Skip to content

1216. Valid Palindrome III πŸ”’

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 $s$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
func isValidPalindrome(s string, k int) bool {
    n := len(s)
    f := make([][]int, n)
    for i := range f