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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] := LPS's length in 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] := LPS's length in 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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