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.HackerRank The Bomberman Game SolutionOne 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 SolutionTaskBomberman 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 i, j, any valid cells (i + 1, j) and (i, j + 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:Initially, Bomberman arbitrarily plants bombs in some of the cells, the initial state.After one second, Bomberman does nothing.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.After one more second, any bombs planted exactly three seconds ago will detonate. Here, Bomberman stands back and observes.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 DescriptionComplete the absolutePermutation function in the editor below.absolutePermutation has the following parameter(s):int n: the upper bound of natural numbers to consider, inclusiveint k: the absolute difference between each element’s value and its indexReturnsint[n]: the lexicographically smallest permutation, or [-1] if there is noneInput FormatThe first line contains an integer t, the number of queries.Each of the next t lines contains 2 space–separated integers, n and k.Constraints1 <= t <= 101 <= n <= 1050 <= k < nSample InputSTDIN 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 Output2 11 2 3-1ExplanationTest Case 0:Test Case 1:Test Case 2:No absolute permutation exists, so we print -1 on a new line.HackerRank The Bomberman Game SolutionThe 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 Javapublic 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 JavaScriptfunction 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(""); } answer(grid, N); } var states = {}, isRepeating = false; function answer(grid, N) { 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 Scalaimport 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 Pascaltype 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 assign(f,''); reset(f); readln(f,r,c,n); 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.Next: HackerRank Emas Supercomputer Solution Post navigationHackerRank Flatland Space Stations Solution HackerRank Emas Supercomputer Solution