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  1. Practice
  2. C++
  3. Inheritance
  4. Magic Spells
  5. Discussions

Magic Spells

  • codeharrier + 4 comments

    What's the point of embedding the LCS problem in this dynamic-cast mystery problem?

    • dwilches + 1 comment

      Exactly my thought, it's two problems in one. I came for inheritance but I found the most complex part was the LCS.

      • yashpalsinghdeo1 + 0 comments

        here is solution of problem Magic Spells in c++ programming https://solution.programmingoneonone.com/2020/06/hackerrank-magic-spells-solution-in-c-plus-plus.html

    • francois_bailly + 0 comments

      Exactly. LCS is the only algorithm to implement here, it has nothing to do with C++ language knowledge. This exercise should be broken into 2 parts.

    • [deleted] + 0 comments

      Exactly. LCS is the only algorithm to implement here, it has nothing to do with C++ language knowledge. This exercise should be broken into 2 parts.

    • Abhishek_G0YAL + 2 comments

      If anyone is wondering about a way to solve this problem without using Dynamic Cast in C++.

      void counterspell(Spell *spell) {
    if(typeid(*spell) == typeid(Fireball)) {
        ((Fireball*)spell)->revealFirepower();
    }
    else if(typeid(*spell) == typeid(Frostbite)) {
        ((Frostbite*)spell)->revealFrostpower();
    }
    else if(typeid(*spell) == typeid(Thunderstorm)) {
        ((Thunderstorm*)spell)->revealThunderpower();
    }
    else if(typeid(*spell) == typeid(Waterbolt)) {
        ((Waterbolt*)spell)->revealWaterpower();
    }
    else {
        string spellName = spell->revealScrollName();
        string spellJournal = SpellJournal::journal;
        // LCS Algorithm
        string &a = spellName, &b = spellJournal;
        vector<vector<int> > dp(a.size() + 1, vector<int>(b.size() + 1));
        for(int i = 1; i <= a.size(); i++) {
            for(int j = 1; j <= b.size(); 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]);
            }
        }
        cout << dp[a.size()][b.size()] << endl;
    }
}

      • chiranjeevising1 + 0 comments
        [deleted]

      • v_k_hungund_two + 0 comments

        exactly my thought , why would anyone want to use dynamic_cast in this question ?