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583. Delete Operation for Two Strings 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minDistance(string word1, string word2) {
    const int k = lcs(word1, word2);
    return (word1.length() - k) + (word2.length() - k);
  }

 private:
  int lcs(const string& a, const string& b) {
    const int m = a.length();
    const int n = b.length();
    // dp[i][j] := the length of LCS(a[0..i), b[0..j))
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        if (a[i - 1] == b[j - 1])
          dp[i][j] = 1 + dp[i - 1][j - 1];
        else
          dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

    return dp[m][n];
  }
};

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class Solution {
  public int minDistance(String word1, String word2) {
    final int k = lcs(word1, word2);
    return (word1.length() - k) + (word2.length() - k);
  }

  private int lcs(final String a, final String b) {
    final int m = a.length();
    final int n = b.length();
    // dp[i][j] := the length of LCS(a[0..i), b[0..j))
    int[][] dp = new int[m + 1][n + 1];

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        if (a.charAt(i - 1) == b.charAt(j - 1))
          dp[i][j] = 1 + dp[i - 1][j - 1];
        else
          dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);

    return dp[m][n];
  }
}

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class Solution:
  def minDistance(self, word1: str, word2: str) -> int:
    k = self._lcs(word1, word2)
    return (len(word1) - k) + (len(word2) - k)

  def _lcs(self, a: str, b: str) -> int:
    m = len(a)
    n = len(b)
    # dp[i][j] := the length of LCS(a[0..i), b[0..j))
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
      for j in range(1, n + 1):
        if a[i - 1] == b[j - 1]:
          dp[i][j] = 1 + dp[i - 1][j - 1]
        else:
          dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    return dp[m][n]