# The Bomberman Game HackerRank Solution

In this The Bomberman Game HackerRank solution, Bomberman lives in a rectangular grid. Each cell in the grid either contains a bomb or nothing at all.

Each bomb can be planted in any cell of the grid but once planted, it will detonate after exactly 3 seconds. Once a bomb detonates, it’s destroyed — along with anything in its four neighboring cells. This means that if a bomb detonates in cell , any valid cells  and  are cleared. If there is a bomb in a neighboring cell, the neighboring bomb is destroyed without detonating, so there’s no chain reaction.

Bomberman is immune to bombs, so he can move freely throughout the grid. Here’s what he does:

1. Initially, Bomberman arbitrarily plants bombs in some of the cells, the initial state.
2. After one second, Bomberman does nothing.
3. After one more second, Bomberman plants bombs in all cells without bombs, thus filling the whole grid with bombs. No bombs detonate at this point.
4. After one more second, any bombs planted exactly three seconds ago will detonate. Here, Bomberman stands back and observes.
5. Bomberman then repeats steps 3 and 4 indefinitely.

Note that during every second Bomberman plants bombs, the bombs are planted simultaneously (i.e., at the exact same moment), and any bombs planted at the same time will detonate at the same time.

Given the initial configuration of the grid with the locations of Bomberman’s first batch of planted bombs, determine the state of the grid after  seconds.

For example, if the initial grid looks like:

...
.O.
...


it looks the same after the first second. After the second second, Bomberman has placed all his charges:

OOO
OOO
OOO


At the third second, the bomb in the middle blows up, emptying all surrounding cells:

O.O
...
O.O


Function Description

Complete the bomberMan function in the editory below.

bomberMan has the following parameter(s):

• int n: the number of seconds to simulate
• string grid[r]: an array of strings that represents the grid

Returns

• string[r]: n array of strings that represent the grid in its final state

Input Format

The first line contains three space-separated integers , , and , The number of rows, columns and seconds to simulate.
Each of the next  lines contains a row of the matrix as a single string of  characters. The . character denotes an empty cell, and the O character (ascii 79) denotes a bomb.

Constraints

•  for  of the maximum score.

Sample Input

STDIN           Function
-----           --------
6 7 3           r = 6, c = 7, n = 3
.......         grid =['.......', '...O...', '....O..',\
...O...                '.......', 'OO.....', 'OO.....']
....O..
.......
OO.....
OO.....


Sample Output

OOO.OOO
OO...OO
OOO...O
..OO.OO
...OOOO
...OOOO


Explanation

The initial state of the grid is:

.......
...O...
....O..
.......
OO.....
OO.....


Bomberman spends the first second doing nothing, so this is the state after 1 second:

.......
...O...
....O..
.......
OO.....
OO.....


Bomberman plants bombs in all the empty cells during his second second, so this is the state after 2 seconds:

OOOOOOO
OOOOOOO
OOOOOOO
OOOOOOO
OOOOOOO
OOOOOOO


In his third second, Bomberman sits back and watches all the bombs he planted 3 seconds ago detonate. This is the final state after  seconds:

OOO.OOO
OO...OO
OOO...O
..OO.OO
...OOOO
...OOOO

## The Bomberman Game HackerRank solution

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in JavaScript

function convertGridToArray(grid){
const arrayGrid = [];
grid.forEach(row=>{
arrayGrid.push(row.split(''))
});
return arrayGrid;
}

function cloneEmptyGrid(grid){
const temp = JSON.parse(JSON.stringify(grid));
temp.forEach((row, idxR)=>{
temp[idxR].fill('O');
});
return temp;
}

function bomberMan(n, grid) {
grid = convertGridToArray(grid);
const rows = grid.length;
const columns = grid[0] ? grid[0].length : 0;

if(n%2===0){
grid.forEach((row, idxR)=>{
row.forEach((column, idxC)=>{
grid[idxR][idxC] = 'O';
})
})
}else{
n /= 2;
const minSeconds = Math.min(n, (n+1)%2+1);
for(let i=1; i<=minSeconds; i++){
const tempGrid = cloneEmptyGrid(grid);
const setGrid = (i, j, ch) =>{
if(i>=0 && i<rows && j>=0 && j < columns){
tempGrid[i][j] = ch;
}
}
grid.forEach((row, idxR)=>{
row.forEach((column, idxC)=>{
if(grid[idxR][idxC] == 'O'){
setGrid(idxR, idxC, '.'); //current row
setGrid(idxR+1, idxC, '.') //down pos
setGrid(idxR-1, idxC, '.') //top pos
setGrid(idxR, idxC+1, '.') //right pos
setGrid(idxR, idxC-1, '.'); //left pos
}
})
});
grid = tempGrid;
}
}

let result = [];
grid.forEach((row, idxR)=>{
result.push(row.join(''));
})
return result;
}

### Problem Solution in C#

public static List<string> bomberMan(int n, List<string> grid)
{
if (n == 1) return grid;
if((n-1) % 4 == 0) return flipped(flipped(grid));
if (n % 2 ==0) return plant_all(grid);
return flipped(grid);
}

private static List<string> plant_all(List<string> grid)
{
var gridCopy = grid;
int r = grid.Count;
int c = grid[0].Length;
string temp = new string('O', c);
for (int i = 0; i < r; i++) gridCopy[i] = temp;
return gridCopy;
}

private static List<string> flipped(List<string> grid)
{
var gridCopy = grid;
int r = grid.Count;
int c = grid[0].Length;

List<string> fgrid = new List<string>();
for(int i=0;i<r;i++)
{
string raw = "";
for(int j=0;j<c;j++)
{
if(grid[i][j] == 'O' || grid[i][j] == '*' || (i-1 >=0 && grid[i-1][j] == 'O') || (i+1 < r && grid[i+1][j] == 'O') || (j-1 >=0 && grid[i][j-1] == 'O') || (j+1 < c && grid[i][j+1] == 'O')) raw += '*';
else raw += '.';
}
}
List<string> dgrid = new List<string>();
for(int i=0;i<r;i++)
{
string raw = "";
for(int j=0;j<c;j++)
{
raw += fgrid[i][j] == '*' ? '.' : 'O';
}
}
return dgrid;
}

### Problem Solution in C++

vector<string> bomberMan(int n, vector<string> grid) {
string full_row (grid[0].length(), 'O');
vector<string> full_map (grid.size(), full_row);
if (n == 1)
return grid;
else if (n % 2 == 0) // seconds is even, map is all bombs
return full_map;

int iters = n;
if (n >= 5) // There are only 2 distinct maps: 3 seconds and 5 seconds
iters = (n % 4 == 3 ? 1 : 2); // Only 1 or 2 iterations are needed

vector<string> map = full_map;
while (iters > 0) {
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[0].length(); j++) {
if (grid[i][j] == 'O') {
map[i][j] = '.'; // Center
map[max(i - 1, 0)][j] = '.'; // UP
map[min(i+1, (int)grid.size() - 1)][j] = '.'; // DOWN
map[i][max(j-1, 0)] = '.'; // LEFT
map[i][min(j+1, (int)grid[0].length() - 1)] = '.'; // RIGHT
}
}
}
grid = map;
map = full_map;
iters--;
}
return grid;
}
Solve original Problem on HackerRank here. Checkout more HackerRank Problems