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1092. Shortest Common Supersequence 👍

  • Time: $O(mn)$
  • Space: $O(mn^2 \cdot |\texttt{string}|)$
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class Solution {
 public:
  string shortestCommonSupersequence(string str1, string str2) {
    string ans;
    int i = 0;  // str1's index
    int j = 0;  // str2's index

    for (const char c : lcs(str1, str2)) {
      // Append the letters that are not part of the LCS.
      while (str1[i] != c)
        ans += str1[i++];
      while (str2[j] != c)
        ans += str2[j++];
      // Append the letter of the LCS and match it with str1 and str2.
      ans += c;
      ++i;
      ++j;
    }

    // Append the remaining letters.
    return ans + str1.substr(i) + str2.substr(j);
  }

 private:
  string 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<string>> dp(m + 1, vector<string>(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] = dp[i - 1][j - 1] + a[i - 1];
        else
          dp[i][j] = dp[i - 1][j].length() > dp[i][j - 1].length()
                         ? dp[i - 1][j]
                         : dp[i][j - 1];

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

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class Solution {
  public String shortestCommonSupersequence(String str1, String str2) {
    StringBuilder sb = new StringBuilder();
    int i = 0; // str1's index
    int j = 0; // str2's index

    for (final char c : lcs(str1, str2).toCharArray()) {
      // Append the letters that are not part of the LCS.
      while (str1.charAt(i) != c)
        sb.append(str1.charAt(i++));
      while (str2.charAt(j) != c)
        sb.append(str2.charAt(j++));
      // Append the letter of the LCS and match it with str1 and str2.
      sb.append(c);
      ++i;
      ++j;
    }

    // Append the remaining letters.
    return sb.toString() + str1.substring(i) + str2.substring(j);
  }

  private String 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))
    StringBuilder[][] dp = new StringBuilder[m + 1][n + 1];

    for (final StringBuilder[] row : dp)
      for (int i = 0; i < row.length; ++i)
        row[i] = new StringBuilder();

    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].append(dp[i - 1][j - 1]).append(a.charAt(i - 1));
        else
          dp[i][j] = dp[i - 1][j].length() > dp[i][j - 1].length() ? dp[i - 1][j] : dp[i][j - 1];

    return dp[m][n].toString();
  }
}