a few solutions
Let A and B be the first and second candidate palindrome strings correspondingly.  Perform DFS + BT considering all possibilities for each character s[i]:\n\n1.
Let `A` and `B` be the first and second candidate palindrome strings correspondingly.  Perform DFS + BT considering all possibilities for each character `s[i]`:\n\n1. `s[i]` is included in `A`\n2. `s[i]` is included in `B`\n3. `s[i]` is *not* included in `A` or `B`\n\n---\n\n*Kotlin*\n```\nclass Solution {\n    fun maxProduct(s: String): Int {\n        var best = 0\n        var ok = { A: MutableList<Char> -> A == A.reversed() }\n        fun go(i: Int = 0, A: MutableList<Char> = mutableListOf<Char>(), B: MutableList<Char> = mutableListOf<Char>()) {\n            if (ok(A) && ok(B))\n                best = Math.max(best, A.size * B.size)\n            if (i == s.length)\n                return\n            A.add(s[i])\n            go(i + 1, A, B)\n            A.removeAt(A.lastIndex)\n            B.add(s[i])\n            go(i + 1, A, B)\n            B.removeAt(B.lastIndex)\n            go(i + 1, A, B)\n        }\n        go()\n        return best\n    }\n}\n```\n\n*Javascript*\n```\nlet maxProduct = (s, best = 0) => {\n    let ok = A => A.join(\'\') == [...A].reverse().join(\'\');\n    let go = (i = 0, A = [], B = []) => {\n        if (ok(A) && ok(B))\n            best = Math.max(best, A.length * B.length);\n        if (i == s.length)\n            return;\n        A.push(s[i]);\n        go(i + 1, A, B);\n        A.pop();\n        B.push(s[i]);\n        go(i + 1, A, B);\n        B.pop();\n        go(i + 1, A, B);\n    };\n    go();\n    return best;\n};\n```\n\n*Python3*\n```\nclass Solution:\n    def maxProduct(self, s: str, best = 0) -> int:\n        ok = lambda A: A == A[::-1]\n        def go(i = 0, A = [], B = []):\n            nonlocal best\n            if ok(A) and ok(B):\n                best = max(best, len(A) * len(B))\n            if i == len(s):\n                return\n            A.append(s[i])\n            go(i + 1, A, B)\n            A.pop()\n            B.append(s[i])\n            go(i + 1, A, B)\n            B.pop()\n            go(i + 1, A, B)\n        go()\n        return best\n```\n\n*C++*\n```\nclass Solution {\npublic:\n    using fun = function<void(int, string&&, string&&)>;\n    int maxProduct(string s, int best = 0) {\n        auto ok = [](auto& s) {\n            int N = s.size(),\n                i = 0,\n                j = N - 1;\n            while (i < j && s[i] == s[j])\n                ++i, --j;\n            return j <= i;\n        };\n        fun go = [&](auto i, auto&& A, auto&& B) {\n            if (ok(A) && ok(B))\n                best = max(best, int(A.size() * B.size()));\n            if (i == s.size())\n                return;\n            A.push_back(s[i]);\n            go(i + 1, move(A), move(B));\n            A.pop_back();\n            B.push_back(s[i]);\n            go(i + 1, move(A), move(B));\n            B.pop_back();\n            go(i + 1, move(A), move(B));\n        };\n        go(0, {}, {});\n        return best;\n    }\n};\n```
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