# HackerRank The Bomberman Game Solution

#### ByBrokenprogrammers

Dec 12, 2022

Hello Programmers, In this post, you will know how to solve the HackerRank The Bomberman Game Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## HackerRank The Bomberman Game 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 ij, any valid cells (i + 1, j) and (ij + 1) 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 N seconds.

Function Description

Complete the absolutePermutation function in the editor below.

absolutePermutation has the following parameter(s):

• int n: the upper bound of natural numbers to consider, inclusive
• int k: the absolute difference between each element’s value and its index

Returns

• int[n]: the lexicographically smallest permutation, or [-1] if there is none

Input Format

The first line contains an integer t, the number of queries.
Each of the next t lines contains 2 spaceseparated integers, n and k.

Constraints

• 1 <= t <= 10
• 1 <= n <= 105
• 0 <= k < n

Sample Input

```STDIN   Function
-----   --------
3       t = 3 (number of queries)
2 1     n = 2, k = 1
3 0     n = 3, k = 0
3 2     n = 3, k = 2
```

Sample Output

2 1
1 2 3
-1

Explanation

Test Case 0:

Test Case 1:

Test Case 2:
No absolute permutation exists, so we print `-1` on a new line.

## HackerRank The Bomberman Game Solution

### The Bomberman Game Solution in C

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int r,c,n;
char set1[201][201];
char set2[201][201];
char set3[201][201];
char set4[201][201];
scanf("%d %d %d",&r,&c,&n);
for(int i=0;i<r;i++)
{
scanf("%s",set1[i]);
for(int j=0;j<c;j++)
{
set2[i][j]=79;
set3[i][j]=79;
set4[i][j]=79;
}
}

for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
if(set1[i][j]==79)
{
set3[i][j]='.';
set3[i][j-1]='.';
set3[i][j+1]='.';
set3[i-1][j]='.';
set3[i+1][j]='.';
}
}
}
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
if(set3[i][j]==79)
{
set4[i][j]='.';
set4[i][j-1]='.';
set4[i][j+1]='.';
set4[i-1][j]='.';
set4[i+1][j]='.';
}
}
}
if(n%2==0)
{
//print set 2
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
printf("%c",set2[i][j]);
}
printf("\n");
}
}
else if(n==1)
{
//print set 1
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
printf("%c",set1[i][j]);
}
printf("\n");
}
}
else if(n%4==1)
{
//print set 1
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
printf("%c",set4[i][j]);
}
printf("\n");
}
}
else
{
//print set 3
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
printf("%c",set3[i][j]);
}
printf("\n");
}
}
return 0;
}```

### The Bomberman Game Solution in Cpp

```#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
char ch[300];
int n,m,T,pd[300][300],bo[300][300];
const int gox[4]={1,-1,0,0},goy[4]={0,0,1,-1};
void print(){
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++) if (pd[i][j]==0) putchar('.'); else putchar('O');
cout<<endl;
}
}
int main(){
scanf("%d%d%d",&n,&m,&T);
for (int i=1;i<=n;i++){
scanf("%s",ch+1);
for (int j=1;j<=m;j++) pd[i][j]=(ch[j]=='O');
}
if (T==1){
print(); return 0;
}
if (T%2==0){
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++) pd[i][j]=1;
print(); return 0;
}
memcpy(bo,pd,sizeof bo);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
for (int k=0;k<4;k++)
pd[i][j]|=bo[i+gox[k]][j+goy[k]];
if ((T/2)%2==0){
memcpy(bo,pd,sizeof bo);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++){
pd[i][j]=bo[i][j];
for (int k=0;k<4;k++)
if (i+gox[k]>0&&i+gox[k]<=n&&j+goy[k]>0&&j+goy[k]<=m&&bo[i+gox[k]][j+goy[k]]==0) pd[i][j]=0;
}
print();
}else {
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++) pd[i][j]^=1;
print();
}
}```

### The Bomberman Game Solution in Java

```public class Solution {
public static void toComplement(char[][] grid, int r, int c){
for(int i=0; i<r; i++){
for(int j=0; j<c; j++){
if(grid[i][j]=='O'){
if(i-1>=0 && grid[i-1][j]!='O') grid[i-1][j]='*';
if(i+1<r && grid[i+1][j]!='O') grid[i+1][j]='*';
if(j-1>=0 && grid[i][j-1]!='O') grid[i][j-1]='*';
if(j+1<c && grid[i][j+1]!='O') grid[i][j+1]='*';
}
}
}
for(int i=0; i<r; i++){
for(int j=0; j<c; j++){
if(grid[i][j]=='.') grid[i][j]='O';
else if(grid[i][j]=='O' || grid[i][j]=='*') grid[i][j]='.';
}
}
}

public static void toFull(char[][] grid, int r, int c){
for(int i=0; i<r; i++){
for(int j=0; j<c; j++){
grid[i][j]='O';
}
}
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int r = in.nextInt();
int c = in.nextInt();
int n = in.nextInt();
in.nextLine();
char[][] grid = new char[r][c];
for(int i=0; i<r; i++){
grid[i] = in.nextLine().toCharArray();
}
if(n%2 == 0) toFull(grid, r, c);
if(n%4 == 3) toComplement(grid, r, c);
if(n%4 == 1 && n!=1){
toComplement(grid, r, c);
toComplement(grid, r, c);
}
for(int i=0; i<r; i++){
for(int j=0; j<c; j++){
System.out.print(grid[i][j]);
}
System.out.println();
}
}
}```

### The Bomberman Game Solution in Python

```# Enter your code here. Read input from STDIN. Print output to STDOUT
def bomb(r, c, grid, t):
complete_grid = [ ["O"]*c for i in range(r) ]
stable_grid = [ ["O"]*c for i in range(r) ]
other_grid = [ ["O"]*c for i in range(r) ]
for i in range(r):
for j in range(c):
if grid[i][j] == "O":
other_grid[i][j] = "."
if i>0:
other_grid[i-1][j] = "."
if j>0:
other_grid[i][j-1] = "."
if i<r-1:
other_grid[i+1][j] = "."
if j<c-1:
other_grid[i][j+1] = "."
for i in range(r):
for j in range(c):
if other_grid[i][j] == "O":
stable_grid[i][j] = "."
if i>0:
stable_grid[i-1][j] = "."
if j>0:
stable_grid[i][j-1] = "."
if i<r-1:
stable_grid[i+1][j] = "."
if j<c-1:
stable_grid[i][j+1] = "."
#print grid
#print complete_grid
#print other_grid
if t==0 or t==1:
return grid
if (t-1)%2==0 and not (t-1)%4==0:
return other_grid
if (t-1)%4==0:
return stable_grid
if t%2==0:
return complete_grid
grid = []
r, c, t = map(int, raw_input().split())
for _ in range(r):
row = [ l for l in raw_input() ]
grid.append(row)
res = "\n".join([ "".join(line) for line in bomb(r, c, grid, t) ])
print res
```

### The Bomberman Game Solution using JavaScript

```function processData(input) {
var split = input.split("\n");
var [R,C,N] = split[0].split(" ");
var grid = split.slice(1);
for(var i=0; i<grid.length; i++) {
grid[i] = grid[i].split("");
}
}
var states = {}, isRepeating = false;
var rowCount = grid.length;
if(N==1) {
return print(grid);
}
else if(N % 2 == 0) {
return printFalseGrid(grid);
}
else {
let loopFor= (N-1)/2 ; //times
for(var i=0; i<loopFor && !isRepeating; i++) {
grid = transformGrid(grid, (2*i)+3);
}
if(isRepeating) {
for(let prop in states) {
let val = states[prop];
if(val.counter >= 2 && (val.N % 4 == N % 4)) {
return print(val.grid);
}
}
}
else {
print(grid);
}
}
}
function transformGrid(initialState, N) {
var grid = initialState;
var rowCount = grid.length;
var colCount = grid[0].length;
var modifiedGrid = [];
var stateObj = {
false: 0,
true: 0
}
for(var i=0; i<rowCount; i++) {
modifiedGrid.push([]);
for(var j=0; j<colCount; j++) {
let cell = grid[i][j];
modifiedGrid[i].push(cell);
if(!isCellFalse(grid, i, j)) {
stateObj.true++;
}
if(isCellFalse(grid, i, j)) {
modifiedGrid[i][j] = ".";
stateObj.false++;
}
else if(!isCellInDangerZone(grid, i, j)) {
//console.log("i:" + i);
//console.log("j:" + j);
modifiedGrid[i][j] = "O";
}
}
}
var stateString = uniqueStateString(stateObj);
if(typeof states[stateString] === 'undefined') {
states[stateString] = {
counter: 0,
grid: modifiedGrid,
N: N
};
}
else {
states[stateString].counter++;
}
if(states[stateString].counter > 2) {
isRepeating = true;
}
return modifiedGrid;

}
function uniqueStateString(stateObj) {
return stateObj.false+","+stateObj.true
}
function isCellFalse(grid, i, j) {
return grid[i][j] == "O";
}
function isCellInDangerZone(grid,i,j) {
if([grid[i][j-1], grid[i][j+1]].indexOf("O") !== -1) return true;
if(grid[i-1] && grid[i-1][j] == "O") return true;
if(grid[i+1] && grid[i+1][j] == "O") return true;
}
function printFalseGrid(grid) {
for(var k=0; k<grid.length; k++) {
console.log("O".repeat(grid[0].length));
}
}
function print(grid) {
for(var i=0; i<grid.length; i++) {
console.log(grid[i].join(""));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});```

### The Bomberman Game Solution in Scala

```import scala.io.StdIn
object Solution {
def main(args: Array[String]) {
val Array(r, c, n) = StdIn.readLine().split("\\s+").map(_.toInt)
var field = (1 to r).map(_ => StdIn.readLine().toArray.map(c => if (c == 'O') '0' else c)).toArray
for (i <- 1 to math.min(n, 200 + n % 4)) {
field = field.map(_.map(c => if (c != '.') {
(c + 1).toChar
} else if (c == '.' && i % 2 == 0) {
'0'
} else {
c
}))
def bomb(x: Int, y: Int) = x >= 0 && x < c && y >= 0 && y < r && field(y)(x) == '3'
if (i % 2 == 1 && i != 1) {
for (y <- 0 until r; x <- 0 until c) {
if (!bomb(x, y) && (bomb(x, y + 1) || bomb(x, y - 1) || bomb(x - 1, y) || bomb(x + 1, y))) {
field(y)(x) = '.'
}
}
for (y <- 0 until r; x <- 0 until c) {
if (bomb(x, y)) {
field(y)(x) = '.'
}
}
}
}
println(field.map(_.map(c => if (c == '.') '.' else 'O').mkString("")).mkString("\n"))
}
}```

### The Bomberman Game Solution in Pascal

```type  grid=array[1..200]of string[200];
var f:text; r,c,i:byte; n:int64;
g:array[0..3]of grid;

function fill:grid;
var g:grid; s:string[200]; i:byte;
begin
s:='';
for i:=1 to c do s:=s+'O';
for i:=1 to r do g[i]:=s;
exit(g)
end;
function invert(g:grid):grid;
var f:grid; i,j:byte;
procedure explode(x,y:byte);
begin
f[x,y]:='.';
if x-1>0 then f[x-1,y]:='.';
if y-1>0 then f[x,y-1]:='.';
if x+1<=r then f[x+1,y]:='.';
if y+1<=c then f[x,y+1]:='.'
end;
begin
f:=fill;
for i:=1 to r do
for j:=1 to c do
if g[i,j]='O' then
explode(i,j);
exit(f)
end;
procedure generate;
begin
g[1]:=g[0]; g[2]:=fill;
g[3]:=invert(g[1]); g[0]:=fill;
g[1]:=invert(g[3])
end;

procedure print(g:grid);
var i:byte;
begin
for i:=1 to r do
writeln(f,g[i])
end;

begin
for i:=1 to r do readln(f,g[0,i]);
close(f);
assign(f,''); rewrite(f);
if n=1 then print(g[0])
else
begin
generate; print(g[n mod 4])
end;
close(f)
end.```

Disclaimer: This problem (The Bomberman Game) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.