LeetCode 1216: Valid Palindrome III Solution

Master LeetCode problem 1216 (Valid Palindrome III), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1216. Valid Palindrome III

Hard

String Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming to determine the minimum number of characters that need to be removed to make a given string a palindrome. We can iterate through the string and keep track of the number of characters that need to be removed to make the substring from i to j a palindrome. We can then use this information to build a dynamic programming table to store the minimum number of characters that need to be removed to make all substrings from i to j a palindrome.

The algorithmic idea involves iterating through the string and filling up the dynamic programming table based on whether the characters at the current indices match or not. We update the table based on three cases:

If the characters at the current indices match, we can ignore them and move to the next indices.
If the characters at the current indices do not match, we have two options: a. We can remove the character at index i and continue with index j. b. We can remove the character at index j and continue with index i.
We choose the minimum of the two options above as the number of characters to be removed.

Finally, we check if the number of characters to be removed is less than or equal to the given k. If it is, then the string can be converted into a palindrome by removing at most k characters.

Time complexity: O(n^2) Space complexity: O(n^2)

Solution Code

class Solution {
    public boolean isValidPalindrome(String s, int k) {
        int n = s.length();
        int[][] dp = new int[n][n];
        
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                int j = i + len - 1;
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i+1][j-1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i+1][j], dp[i][j-1]);
                }
            }
        }
        
        return dp[0][n-1] <= k;
    }
}

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