Longest Palindrome After Substring Concatenation II
<p>You are given two strings, <code>s</code> and <code>t</code>.</p>

<p>You can create a new string by selecting a <span data-keyword="substring">substring</span> from <code>s</code> (possibly empty) and a substring from <code>t</code> (possibly empty), then concatenating them <strong>in order</strong>.</p>

<p>Return the length of the <strong>longest</strong> <span data-keyword="palindrome-string">palindrome</span> that can be formed this way.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;a&quot;, t = &quot;a&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>Concatenating <code>&quot;a&quot;</code> from <code>s</code> and <code>&quot;a&quot;</code> from <code>t</code> results in <code>&quot;aa&quot;</code>, which is a palindrome of length 2.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abc&quot;, t = &quot;def&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p>Since all characters are different, the longest palindrome is any single character, so the answer is 1.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;b&quot;, t = &quot;aaaa&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>Selecting &quot;<code>aaaa</code>&quot; from <code>t</code> is the longest palindrome, so the answer is 4.</p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abcde&quot;, t = &quot;ecdba&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p>Concatenating <code>&quot;abc&quot;</code> from <code>s</code> and <code>&quot;ba&quot;</code> from <code>t</code> results in <code>&quot;abcba&quot;</code>, which is a palindrome of length 5.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length, t.length &lt;= 1000</code></li>
	<li><code>s</code> and <code>t</code> consist of lowercase English letters.</li>
</ul>
Two Pointers; String; Dynamic Programming
class Solution:
    def longestPalindrome(self, s: str, t: str) -> int:
        def expand(s: str, g: List[int], l: int, r: int):
            while l >= 0 and r < len(s) and s[l] == s[r]:
                g[l] = max(g[l], r - l + 1)
                l, r = l - 1, r + 1

        def calc(s: str) -> List[int]:
            n = len(s)
            g = [0] * n
            for i in range(n):
                expand(s, g, i, i)
                expand(s, g, i, i + 1)
            return g

        m, n = len(s), len(t)
        t = t[::-1]
        g1, g2 = calc(s), calc(t)
        ans = max(*g1, *g2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i, a in enumerate(s, 1):
            for j, b in enumerate(t, 1):
                if a == b:
                    f[i][j] = f[i - 1][j - 1] + 1
                    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]))
        return ans
