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

The Bomberman Game

  • kris_tobiasson
     + 0 comments

    Great problem. I did the following:

    1. if n%2 == 0 then we return all "O"s.
    2. And if n == 1 of course we return the original grid.
    3. if (n-1)%4 == 0 we do 2 bombings of the grid.
    4. otherwise (if (n-3)%4 == 0) we just do one bombing of the grid.

    My JS solution is below:

        if(n === 1){ return grid; }

    if((n-1)%4 === 0){
        let grid2 = bombGrid(grid);
        return bombGrid(grid2);
    }

    if(n%2 === 0){
        let str = ``;

        for(let i = 0; i < grid[0].length; i++){
            str += `O`;
        }
        for(let i = 0; i < grid.length; i++){
            grid[i] = str;
        }
        return grid;
    }

    return bombGrid(grid);

    function bombGrid(g){
        
        let g2 = new Array (g.length);
        g2[0] = new Array (g[0].length);

        for(let i = 0; i < g.length; i++){
            if(i !== g.length-1){ g2[i+1] = new Array(g[i].length); }

            for(let j = 0; j < g[i].length; j++){

                if(g[i].charAt(j) === `O`){
                    g2[i][j] = `.`;
                    if(j !== g[i].length-1){ g2[i][j+1] = `.`; }
                    if(j !== 0){ g2[i][j-1] = `.`; }
                    if(i !== 0){ g2[i-1][j] = `.`; }
                    if(i !== g.length-1){ g2[i+1][j] = `.`; }
                }
                else if(g2[i][j] === undefined){
                    g2[i][j] = `O`;
                }
            }
            if(i !== 0){ g2[i-1] = g2[i-1].join(``); }
            if(i === g.length-1){ g2[i] = g2[i].join(``); }
        }
        return g2;
    }