2002. Maximum Product of The Length of Two Palindromic Subsequences - Explanation

Prerequisites

Before attempting this problem, you should be comfortable with:

Recursion and Backtracking - Used to explore all possible ways to partition characters into two subsequences
Palindrome Checking - Understanding how to verify if a string reads the same forwards and backwards using two pointers
Bitmask/Bit Manipulation - Representing subsets of characters as binary numbers for efficient enumeration
Dynamic Programming (Longest Palindromic Subsequence) - Computing the LPS of a string using DP to optimize the solution

1. Recursion (Backtracking)

Intuition

We need two disjoint palindromic subsequences with maximum product of their lengths. For each character, we have three choices: add it to the first subsequence, add it to the second subsequence, or skip it entirely. At the end, if both subsequences are palindromes, we compute their product.

Algorithm

Use recursion with three branches per character: skip, add to seq1, or add to seq2.
At the end of the string, check if both subsequences are palindromes.
If both are palindromes, update the result with their length product.
Return the maximum product found.

class Solution:
    def maxProduct(self, s: str) -> int:
        def isPal(s):
            i, j = 0, len(s) - 1
            while i < j:
                if s[i] != s[j]:
                    return False
                i += 1
                j -= 1
            return True

        res = 0
        def rec(i, seq1, seq2):
            nonlocal res
            if i == len(s):
                if isPal(seq1) and isPal(seq2):
                    res = max(res, len(seq1) * len(seq2))
                return

            rec(i + 1, seq1, seq2)
            rec(i + 1, seq1 + s[i], seq2)
            rec(i + 1, seq1, seq2 + s[i])

        rec(0, "", "")
        return res

public class Solution {
    private boolean isPal(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) return false;
            i++;
            j--;
        }
        return true;
    }

    public void rec(int i, String s, StringBuilder seq1, StringBuilder seq2, int[] res) {
        if (i == s.length()) {
            if (isPal(seq1.toString()) && isPal(seq2.toString())) {
                res[0] = Math.max(res[0], seq1.length() * seq2.length());
            }
            return;
        }

        rec(i + 1, s, seq1, seq2, res);
        seq1.append(s.charAt(i));
        rec(i + 1, s, seq1, seq2, res);
        seq1.deleteCharAt(seq1.length() - 1);

        seq2.append(s.charAt(i));
        rec(i + 1, s, seq1, seq2, res);
        seq2.deleteCharAt(seq2.length() - 1);
    }

    public int maxProduct(String s) {
        int[] res = new int[1];
        rec(0, s, new StringBuilder(), new StringBuilder(), res);
        return res[0];
    }
}

class Solution {
public:
    bool isPal(const string &s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }

    void rec(int i, string& s, string& seq1, string& seq2, int &res) {
        if (i == s.length()) {
            if (isPal(seq1) && isPal(seq2)) {
                res = max(res, (int)seq1.length() * (int)seq2.length());
            }
            return;
        }

        rec(i + 1, s, seq1, seq2, res);
        seq1.push_back(s[i]);
        rec(i + 1, s, seq1, seq2, res);
        seq1.pop_back();
        seq2.push_back(s[i]);
        rec(i + 1, s, seq1, seq2, res);
        seq2.pop_back();
    }

    int maxProduct(string s) {
        int res = 0;
        string seq1 = "", seq2 = "";
        rec(0, s, seq1, seq2, res);
        return res;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    maxProduct(s) {
        const isPal = (str) => {
            let i = 0,
                j = str.length - 1;
            while (i < j) {
                if (str[i] !== str[j]) return false;
                i++;
                j--;
            }
            return true;
        };

        let res = 0;

        const rec = (i, seq1, seq2) => {
            if (i === s.length) {
                if (isPal(seq1) && isPal(seq2)) {
                    res = Math.max(res, seq1.length * seq2.length);
                }
                return;
            }

            rec(i + 1, seq1, seq2);
            rec(i + 1, seq1 + s[i], seq2);
            rec(i + 1, seq1, seq2 + s[i]);
        };

        rec(0, '', '');
        return res;
    }
}

public class Solution {
    public int MaxProduct(string s) {
        int res = 0;
        Rec(0, s, "", "", ref res);
        return res;
    }

    private void Rec(int i, string s, string seq1, string seq2, ref int res) {
        if (i == s.Length) {
            if (IsPal(seq1) && IsPal(seq2)) {
                res = Math.Max(res, seq1.Length * seq2.Length);
            }
            return;
        }

        Rec(i + 1, s, seq1, seq2, ref res);
        Rec(i + 1, s, seq1 + s[i], seq2, ref res);
        Rec(i + 1, s, seq1, seq2 + s[i], ref res);
    }

    private bool IsPal(string s) {
        int i = 0, j = s.Length - 1;
        while (i < j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
}

func maxProduct(s string) int {
    res := 0

    var isPal func(str string) bool
    isPal = func(str string) bool {
        i, j := 0, len(str)-1
        for i < j {
            if str[i] != str[j] {
                return false
            }
            i++
            j--
        }
        return true
    }

    var rec func(i int, seq1, seq2 string)
    rec = func(i int, seq1, seq2 string) {
        if i == len(s) {
            if isPal(seq1) && isPal(seq2) {
                if len(seq1)*len(seq2) > res {
                    res = len(seq1) * len(seq2)
                }
            }
            return
        }

        rec(i+1, seq1, seq2)
        rec(i+1, seq1+string(s[i]), seq2)
        rec(i+1, seq1, seq2+string(s[i]))
    }

    rec(0, "", "")
    return res
}

class Solution {
    fun maxProduct(s: String): Int {
        var res = 0

        fun isPal(str: String): Boolean {
            var i = 0
            var j = str.length - 1
            while (i < j) {
                if (str[i] != str[j]) return false
                i++
                j--
            }
            return true
        }

        fun rec(i: Int, seq1: String, seq2: String) {
            if (i == s.length) {
                if (isPal(seq1) && isPal(seq2)) {
                    res = maxOf(res, seq1.length * seq2.length)
                }
                return
            }

            rec(i + 1, seq1, seq2)
            rec(i + 1, seq1 + s[i], seq2)
            rec(i + 1, seq1, seq2 + s[i])
        }

        rec(0, "", "")
        return res
    }
}

class Solution {
    func maxProduct(_ s: String) -> Int {
        let chars = Array(s)
        var res = 0

        func isPal(_ str: [Character]) -> Bool {
            var i = 0, j = str.count - 1
            while i < j {
                if str[i] != str[j] { return false }
                i += 1
                j -= 1
            }
            return true
        }

        func rec(_ i: Int, _ seq1: [Character], _ seq2: [Character]) {
            if i == chars.count {
                if isPal(seq1) && isPal(seq2) {
                    res = max(res, seq1.count * seq2.count)
                }
                return
            }

            rec(i + 1, seq1, seq2)
            rec(i + 1, seq1 + [chars[i]], seq2)
            rec(i + 1, seq1, seq2 + [chars[i]])
        }

        rec(0, [], [])
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n * 3 ^ n)$
Space complexity: $O(n)$

2. Bit Mask

Intuition

Since the string length is small (up to 12), we can represent each subsequence as a bitmask. We first enumerate all possible subsequences, check which ones are palindromes, and store them. Then we check all pairs of palindromic subsequences. Two subsequences are disjoint if their masks have no overlapping bits (bitwise AND equals 0).

Algorithm

Enumerate all bitmasks from 1 to $2^n - 1$ .
For each mask, extract the corresponding subsequence and check if it is a palindrome.
Store palindromic masks with their lengths in a map.
For each pair of palindromic masks, if they are disjoint (AND equals 0), compute the product.
Return the maximum product.

class Solution:
    def maxProduct(self, s: str) -> int:
        def isPal(s):
            i, j = 0, len(s) - 1
            while i < j:
                if s[i] != s[j]:
                    return False
                i += 1
                j -= 1
            return True

        N, pali = len(s), {}

        for mask in range(1, 1 << N):
            subseq = ""
            for i in range(N):
                if mask & (1 << i):
                    subseq += s[i]

            if isPal(subseq):
                pali[mask] = len(subseq)

        res = 0
        for m1 in pali:
            for m2 in pali:
                if m1 & m2 == 0:
                    res = max(res, pali[m1] * pali[m2])

        return res

public class Solution {
    public int maxProduct(String s) {
        int N = s.length();
        int res = 0;
        Map<Integer, Integer> pali = new HashMap<>();

        for (int mask = 1; mask < (1 << N); mask++) {
            StringBuilder subseq = new StringBuilder();
            for (int i = 0; i < N; i++) {
                if ((mask & (1 << i)) != 0) {
                    subseq.append(s.charAt(i));
                }
            }

            if (isPal(subseq.toString())) {
                pali.put(mask, subseq.length());
            }
        }

        for (int m1 : pali.keySet()) {
            for (int m2 : pali.keySet()) {
                if ((m1 & m2) == 0) {
                    res = Math.max(res, pali.get(m1) * pali.get(m2));
                }
            }
        }

        return res;
    }

    private boolean isPal(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

class Solution {
public:
    int maxProduct(string s) {
        int N = s.length();
        int res = 0;
        unordered_map<int, int> pali;

        for (int mask = 1; mask < (1 << N); mask++) {
            string subseq = "";
            for (int i = 0; i < N; i++) {
                if (mask & (1 << i)) {
                    subseq += s[i];
                }
            }

            if (isPal(subseq)) {
                pali[mask] = subseq.length();
            }
        }

        for (auto& m1 : pali) {
            for (auto& m2 : pali) {
                if ((m1.first & m2.first) == 0) {
                    res = max(res, m1.second * m2.second);
                }
            }
        }

        return res;
    }

private:
    bool isPal(const string& s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s[i] != s[j]) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    maxProduct(s) {
        const isPal = (str) => {
            let i = 0,
                j = str.length - 1;
            while (i < j) {
                if (str[i] !== str[j]) return false;
                i++;
                j--;
            }
            return true;
        };

        const N = s.length;
        let res = 0;
        const pali = new Map();

        for (let mask = 1; mask < 1 << N; mask++) {
            let subseq = '';
            for (let i = 0; i < N; i++) {
                if ((mask & (1 << i)) !== 0) {
                    subseq += s[i];
                }
            }

            if (isPal(subseq)) {
                pali.set(mask, subseq.length);
            }
        }

        for (let [m1, len1] of pali) {
            for (let [m2, len2] of pali) {
                if ((m1 & m2) === 0) {
                    res = Math.max(res, len1 * len2);
                }
            }
        }

        return res;
    }
}

public class Solution {
    public int MaxProduct(string s) {
        int N = s.Length;
        int res = 0;
        var pali = new Dictionary<int, int>();

        for (int mask = 1; mask < (1 << N); mask++) {
            var subseq = new System.Text.StringBuilder();
            for (int i = 0; i < N; i++) {
                if ((mask & (1 << i)) != 0) {
                    subseq.Append(s[i]);
                }
            }

            if (IsPal(subseq.ToString())) {
                pali[mask] = subseq.Length;
            }
        }

        foreach (var m1 in pali) {
            foreach (var m2 in pali) {
                if ((m1.Key & m2.Key) == 0) {
                    res = Math.Max(res, m1.Value * m2.Value);
                }
            }
        }

        return res;
    }

    private bool IsPal(string s) {
        int i = 0, j = s.Length - 1;
        while (i < j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
}

func maxProduct(s string) int {
    isPal := func(str string) bool {
        i, j := 0, len(str)-1
        for i < j {
            if str[i] != str[j] {
                return false
            }
            i++
            j--
        }
        return true
    }

    N := len(s)
    res := 0
    pali := make(map[int]int)

    for mask := 1; mask < (1 << N); mask++ {
        subseq := ""
        for i := 0; i < N; i++ {
            if mask&(1<<i) != 0 {
                subseq += string(s[i])
            }
        }

        if isPal(subseq) {
            pali[mask] = len(subseq)
        }
    }

    for m1, len1 := range pali {
        for m2, len2 := range pali {
            if m1&m2 == 0 {
                if len1*len2 > res {
                    res = len1 * len2
                }
            }
        }
    }

    return res
}

class Solution {
    fun maxProduct(s: String): Int {
        fun isPal(str: String): Boolean {
            var i = 0
            var j = str.length - 1
            while (i < j) {
                if (str[i] != str[j]) return false
                i++
                j--
            }
            return true
        }

        val N = s.length
        var res = 0
        val pali = mutableMapOf<Int, Int>()

        for (mask in 1 until (1 shl N)) {
            val subseq = StringBuilder()
            for (i in 0 until N) {
                if ((mask and (1 shl i)) != 0) {
                    subseq.append(s[i])
                }
            }

            if (isPal(subseq.toString())) {
                pali[mask] = subseq.length
            }
        }

        for ((m1, len1) in pali) {
            for ((m2, len2) in pali) {
                if ((m1 and m2) == 0) {
                    res = maxOf(res, len1 * len2)
                }
            }
        }

        return res
    }
}

class Solution {
    func maxProduct(_ s: String) -> Int {
        func isPal(_ str: String) -> Bool {
            let chars = Array(str)
            var i = 0, j = chars.count - 1
            while i < j {
                if chars[i] != chars[j] { return false }
                i += 1
                j -= 1
            }
            return true
        }

        let chars = Array(s)
        let N = chars.count
        var res = 0
        var pali = [Int: Int]()

        for mask in 1..<(1 << N) {
            var subseq = ""
            for i in 0..<N {
                if (mask & (1 << i)) != 0 {
                    subseq.append(chars[i])
                }
            }

            if isPal(subseq) {
                pali[mask] = subseq.count
            }
        }

        for (m1, len1) in pali {
            for (m2, len2) in pali {
                if (m1 & m2) == 0 {
                    res = max(res, len1 * len2)
                }
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(4 ^ n)$
Space complexity: $O(2 ^ n)$

3. Bit Mask + Longest Palindromic Subsequence

Intuition

Instead of checking all pairs of palindromic masks, we can optimize by iterating through palindromic masks for the first subsequence and computing the longest palindromic subsequence (LPS) in the remaining characters. This avoids redundant pair comparisons while still finding the optimal solution.

Algorithm

Enumerate all bitmasks from 1 to $2^n - 1$ .
For each mask, check if the corresponding subsequence is a palindrome.
If it is, compute the LPS of the remaining characters (those not in the mask) using dynamic programming.
Update the result with the product of the palindrome length and the LPS length.
Return the maximum product.

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s)
        if n == 0:
            return 0

        dp = [1] * n
        for i in range(n - 1, -1, -1):
            prev = 0
            for j in range(i + 1, n):
                tmp = dp[j]
                if s[i] == s[j]:
                    dp[j] = 2 + prev
                else:
                    dp[j] = max(dp[j - 1], dp[j])
                prev = tmp
        return dp[n - 1]

    def maxProduct(self, s: str) -> int:
        n = len(s)
        res = 0

        for i in range(1, 1 << n):
            seq1, seq2 = [], []

            for j in range(n):
                if (i & (1 << j)) != 0:
                    seq1.append(s[j])
                else:
                    seq2.append(s[j])

            if not self.isPal(seq1):
                continue

            lps = self.longestPalindromeSubseq(''.join(seq2))
            res = max(res, len(seq1) * lps)

        return res

    def isPal(self, s: str) -> bool:
        i, j = 0, len(s) - 1
        while i < j:
            if s[i] != s[j]:
                return False
            i += 1
            j -= 1
        return True

public class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        if (n == 0) return 0;

        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        for (int i = n - 1; i >= 0; i--) {
            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if (s.charAt(i) == s.charAt(j)) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    public int maxProduct(String s) {
        int n = s.length();
        int res = 0;

        for (int i = 1; i < (1 << n); i++) {
            StringBuilder seq1 = new StringBuilder(), seq2 = new StringBuilder();

            for (int j = 0; j < n; j++) {
                char c = s.charAt(j);
                if ((i & (1 << j)) != 0) seq1.append(c);
                else seq2.append(c);
            }

            if (!isPal(seq1.toString())) continue;
            int lps = longestPalindromeSubseq(seq2.toString());
            res = Math.max(res, seq1.length() * lps);
        }

        return res;
    }

    private boolean isPal(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

class Solution {
public:
    int longestPalindromeSubseq(string& s) {
        int n = s.length();
        if (n == 0) return 0;

        vector<int> dp(n, 1);
        for (int i = n - 1; i >= 0; i--) {
            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if (s[i] == s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    int maxProduct(string s) {
        int n = s.length();
        int res = 0;

        for (int i = 1; i < (1 << n); i++) {
            string seq1 = "", seq2 = "";

            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0) seq1 += s[j];
                else seq2 += s[j];
            }

            if (!isPal(seq1)) continue;

            int lps = longestPalindromeSubseq(seq2);
            res = max(res, int(seq1.length()) * lps);
        }

        return res;
    }

    bool isPal(string& s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s[i] != s[j]) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    longestPalindromeSubseq(s) {
        const n = s.length;
        if (n === 0) return 0;

        const dp = new Array(n).fill(1);
        for (let i = n - 1; i >= 0; i--) {
            let prev = 0;
            for (let j = i + 1; j < n; j++) {
                let tmp = dp[j];
                if (s[i] === s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    /**
     * @param {string} s
     * @return {number}
     */
    maxProduct(s) {
        const n = s.length;
        let res = 0;

        for (let i = 1; i < 1 << n; i++) {
            let seq1 = '',
                seq2 = '';

            for (let j = 0; j < n; j++) {
                if ((i & (1 << j)) !== 0) seq1 += s[j];
                else seq2 += s[j];
            }

            if (!this.isPal(seq1)) continue;

            const lps = this.longestPalindromeSubseq(seq2);
            res = Math.max(res, seq1.length * lps);
        }

        return res;
    }

    /**
     * @param {string} s
     * @return {boolean}
     */
    isPal(s) {
        let i = 0,
            j = s.length - 1;
        while (i < j) {
            if (s[i] !== s[j]) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

public class Solution {
    public int LongestPalindromeSubseq(string s) {
        int n = s.Length;
        if (n == 0) return 0;

        int[] dp = new int[n];
        Array.Fill(dp, 1);
        for (int i = n - 1; i >= 0; i--) {
            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if (s[i] == s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.Max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    public int MaxProduct(string s) {
        int n = s.Length;
        int res = 0;

        for (int i = 1; i < (1 << n); i++) {
            var seq1 = new System.Text.StringBuilder();
            var seq2 = new System.Text.StringBuilder();

            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0) seq1.Append(s[j]);
                else seq2.Append(s[j]);
            }

            if (!IsPal(seq1.ToString())) continue;
            int lps = LongestPalindromeSubseq(seq2.ToString());
            res = Math.Max(res, seq1.Length * lps);
        }

        return res;
    }

    private bool IsPal(string s) {
        int i = 0, j = s.Length - 1;
        while (i < j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
}

func maxProduct(s string) int {
    longestPalindromeSubseq := func(str string) int {
        n := len(str)
        if n == 0 {
            return 0
        }

        dp := make([]int, n)
        for i := range dp {
            dp[i] = 1
        }
        for i := n - 1; i >= 0; i-- {
            prev := 0
            for j := i + 1; j < n; j++ {
                tmp := dp[j]
                if str[i] == str[j] {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = max(dp[j-1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n-1]
    }

    isPal := func(str string) bool {
        i, j := 0, len(str)-1
        for i < j {
            if str[i] != str[j] {
                return false
            }
            i++
            j--
        }
        return true
    }

    n := len(s)
    res := 0

    for i := 1; i < (1 << n); i++ {
        seq1, seq2 := "", ""

        for j := 0; j < n; j++ {
            if i&(1<<j) != 0 {
                seq1 += string(s[j])
            } else {
                seq2 += string(s[j])
            }
        }

        if !isPal(seq1) {
            continue
        }

        lps := longestPalindromeSubseq(seq2)
        res = max(res, len(seq1)*lps)
    }

    return res
}

class Solution {
    fun longestPalindromeSubseq(s: String): Int {
        val n = s.length
        if (n == 0) return 0

        val dp = IntArray(n) { 1 }
        for (i in n - 1 downTo 0) {
            var prev = 0
            for (j in i + 1 until n) {
                val tmp = dp[j]
                if (s[i] == s[j]) {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = maxOf(dp[j - 1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n - 1]
    }

    fun maxProduct(s: String): Int {
        val n = s.length
        var res = 0

        for (i in 1 until (1 shl n)) {
            val seq1 = StringBuilder()
            val seq2 = StringBuilder()

            for (j in 0 until n) {
                if ((i and (1 shl j)) != 0) seq1.append(s[j])
                else seq2.append(s[j])
            }

            if (!isPal(seq1.toString())) continue
            val lps = longestPalindromeSubseq(seq2.toString())
            res = maxOf(res, seq1.length * lps)
        }

        return res
    }

    private fun isPal(s: String): Boolean {
        var i = 0
        var j = s.length - 1
        while (i < j) {
            if (s[i] != s[j]) return false
            i++
            j--
        }
        return true
    }
}

class Solution {
    func longestPalindromeSubseq(_ s: String) -> Int {
        let chars = Array(s)
        let n = chars.count
        if n == 0 { return 0 }

        var dp = [Int](repeating: 1, count: n)
        for i in stride(from: n - 1, through: 0, by: -1) {
            var prev = 0
            for j in (i + 1)..<n {
                let tmp = dp[j]
                if chars[i] == chars[j] {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = max(dp[j - 1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n - 1]
    }

    func maxProduct(_ s: String) -> Int {
        let chars = Array(s)
        let n = chars.count
        var res = 0

        for i in 1..<(1 << n) {
            var seq1 = ""
            var seq2 = ""

            for j in 0..<n {
                if (i & (1 << j)) != 0 {
                    seq1.append(chars[j])
                } else {
                    seq2.append(chars[j])
                }
            }

            if !isPal(seq1) { continue }
            let lps = longestPalindromeSubseq(seq2)
            res = max(res, seq1.count * lps)
        }

        return res
    }

    private func isPal(_ s: String) -> Bool {
        let chars = Array(s)
        var i = 0, j = chars.count - 1
        while i < j {
            if chars[i] != chars[j] { return false }
            i += 1
            j -= 1
        }
        return true
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2 * 2 ^ n)$
Space complexity: $O(n)$

4. Bit Mask + LPS (Optimal)

Intuition

We can further optimize by computing the LPS directly on the original string while using the mask to skip characters that belong to the first subsequence. This avoids creating new strings and reduces overhead. We also validate palindromes efficiently using two pointers with mask-based skipping.

Algorithm

For each bitmask, use a two-pointer approach to check if the selected characters form a palindrome and compute its length.
If it is a palindrome, run the LPS algorithm on the original string, but skip indices that are set in the mask.
Reset the DP array appropriately before each LPS computation.
Track and return the maximum product.

class Solution:
    def longestPalindromeSubseq(self, s: str, mask: int, dp: List[int]) -> int:
        n = len(s)
        for i in range(n - 1, -1, -1):
            if (mask & (1 << i)) != 0:
                continue

            prev = 0
            for j in range(i + 1, n):
                tmp = dp[j]
                if (mask & (1 << j)) == 0 and s[i] == s[j]:
                    dp[j] = 2 + prev
                else:
                    dp[j] = max(dp[j - 1], dp[j])
                prev = tmp
        return dp[-1]

    def maxProduct(self, s: str) -> int:
        n = len(s)
        res = 0
        dp = [1] * n

        for i in range(1, 1 << n):
            m1 = self.palsize(s, i)
            if m1 == 0:
                continue

            for j in range(n):
                if (i & (1 << j)) == 0:
                    dp[j] = 1
                else:
                    dp[j] = 0

            m2 = self.longestPalindromeSubseq(s, i, dp)
            res = max(res, m1 * m2)

        return res

    def palsize(self, s: str, mask: int) -> int:
        i, j = 0, len(s) - 1
        res = 0
        while i <= j:
            if (mask & (1 << i)) == 0:
                i += 1
            elif (mask & (1 << j)) == 0:
                j -= 1
            else:
                if s[i] != s[j]:
                    return 0
                res += 1 if i == j else 2
                i += 1
                j -= 1
        return res

public class Solution {
    public int longestPalindromeSubseq(String s, int mask, int[] dp) {
        int n = s.length();
        for (int i = n - 1; i >= 0; i--) {
            if ((mask & (1 << i)) != 0) continue;

            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if ((mask & (1 << j)) == 0 && s.charAt(i) == s.charAt(j)) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    public int maxProduct(String s) {
        int n = s.length();
        int res = 0;
        int[] dp = new int[n];

        for (int i = 1; i < (1 << n); i++) {
            int m1 = palsize(s, i);
            if (m1 == 0) continue;

            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) == 0) {
                    dp[j] = 1;
                } else {
                    dp[j] = 0;
                }
            }
            int m2 = longestPalindromeSubseq(s, i, dp);
            res = Math.max(res, m1 * m2);
        }
        return res;
    }

    public int palsize(String s, int mask) {
        int i = 0, j = s.length() - 1;
        int res = 0;
        while (i <= j) {
            if ((mask & (1 << i)) == 0) i++;
            else if ((mask & (1 << j)) == 0) j--;
            else {
                if (s.charAt(i) != s.charAt(j)) return 0;
                res += (i == j) ? 1 : 2;
                i++;
                j--;
            }
        }
        return res;
    }
}

class Solution {
public:
    int longestPalindromeSubseq(string& s, int mask, vector<int>& dp) {
        int n = s.length();
        for (int i = n - 1; i >= 0; i--) {
            if ((mask & (1 << i)) != 0) continue;

            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if ((mask & (1 << j)) == 0 && s[i] == s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    int maxProduct(string s) {
        int n = s.length();
        int res = 0;
        vector<int> dp(n, 1);

        for (int i = 1; i < (1 << n); i++) {
            int m1 = palsize(s, i);
            if (m1 == 0) continue;

            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) == 0) {
                    dp[j] = 1;
                } else {
                    dp[j] = 0;
                }
            }
            int m2 = longestPalindromeSubseq(s, i, dp);
            res = max(res, m1 * m2);
        }

        return res;
    }

    int palsize(string& s, int mask) {
        int i = 0, j = s.length() - 1;
        int res = 0;
        while (i <= j) {
            if ((mask & (1 << i)) == 0) i++;
            else if ((mask & (1 << j)) == 0) j--;
            else {
                if (s[i] != s[j]) return 0;
                res += (i == j) ? 1 : 2;
                i++;
                j--;
            }
        }
        return res;
    }
};

class Solution {
    /**
     * @param {string} s
     * @param {number} mask
     * @param {number[]} dp
     * @return {number}
     */
    longestPalindromeSubseq(s, mask, dp) {
        const n = s.length;
        for (let i = n - 1; i >= 0; i--) {
            if ((mask & (1 << i)) !== 0) continue;

            let prev = 0;
            for (let j = i + 1; j < n; j++) {
                const tmp = dp[j];
                if ((mask & (1 << j)) === 0 && s[i] === s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    /**
     * @param {string} s
     * @return {boolean}
     */
    maxProduct(s) {
        const n = s.length;
        let res = 0;
        const dp = Array(n).fill(1);

        for (let i = 1; i < 1 << n; i++) {
            const m1 = this.palsize(s, i);
            if (m1 === 0) continue;

            for (let j = 0; j < n; j++) {
                if ((i & (1 << j)) === 0) {
                    dp[j] = 1;
                } else {
                    dp[j] = 0;
                }
            }
            const m2 = this.longestPalindromeSubseq(s, i, dp);
            res = Math.max(res, m1 * m2);
        }
        return res;
    }

    /**
     * @param {string} s
     * @param {number} mask
     * @return {number}
     */
    palsize(s, mask) {
        let i = 0,
            j = s.length - 1;
        let res = 0;
        while (i <= j) {
            if ((mask & (1 << i)) === 0) i++;
            else if ((mask & (1 << j)) === 0) j--;
            else {
                if (s[i] !== s[j]) return 0;
                res += i === j ? 1 : 2;
                i++;
                j--;
            }
        }
        return res;
    }
}

public class Solution {
    public int LongestPalindromeSubseq(string s, int mask, int[] dp) {
        int n = s.Length;
        for (int i = n - 1; i >= 0; i--) {
            if ((mask & (1 << i)) != 0) continue;

            int prev = 0;
            for (int j = i + 1; j < n; j++) {
                int tmp = dp[j];
                if ((mask & (1 << j)) == 0 && s[i] == s[j]) {
                    dp[j] = 2 + prev;
                } else {
                    dp[j] = Math.Max(dp[j - 1], dp[j]);
                }
                prev = tmp;
            }
        }
        return dp[n - 1];
    }

    public int MaxProduct(string s) {
        int n = s.Length;
        int res = 0;
        int[] dp = new int[n];

        for (int i = 1; i < (1 << n); i++) {
            int m1 = Palsize(s, i);
            if (m1 == 0) continue;

            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) == 0) {
                    dp[j] = 1;
                } else {
                    dp[j] = 0;
                }
            }
            int m2 = LongestPalindromeSubseq(s, i, dp);
            res = Math.Max(res, m1 * m2);
        }
        return res;
    }

    public int Palsize(string s, int mask) {
        int i = 0, j = s.Length - 1;
        int res = 0;
        while (i <= j) {
            if ((mask & (1 << i)) == 0) i++;
            else if ((mask & (1 << j)) == 0) j--;
            else {
                if (s[i] != s[j]) return 0;
                res += (i == j) ? 1 : 2;
                i++;
                j--;
            }
        }
        return res;
    }
}

func maxProduct(s string) int {
    n := len(s)
    res := 0
    dp := make([]int, n)

    longestPalindromeSubseq := func(mask int) int {
        for i := n - 1; i >= 0; i-- {
            if (mask & (1 << i)) != 0 {
                continue
            }

            prev := 0
            for j := i + 1; j < n; j++ {
                tmp := dp[j]
                if (mask&(1<<j)) == 0 && s[i] == s[j] {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = max(dp[j-1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n-1]
    }

    palsize := func(mask int) int {
        i, j := 0, n-1
        res := 0
        for i <= j {
            if (mask & (1 << i)) == 0 {
                i++
            } else if (mask & (1 << j)) == 0 {
                j--
            } else {
                if s[i] != s[j] {
                    return 0
                }
                if i == j {
                    res += 1
                } else {
                    res += 2
                }
                i++
                j--
            }
        }
        return res
    }

    for i := 1; i < (1 << n); i++ {
        m1 := palsize(i)
        if m1 == 0 {
            continue
        }

        for j := 0; j < n; j++ {
            if (i & (1 << j)) == 0 {
                dp[j] = 1
            } else {
                dp[j] = 0
            }
        }
        m2 := longestPalindromeSubseq(i)
        res = max(res, m1*m2)
    }

    return res
}

class Solution {
    fun longestPalindromeSubseq(s: String, mask: Int, dp: IntArray): Int {
        val n = s.length
        for (i in n - 1 downTo 0) {
            if ((mask and (1 shl i)) != 0) continue

            var prev = 0
            for (j in i + 1 until n) {
                val tmp = dp[j]
                if ((mask and (1 shl j)) == 0 && s[i] == s[j]) {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = maxOf(dp[j - 1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n - 1]
    }

    fun maxProduct(s: String): Int {
        val n = s.length
        var res = 0
        val dp = IntArray(n) { 1 }

        for (i in 1 until (1 shl n)) {
            val m1 = palsize(s, i)
            if (m1 == 0) continue

            for (j in 0 until n) {
                if ((i and (1 shl j)) == 0) {
                    dp[j] = 1
                } else {
                    dp[j] = 0
                }
            }
            val m2 = longestPalindromeSubseq(s, i, dp)
            res = maxOf(res, m1 * m2)
        }
        return res
    }

    fun palsize(s: String, mask: Int): Int {
        var i = 0
        var j = s.length - 1
        var res = 0
        while (i <= j) {
            if ((mask and (1 shl i)) == 0) i++
            else if ((mask and (1 shl j)) == 0) j--
            else {
                if (s[i] != s[j]) return 0
                res += if (i == j) 1 else 2
                i++
                j--
            }
        }
        return res
    }
}

class Solution {
    func longestPalindromeSubseq(_ s: [Character], _ mask: Int, _ dp: inout [Int]) -> Int {
        let n = s.count
        for i in stride(from: n - 1, through: 0, by: -1) {
            if (mask & (1 << i)) != 0 { continue }

            var prev = 0
            for j in (i + 1)..<n {
                let tmp = dp[j]
                if (mask & (1 << j)) == 0 && s[i] == s[j] {
                    dp[j] = 2 + prev
                } else {
                    dp[j] = max(dp[j - 1], dp[j])
                }
                prev = tmp
            }
        }
        return dp[n - 1]
    }

    func maxProduct(_ s: String) -> Int {
        let chars = Array(s)
        let n = chars.count
        var res = 0
        var dp = [Int](repeating: 1, count: n)

        for i in 1..<(1 << n) {
            let m1 = palsize(chars, i)
            if m1 == 0 { continue }

            for j in 0..<n {
                if (i & (1 << j)) == 0 {
                    dp[j] = 1
                } else {
                    dp[j] = 0
                }
            }
            let m2 = longestPalindromeSubseq(chars, i, &dp)
            res = max(res, m1 * m2)
        }
        return res
    }

    func palsize(_ s: [Character], _ mask: Int) -> Int {
        var i = 0, j = s.count - 1
        var res = 0
        while i <= j {
            if (mask & (1 << i)) == 0 { i += 1 }
            else if (mask & (1 << j)) == 0 { j -= 1 }
            else {
                if s[i] != s[j] { return 0 }
                res += (i == j) ? 1 : 2
                i += 1
                j -= 1
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2 * 2 ^ n)$
Space complexity: $O(n)$

Common Pitfalls

Allowing Overlapping Characters Between Subsequences

The two palindromic subsequences must be disjoint, meaning they cannot share any character. When using bitmasks, ensure that (mask1 & mask2) == 0 before considering a pair. Forgetting this check leads to invalid solutions where the same character is counted in both subsequences.

Not Validating Palindrome Before Computing Product

When iterating through bitmasks, some masks represent subsequences that are not palindromes. Computing the product of lengths without first verifying that both subsequences are actually palindromes leads to incorrect maximum values.

Inefficient Palindrome Checking in Tight Loops

Constructing new strings for each subsequence and then checking if they are palindromes is expensive. For optimal solutions, use two-pointer validation directly on the original string with mask-based index skipping to avoid repeated string allocations inside the exponential loop.