Welcome to Subscribe On Youtube

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
        }
    }
    
    

All Problems

All Solutions