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 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.
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:
... ... ...
Function Description
Complete the bomberMan function in the editory below. It should return an array of strings that represent the grid in its final state.
bomberMan has the following parameter(s):
n: an integer, the number of seconds to simulate
grid: an array of strings that represents the grid
Input Format
The first line contains three space-separated integers r, c, and n, The number of rows, columns and seconds to simulate.
Each of the next r lines contains a row of the matrix as a single string of c characters. The . character denotes an empty cell, and the O character (ascii 79) denotes a bomb.
Output Format
Print the grid’s final state. This means R lines where each line contains C characters, and each character is either a . or an O (ascii 79). This grid must represent the state of the grid after n seconds.
Sample Input
6 7 3 ....... ...O... ....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 in C
#include <stdio.h> #define MAX_ROW_COL (201) #define BOMB (79) #define EMPTY ('.') #define TEMP ('A') typedef int INT32; char gGrid[MAX_ROW_COL][MAX_ROW_COL]; int main(void) { INT32 nLpCnt1, nR, nC, nS, nLpCnt2, ndCnt; //read nN, nK scanf("%d%d%d", &nR, &nC, &nS); //read the matrix for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) scanf("%s", gGrid[nLpCnt1]); if(nS < 2) { for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) printf("%s\n", gGrid[nLpCnt1]); return 0; } nS %= 4; if((nS == 0) || (nS == 2)) //all bombs { for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) for(nLpCnt2 = 0; nLpCnt2 < nC; nLpCnt2++) gGrid[nLpCnt1][nLpCnt2] = BOMB; } else //bombs detonated once or twice { if(nS == 3) nS = 1; else nS = 2; for(ndCnt = 0; ndCnt < nS; ndCnt++) { for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) { //form the array with temp for(nLpCnt2 = 0; nLpCnt2 < nC; nLpCnt2++) { if(gGrid[nLpCnt1][nLpCnt2] == BOMB) { gGrid[nLpCnt1][nLpCnt2] = TEMP; if(nLpCnt1 > 0) { gGrid[nLpCnt1 - 1][nLpCnt2] = TEMP; } if(nLpCnt2 > 0) { gGrid[nLpCnt1][nLpCnt2 - 1] = TEMP; } if(((nLpCnt1 + 1) < nR) && (gGrid[nLpCnt1 + 1][nLpCnt2] != BOMB)) { gGrid[nLpCnt1 + 1][nLpCnt2] = TEMP; } if(((nLpCnt2 + 1) < nC) && (gGrid[nLpCnt1][nLpCnt2 + 1] != BOMB)) { gGrid[nLpCnt1][nLpCnt2 + 1] = TEMP; } } } } //convert empty to bombs and temp to empty for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) { //form the array with temp for(nLpCnt2 = 0; nLpCnt2 < nC; nLpCnt2++) { if(gGrid[nLpCnt1][nLpCnt2] == EMPTY) gGrid[nLpCnt1][nLpCnt2] = BOMB; else gGrid[nLpCnt1][nLpCnt2] = EMPTY; } } } } for(nLpCnt1 = 0; nLpCnt1 < nR; nLpCnt1++) printf("%s\n", gGrid[nLpCnt1]); return 0; }
The Bomberman Game HackerRank Solution in C++
#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 HackerRank Solution in Java
import java.awt.*; import java.awt.event.*; import java.awt.geom.*; import java.io.*; import java.math.*; import java.text.*; import java.util.*; import java.util.regex.*; public class Solution { private static BufferedReader br; private static StringTokenizer st; private static PrintWriter pw; public static void main(String[] args) throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out))); int qq = 1; //int qq = Integer.MAX_VALUE; //int qq = readInt(); for(int casenum = 1; casenum <= qq; casenum++) { int r = readInt(); int c = readInt(); int n = readInt(); n--; int[][] bomb = new int[r][c]; for(int i = 0; i < r; i++) { String s = nextToken(); for(int j = 0; j < c; j++) { if(s.charAt(j) == 'O') { bomb[i][j] = 2; } } } n %= 100; int[] dx = new int[]{-1,1,0,0}; int[] dy = new int[]{0,0,-1,1}; for(int i = 1; i <= n; i++) { if(i%2 == 1) { for(int a = 0; a < r; a++) { for(int b = 0; b < c; b++) { if(bomb[a][b] == 0) { bomb[a][b] = 3; } else if(bomb[a][b] > 0) { bomb[a][b]--; } } } } else { boolean[][] dead = new boolean[r][c]; for(int a = 0; a < r; a++) { for(int b = 0; b < c; b++) { if(bomb[a][b] == 1) { dead[a][b] = true; for(int k = 0; k < dx.length; k++) { int nx = a + dx[k]; int ny = b + dy[k]; if(nx >= 0 && nx < r && ny >= 0 && ny < c) { dead[nx][ny] = true; } } } } } for(int a = 0; a < r; a++) { for(int b = 0; b < c; b++) { if(dead[a][b]) bomb[a][b] = 0; else if(bomb[a][b] > 0) { bomb[a][b]--; } } } } } for(int[] out: bomb) { for(int out2: out) { if(out2 > 0) { pw.print('O'); } else { pw.print('.'); } } pw.println(); } } exitImmediately(); } private static void exitImmediately() { pw.close(); System.exit(0); } private static long readLong() throws IOException { return Long.parseLong(nextToken()); } private static double readDouble() throws IOException { return Double.parseDouble(nextToken()); } private static int readInt() throws IOException { return Integer.parseInt(nextToken()); } private static String nextLine() throws IOException { if(!br.ready()) { exitImmediately(); } st = null; return br.readLine(); } private static String nextToken() throws IOException { while(st == null || !st.hasMoreTokens()) { if(!br.ready()) { exitImmediately(); } st = new StringTokenizer(br.readLine().trim()); } return st.nextToken(); } }
The Bomberman Game HackerRank Solution in Python
#!/usr/bin/python import os import sys import itertools import copy def solve(f): r, c, n = f.read_int_list() m = [ list(f.read_str()) for _ in xrange(r) ] s = [[0]*c for _ in xrange(r)] t = 1 for i in xrange(r): for j in xrange(c): if m[i][j] == 'O': s[i][j] = 2 if n > 5: n -= (n-8)/4 * 4 while t < n: t += 1 if t%2 == 0: for i in xrange(r): for j in xrange(c): if s[i][j] > 0: s[i][j] -= 1 else: s[i][j] = 3 else: b = [[False]*c for _ in xrange(r)] for i in xrange(r): for j in xrange(c): s[i][j] -= 1 if s[i][j] == 0: b[i][j] = True if i > 0: b[i-1][j] = True if j > 0: b[i][j-1] = True if i < r-1: b[i+1][j] = True if j < c-1: b[i][j+1] = True for i in xrange(r): for j in xrange(c): if b[i][j]: s[i][j] = 0 if all(map(lambda x: all([i==0 for i in x]), s)): break for i in xrange(r): for j in xrange(c): m[i][j] = '.' if s[i][j] == 0 else 'O' return '\n'.join(map(lambda x: ''.join(x),m)) class Reader(object): def __init__(self, filename=None): self.test_mode = filename is not None self.cases = 1 self.buffer = [] if self.test_mode: with open(filename) as f: blank_flg = False for line in f: line = line.strip() if line: self.buffer.append(line) blank_flg = False else: if not blank_flg: self.cases += 1 blank_flg = True def __readline(self): return self.buffer.pop(0) if self.test_mode else raw_input() def read_int(self): return int(self.__readline()) def read_float(self): return float(self.__readline()) def read_long(self): return long(self.__readline()) def read_str(self): return self.__readline() def read_int_list(self): return [int(item) for item in self.__readline().split()] def read_float_list(self): return [float(item) for item in self.__readline().split()] def read_long_list(self): return [long(item) for item in self.__readline().split()] def read_str_list(self): return self.__readline().split() if __name__ == '__main__': filename = sys.argv[1] if len(sys.argv)>1 else None f = Reader(filename) if f.test_mode: for c in xrange(f.cases): print "Case #%d"%(c+1) print solve(f) else: print solve(f)
The Bomberman Game HackerRank Solution in C#
using System; using System.Collections.Generic; using System.IO; using System.Text; class Solution { static void Main(String[] args) { string[] tokens = Console.ReadLine().Split(' '); int rows = Convert.ToInt32(tokens[0]); int columns = Convert.ToInt32(tokens[1]); int seconds = Convert.ToInt32(tokens[2]); List<string> steps = new List<string>(); char[][] map = new char[rows][]; for (int i = 0; i < rows; ++i) { map[i] = Console.ReadLine().ToCharArray(); } if (seconds == 1) { DumpMap(map); return; } if (seconds % 2 == 0) { for (int i = 0; i < rows; ++i) for (int j = 0; j < columns; ++j) map[i][j] = 'O'; DumpMap(map); return; } IterateMap(map); if (seconds % 4 == 1) IterateMap(map); DumpMap(map); } private static void IterateMap(char[][] map) { // Blow up the bombs for (int i = 0; i < map.Length; ++i) { for (int j = 0; j < map[i].Length; ++j) { if (map[i][j] == 'O') { map[i][j] = ' '; ClearCell(map, i - 1, j); ClearCell(map, i + 1, j); ClearCell(map, i, j - 1); ClearCell(map, i, j + 1); } } } for (int i = 0; i < map.Length; ++i) { for (int j = 0; j < map[i].Length; ++j) { if (map[i][j] == '.') map[i][j] = 'O'; else map[i][j] = '.'; } } } private static void DumpMap(char[][] map) { for (int i = 0; i < map.Length; ++i) { Console.Out.WriteLine(String.Join("", map[i])); } } private static void ClearCell(char[][] map,int r,int c) { if (r < 0) return; if (r >= map.Length) return; if (c < 0) return; if (c >= map[r].Length) return; if (map[r][c] == 'O') return; map[r][c] = ' '; } }
Attempt The Bomberman Game HackerRank Challenge
Link – https://www.hackerrank.com/challenges/bomber-man/
Next HackerRank Challenge Solution
Link – https://exploringbits.com/emas-supercomputer-hackerrank-solution/
Aayush Kumar Gupta is the founder and creator of ExploringBits, a website dedicated to providing useful content for people passionate about Engineering and Technology. Aayush has completed his Bachelor of Technology (Computer Science & Engineering) from 2018-2022. From July 2022, Aayush has been working as a full-time Devops Engineer.