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1143. Longest Common Subsequence 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int longestCommonSubsequence(string text1, string text2) {
    const int m = text1.length();
    const int n = text2.length();
    // dp[i][j] := the length of LCS(text1[0..i), text2[0..j))
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        dp[i + 1][j + 1] = text1[i] == text2[j]
                               ? 1 + dp[i][j]
                               : max(dp[i][j + 1], dp[i + 1][j]);

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

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class Solution {
  public int longestCommonSubsequence(String text1, String text2) {
    final int m = text1.length();
    final int n = text2.length();
    // dp[i][j] := the length of LCS(text1[0..i), text2[0..j))
    int[][] dp = new int[m + 1][n + 1];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        dp[i + 1][j + 1] = text1.charAt(i) == text2.charAt(j)
                               ? 1 + dp[i][j]
                               : Math.max(dp[i][j + 1], dp[i + 1][j]);

    return dp[m][n];
  }
}

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class Solution:
  def longestCommonSubsequence(self, text1: str, text2: str) -> int:
    m = len(text1)
    n = len(text2)
    # dp[i][j] := the length of LCS(text1[0..i), text2[0..j))
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(m):
      for j in range(n):
        dp[i + 1][j + 1] = (1 + dp[i][j] if text1[i] == text2[j]
                            else max(dp[i][j + 1], dp[i + 1][j]))

    return dp[m][n]