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3503. Longest Palindrome After Substring Concatenation I

Description

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.

Solutions

Solution 1: Enumerate Palindrome Centers + Dynamic Programming

According to the problem description, the concatenated palindrome string can be composed entirely of string \(s\), entirely of string \(t\), or a combination of both strings \(s\) and \(t\). Additionally, there may be extra palindromic substrings in either string \(s\) or \(t\).

Therefore, we first reverse string \(t\) and preprocess arrays \(\textit{g1}\) and \(\textit{g2}\), where \(\textit{g1}[i]\) represents the length of the longest palindromic substring starting at index \(i\) in string \(s\), and \(\textit{g2}[i]\) represents the length of the longest palindromic substring starting at index \(i\) in string \(t\).

We can initialize the answer \(\textit{ans}\) as the maximum value in \(\textit{g1}\) and \(\textit{g2}\).

Next, we define \(\textit{f}[i][j]\) as the length of the palindromic substring ending at the \(i\)-th character of string \(s\) and the \(j\)-th character of string \(t\).

For \(\textit{f}[i][j]\), if \(s[i - 1]\) equals \(t[j - 1]\), then \(\textit{f}[i][j] = \textit{f}[i - 1][j - 1] + 1\). We then update the answer:

\[ \textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } i \geq m \text{ else } \textit{g1}[i])) \\ \textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } j \geq n \text{ else } \textit{g2}[j])) \]

Finally, we return the answer \(\textit{ans}\).

The time complexity is \(O(m \times (m + n))\), and the space complexity is \(O(m \times n)\), where \(m\) and \(n\) are the lengths of strings \(s\) and \(t\), respectively.

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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

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class Solution {
    public int longestPalindrome(String S, String T) {
        char[] s = S.toCharArray();
        char[] t = new StringBuilder(T).reverse().toString().toCharArray();
        int m = s.length, n = t.length;
        int[] g1 = calc(s), g2 = calc(t);
        int ans = Math.max(Arrays.stream(g1).max().getAsInt(), Arrays.stream(g2).max().getAsInt());
        int[][] f = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s[i - 1] == t[j - 1]) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                    ans = Math.max(ans, f[i][j] * 2 + (i < m ? g1[i] : 0));
                    ans = Math.max(ans, f[i][j] * 2 + (j < n ? g2[j] : 0));
                }
            }
        }
        return ans;
    }

    private void expand(char[] s, int[] g, int l, int r) {
        while (l >= 0 && r < s.length && s[l] == s[r]) {
            g[l] = Math.max(g[l], r - l + 1);
            --l;
            ++r;
        }
    }

    private int[] calc(char[] s) {
        int n = s.length;
        int[] g = new int[n];
        for (int i = 0; i < n; ++i) {
            expand(s, g, i, i);
            expand(s, g, i, i + 1);
        }
        return g;
    }
}

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class Solution {
public:
    int longestPalindrome(string s, string t) {
        int m = s.size(), n = t.size();
        ranges::reverse(t);
        vector<int> g1 = calc(s), g2 = calc(t);
        int ans = max(ranges::max(g1), ranges::max(g2));
        vector<vector<int>> f(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s[i - 1] == t[j - 1]) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                    ans = max(ans, f[i][j] * 2 + (i < m ? g1[i] : 0));
                    ans = max(ans, f[i][j] * 2 + (j < n ? g2[j] : 0));
                }
            }
        }
        return ans;
    }

private:
    void expand(const string& s, vector<int>& g, int l, int r) {
        while (l >= 0 && r < s.size() && s[l] == s[r]) {
            g[l] = max(g[l], r - l + 1);
            --l;
            ++r;
        }
    }

    vector<int> calc(const string& s) {
        int n = s.size();
        vector<int> g(n, 0);
        for (int i = 0; i < n; ++i) {
            expand(s, g, i, i);
            expand(s, g, i, i + 1);
        }
        return g;
    }
};

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func longestPalindrome(s, t string) int {
    m, n := len(s), len(t)
    t = reverse(t)

    g1, g2 := calc(s), calc(t)
    ans := max(slices.Max(g1), slices.Max(g2))

    f := make([][]int, m+1)
    for i := range f {
        f[i] = make([]int, n+1)
    }

    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            if s[i-1] == t[j-1] {
                f[i][j] = f[i-1][j-1] + 1
                a, b := 0, 0
                if i < m {
                    a = g1[i]
                }
                if j < n {
                    b = g2[j]
                }
                ans = max(ans, f[i][j]*2+a)
                ans = max(ans, f[i][j]*2+b)
            }
        }
    }
    return ans
}

func calc(s string) []int {
    n, g := len(s), make([]int, len(s))
    for i := 0; i < n; i++ {
        expand(s, g, i, i)
        expand(s, g, i, i+1)
    }
    return g
}

func expand(s string, g []int, l, r int) {
    for l >= 0 && r < len(s) && s[l] == s[r] {
        g[l] = max(g[l], r-l+1)
        l, r = l-1, r+1
    }
}

func reverse(s string) string {
    r := []rune(s)
    slices.Reverse(r)
    return string(r)
}

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function longestPalindrome(s: string, t: string): number {
    function expand(s: string, g: number[], l: number, r: number): void {
        while (l >= 0 && r < s.length && s[l] === s[r]) {
            g[l] = Math.max(g[l], r - l + 1);
            l--;
            r++;
        }
    }

    function calc(s: string): number[] {
        const n = s.length;
        const g: number[] = Array(n).fill(0);
        for (let i = 0; i < n; i++) {
            expand(s, g, i, i);
            expand(s, g, i, i + 1);
        }
        return g;
    }

    const m = s.length,
        n = t.length;
    t = t.split('').reverse().join('');
    const g1 = calc(s);
    const g2 = calc(t);
    let ans = Math.max(...g1, ...g2);

    const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));

    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (s[i - 1] === t[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
                ans = Math.max(ans, f[i][j] * 2 + (i >= m ? 0 : g1[i]));
                ans = Math.max(ans, f[i][j] * 2 + (j >= n ? 0 : g2[j]));
            }
        }
    }

    return ans;
}

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