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  1. Practice
  2. Algorithms
  3. Implementation
  4. The Bomberman Game
  5. Discussions

The Bomberman Game

  • sarathy_v_krish1 + 4 comments

    C++ solution :

    vector<string> bomberMan(int n, vector<string> grid) {
    vector<string> set1=grid, set2, set3;

    for (int i=0;i<grid.size();i++)
        for (int j=0;j<grid[0].size();j++)
            set1[i][j]='O';
    set2=set3=set1;
    for (int i=0;i<grid.size();i++)
        for (int j=0;j<grid[0].size();j++)
            if (grid[i][j]=='O')
            {
                if (i>0)    set2[i-1][j]='.';
                if (i<grid.size()-1)    set2[i+1][j]='.';
                if (j>0)    set2[i][j-1]='.';
                if (j<grid[0].size()-1) set2[i][j+1]='.';
                set2[i][j]='.';
            }
    for (int i=0;i<grid.size();i++)
        for (int j=0;j<grid[0].size();j++)
            if (set2[i][j]=='O')
            {
                if (i>0)    set3[i-1][j]='.';
                if (i<set2.size()-1)    set3[i+1][j]='.';
                if (j>0)    set3[i][j-1]='.';
                if (j<set2[0].size()-1) set3[i][j+1]='.';
                set3[i][j]='.';
            }
    /*cout << "Printing grid, set1, set2 : \n";
    for (int i=0;i<grid.size();i++) cout << grid[i] << endl;
    cout << endl << endl;
    for (int i=0;i<grid.size();i++) cout << set1[i] << endl;
    cout << endl << endl;
    for (int i=0;i<grid.size();i++) cout << set2[i] << endl;*/

    if (n%2==0) return set1;
    if (n==1) return grid;
    if (n%4==3) return set2;
    return set3;
}

    • bng442 + 1 comment

      whats your logic behind this:

      
if (n%2==0) return set1;
    if (n==1) return grid;
    if (n%4==3) return set2;
    return set3;

      • sarathy_v_krish1 + 2 comments

        The idea to grasp is that after the first time instant (n=1), the grid keeps toggling between 3 possible configurations.

        So after generating the 3 patterns, we need to notice that the patterns keep repeating in the order(set no.) : . . . . . . . . 3 1 2 1 3 1 2 1 3 1 2 1 . . . . . . . . with a net frequency of 4

        So if n is even, print set 1.

        If n is odd and n%4 is 1, print set 3.

        If n is odd and n%4 is 3, print set 2.

        • niwanshu16 + 0 comments

          Great work man

        • nencybansal42 + 0 comments

          Awesome work!!

    • Abe10 + 0 comments

      Brilliant dude!!

    • balla_b170115cs + 0 comments

      hahah samelogic bro . to get the grid at 5 ,9,13 we apply the same method used to caluclate the grid for 3,7,11 but using set 2 as grid for calculating set 3 .But iam getting wrong answer for some cases.But glad to see someone came up with same idea

    • manojkumar_g2 + 0 comments

      You are amazing!! I was building the exact same code(return conditions, chunks ), but lost the courage to proceed when i needed to check for the upper and lower bounds.. I felt that im making the code too heavy(time complexity wise)..