taking substring "ab" from s
• Taking substring "ba" from t (which is "ab" in reversed t)
• This forms "abba" (length 4)
• Plus extending with one more character 'c' from s, giving "abbac" which... wait, that's not a palindrome.

Let me recalculate: Actually, the matching "ab" sequence means we take "ab" from position 0-1 in s and the corresponding reverse "ba" from the original t (positions 1-2). This forms "abba" with length 4. The extension g1[2] would add characters starting from position 2 in s, but this doesn't maintain the palindrome property in this case.

The actual maximum is 4 from the palindrome "abba" formed by taking "ab" from s and "ba" from t.

Solution Implementation

Time and Space Complexity

The time complexity of this algorithm is O(m² + n² + m × n), where m is the length of string s and n is the length of string t.

Breaking down the time complexity:

• The calc function is called twice - once for string s with complexity O(m²) and once for reversed string t with complexity O(n²). This is because for each position i in the string, the expand function potentially examines all characters, leading to quadratic complexity.
• The nested loops for dynamic programming iterate through m × n combinations, with each iteration performing constant time operations, contributing O(m × n).
• Therefore, the total time complexity is O(m² + n² + m × n).

However, the reference answer states the time complexity as O(m × (m + n)). This could be a simplified or different interpretation where:

• If we consider m ≈ n or focus on the dominant term when one string is significantly longer
• Or if there's an optimization in the palindrome calculation that bounds it differently

The space complexity is O(m × n), which matches the reference answer. This is due to:

• The 2D DP array f of size (m + 1) × (n + 1) which requires O(m × n) space
• The arrays g1 and g2 require O(m) and O(n) space respectively
• The reversed string t requires O(n) space
• The dominant term is O(m × n) from the DP array

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Problem Requirements

Pitfall: Many developers initially interpret this problem as finding any palindrome that can be formed by rearranging characters from both strings, or concatenating entire strings rather than substrings.

Why it happens: The problem statement mentions "selecting a substring" and "concatenating them," which might be misread as selecting characters or concatenating full strings.

Solution: Carefully note that:

• We must select contiguous substrings (not subsequences or individual characters)
• The order is fixed: substring from s comes first, then substring from t
• Either substring can be empty (allowing palindromes from just one string)

2. Incorrect Palindrome Extension Logic

Pitfall: When calculating palindrome lengths starting at each position, developers often forget to handle both odd and even-length palindromes, or incorrectly update the palindrome length array.

Example of incorrect code:

# Wrong: Only checks odd-length palindromes
for i in range(n):
    expand_around_center(string, palindrome_lengths, i, i)
    # Missing even-length palindrome check

Solution: Always check both cases:

for i in range(n):
    expand_around_center(string, palindrome_lengths, i, i)      # Odd-length
    expand_around_center(string, palindrome_lengths, i, i + 1)  # Even-length

3. Off-by-One Errors in DP Table

Pitfall: Confusion between 0-indexed strings and 1-indexed DP table leads to accessing wrong characters or array indices.

Example of incorrect indexing:

# Wrong: Using wrong indices
if s[i] == t_reversed[j]:  # Should be s[i-1] == t_reversed[j-1]
    dp[i][j] = dp[i-1][j-1] + 1

Solution: Use enumerate with proper starting index:

for i, char_s in enumerate(s, 1):      # Start from 1
    for j, char_t in enumerate(t_reversed, 1):  # Start from 1
        if char_s == char_t:            # Direct character comparison
            dp[i][j] = dp[i-1][j-1] + 1

4. Forgetting to Reverse String t

Pitfall: Working with the original string t instead of its reverse breaks the palindrome formation logic.

Why it's critical: When we concatenate substring from s with substring from t, for them to form a palindrome, the end of the s substring must match the beginning of the t substring in reverse order.

Solution: Always reverse t before processing:

t_reversed = t[::-1]  # Critical step

5. Incorrect Answer Initialization

Pitfall: Initializing the answer to 0 or forgetting to consider single-string palindromes.

Example of incorrect initialization:

# Wrong: Misses palindromes that come from just one string
max_palindrome_length = 0

Solution: Initialize with the longest palindrome from either string alone:

max_palindrome_length = max(*s_palindrome_lengths, *t_palindrome_lengths)

6. Boundary Check Errors When Extending Palindromes

Pitfall: Not checking if there are remaining characters after the matching prefix when trying to extend with additional palindromes.

Example of incorrect boundary checking:

# Wrong: Always tries to add palindrome extension
max_palindrome_length = max(max_palindrome_length, 
                           dp[i][j] * 2 + s_palindrome_lengths[i])
# This causes index out of bounds when i == m

Solution: Always verify boundaries before accessing extension arrays:

if i < m:  # Check boundary
    max_palindrome_length = max(max_palindrome_length, 
                               dp[i][j] * 2 + s_palindrome_lengths[i])
else:
    max_palindrome_length = max(max_palindrome_length, dp[i][j] * 2)