# 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 $s$.

• 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;
}
};

• 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


• 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
}
}