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1216. Valid Palindrome III 👍

  • Time: $O(n^2)$
  • Space: $O(n^2)$
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class Solution {
 public:
  bool isValidPalindrome(string s, int k) {
    return s.length() - longestPalindromeSubseq(s) <= k;
  }

 private:
  // Same as 516. Longest Palindromic Subsequence
  int longestPalindromeSubseq(const string& s) {
    const int n = s.length();
    // dp[i][j] := the length of LPS(s[i..j])
    vector<vector<int>> dp(n, vector<int>(n));

    for (int i = 0; i < n; ++i)
      dp[i][i] = 1;

    for (int d = 1; d < n; ++d)
      for (int i = 0; i + d < n; ++i) {
        const int j = i + d;
        if (s[i] == s[j])
          dp[i][j] = 2 + dp[i + 1][j - 1];
        else
          dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
      }

    return dp[0][n - 1];
  }
};

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class Solution {
  public boolean isValidPalindrome(String s, int k) {
    return s.length() - longestPalindromeSubseq(s) <= k;
  }

  // Same as 516. Longest Palindromic Subsequence
  private int longestPalindromeSubseq(final String s) {
    final int n = s.length();
    // dp[i][j] := the length of LPS(s[i..j])
    int[][] dp = new int[n][n];

    for (int i = 0; i < n; ++i)
      dp[i][i] = 1;

    for (int d = 1; d < n; ++d)
      for (int i = 0; i + d < n; ++i) {
        final int j = i + d;
        if (s.charAt(i) == s.charAt(j))
          dp[i][j] = 2 + dp[i + 1][j - 1];
        else
          dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
      }

    return dp[0][n - 1];
  }
}

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class Solution:
  def isValidPalindrome(self, s: str, k: int) -> bool:
    return len(s) - self._longestPalindromeSubseq(s) <= k

  # Same as 516. Longest Palindromic Subsequence
  def _longestPalindromeSubseq(self, s: str) -> int:
    n = len(s)
    # dp[i][j] := the length of LPS(s[i..j])
    dp = [[0] * n for _ in range(n)]

    for i in range(n):
      dp[i][i] = 1

    for d in range(1, n):
      for i in range(n - d):
        j = i + d
        if s[i] == s[j]:
          dp[i][j] = 2 + dp[i + 1][j - 1]
        else:
          dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])

    return dp[0][n - 1]