Given a string which we can delete at most k, return whether you can make a palindrome.

Python:

def can_make_palindrome(s, k):
    def is_palindrome(sub):
        return sub == sub[::-1]

    n = len(s)
    deletions_needed = n - sum([1 for i, c in enumerate(s) if c != s[n - 1 - i]])

    return deletions_needed <= k or is_palindrome(s)

# Example usage:
s_example = "abca"
k_example = 1
result = can_make_palindrome(s_example, k_example)
print(result)

JavaScript:

function canMakePalindrome(s, k) {
    const isPalindrome = (sub) => sub === sub.split('').reverse().join('');

    const n = s.length;
    const deletionsNeeded = n - [...s].reduce((count, c, i) => (c !== s[n - 1 - i]) ? count + 1 : count, 0);

    return deletionsNeeded <= k || isPalindrome(s);
}

// Example usage:
const sExample = "abca";
const kExample = 1;
const result = canMakePalindrome(sExample, kExample);
console.log(result);

Java:

public class PalindromeChecker {
    public static boolean canMakePalindrome(String s, int k) {
        int deletionsNeeded = s.length() - s.chars().mapToObj(c -> (char) c)
                .filter(c -> c != s.charAt(s.length() - 1 - s.indexOf(c)))
                .count();

        return deletionsNeeded <= k || isPalindrome(s);
    }

    private static boolean isPalindrome(String sub) {
        return sub.equals(new StringBuilder(sub).reverse().toString());
    }

    public static void main(String[] args) {
        // Example usage:
        String sExample = "abca";
        int kExample = 1;
        boolean result = canMakePalindrome(sExample, kExample);
        System.out.println(result);
    }
}

These solutions check if a string can be converted into a palindrome by allowing at most k deletions. The is_palindrome function is used to check if a given substring is a palindrome. The Java solution includes a main method for example usage.

Asked By: Google