1216. Valid Palindrome III 🔒

Description

Given a string s and an integer k, return true if s is a k-palindrome.

A string is k-palindrome if it can be transformed into a palindrome by removing at most k characters from it.

Example 1:

Input: s = "abcdeca", k = 2
Output: true
Explanation: Remove 'b' and 'e' characters.

Example 2:

Input: s = "abbababa", k = 1
Output: true

Constraints:

1 <= s.length <= 1000
s consists of only lowercase English letters.
1 <= k <= s.length

Solutions

Solution 1: Dynamic Programming

The problem requires us to remove at most \(k\) characters to make the remaining string a palindrome. This can be transformed into finding the longest palindromic subsequence.

We define \(f[i][j]\) as the length of the longest palindromic subsequence in the substring \(s[i..j]\). Initially, we have \(f[i][i] = 1\) for all \(i\), since each single character is a palindrome.

If \(s[i] = s[j]\), then we have \(f[i][j] = f[i+1][j-1] + 2\), since we can add both \(s[i]\) and \(s[j]\) to the longest palindromic subsequence of \(s[i+1..j-1]\).

If \(s[i] \neq s[j]\), then we have \(f[i][j] = \max(f[i+1][j], f[i][j-1])\), since we need to remove either \(s[i]\) or \(s[j]\) to make the remaining substring a palindrome.

Finally, we check whether there exists \(f[i][j] + k \geq n\), where \(n\) is the length of the string \(s\). If so, it means that we can remove at most \(k\) characters to make the remaining string a palindrome.

The time complexity is \(O(n^2)\), and the space complexity is \(O(n^2)\). Here, \(n\) is the length of the string \(s\).

Python3JavaC++GoTypeScriptRust

class Solution:
    def isValidPalindrome(self, s: str, k: int) -> bool:
        n = len(s)
        f = [[0] * n for _ in range(n)]
        for i in range(n):
            f[i][i] = 1
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    f[i][j] = f[i + 1][j - 1] + 2
                else:
                    f[i][j] = max(f[i + 1][j], f[i][j - 1])
                if f[i][j] + k >= n:
                    return True
        return False

class Solution {
    public boolean isValidPalindrome(String s, int k) {
        int n = s.length();
        int[][] f = new int[n][n];
        for (int i = 0; i < n; ++i) {
            f[i][i] = 1;
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    f[i][j] = f[i + 1][j - 1] + 2;
                } else {
                    f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
                }
                if (f[i][j] + k >= n) {
                    return true;
                }
            }
        }
        return false;
    }
}

class Solution {
public:
    bool isValidPalindrome(string s, int k) {
        int n = s.length();
        int f[n][n];
        memset(f, 0, sizeof f);
        for (int i = 0; i < n; ++i) {
            f[i][i] = 1;
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    f[i][j] = f[i + 1][j - 1] + 2;
                } else {
                    f[i][j] = max(f[i + 1][j], f[i][j - 1]);
                }
                if (f[i][j] + k >= n) {
                    return true;
                }
            }
        }
        return false;
    }
};

func isValidPalindrome(s string, k int) bool {
    n := len(s)
    f := make([][]int, n)
    for i := range f {
        f[i] = make([]int, n)
        f[i][i] = 1
    }
    for i := n - 2; i >= 0; i-- {
        for j := i + 1; j < n; j++ {
            if s[i] == s[j] {
                f[i][j] = f[i+1][j-1] + 2
            } else {
                f[i][j] = max(f[i+1][j], f[i][j-1])
            }
            if f[i][j]+k >= n {
                return true
            }
        }
    }
    return false
}

function isValidPalindrome(s: string, k: number): boolean {
    const n = s.length;
    const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
    for (let i = 0; i < n; ++i) {
        f[i][i] = 1;
    }
    for (let i = n - 2; ~i; --i) {
        for (let j = i + 1; j < n; ++j) {
            if (s[i] === s[j]) {
                f[i][j] = f[i + 1][j - 1] + 2;
            } else {
                f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
            }
            if (f[i][j] + k >= n) {
                return true;
            }
        }
    }
    return false;
}

impl Solution {
    pub fn is_valid_palindrome(s: String, k: i32) -> bool {
        let s = s.as_bytes();
        let n = s.len();
        let mut f = vec![vec![0; n]; n];

        for i in 0..n {
            f[i][i] = 1;
        }

        for i in (0..n - 2).rev() {
            for j in i + 1..n {
                if s[i] == s[j] {
                    f[i][j] = f[i + 1][j - 1] + 2;
                } else {
                    f[i][j] = std::cmp::max(f[i + 1][j], f[i][j - 1]);
                }

                if f[i][j] + k >= (n as i32) {
                    return true;
                }
            }
        }

        false
    }
}

1216. Valid Palindrome III 🔒

Description

Solutions

Solution 1: Dynamic Programming

Comments