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 <= 1000s 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] ≠ 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 ≥ 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(n2), and the space complexity is O(n2). Here, n is the length of the string s.
PythonJavaC++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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
