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// Complete the bomberMan function below.
static String[] bomberMan(int n, String[] grid) {
String[] ans = new String[grid.length];
int aa = grid[0].length();
String ab = "";
//String[] grd = new String(grid.length);
for(int i=0;i<aa;i++){
ab = ab.concat("0");
}
if(n%2 == 0){
for(int i=0;i<grid.length;i++){
ans[i] = ab;
}
}
else{
if(n==1){
ans = grid;
}
else{
int count = 0;
while(count<2){
for(int i=0;i<grid.length;i++){
char[] tmp1 = grid[i].toCharArray();
for(int j=0;j<aa;j++){
if(tmp1[j] == 'O'){
tmp1[j] = '1';
}
else if(tmp1[j] == '.'){
tmp1[j] = 'O';
}
}
String ala = new String(tmp1);
grid[i]=ala;
}
int flag = 0;
int flag1 = 0;
for(int i=0;i<grid.length;i++){
char[] tmp = grid[i].toCharArray();
char[] tm2 = grid[i].toCharArray();
char[] tm3 = grid[i].toCharArray();
if(i-1 != -1){
tm2 = grid[i-1].toCharArray();
}
else{
flag = 1;
}
if(i+1 != grid.length){
tm3 = grid[i+1].toCharArray();
}
else{
flag1 = 1;
}
for(int j=0;j<aa;j++){
if(flag == 0 && flag1 == 0){
if(tmp[j] == '1'){
tmp[j] = '.';
if(j!=0){
tmp[j-1] = '.';
}
if(j!=tmp.length-1){
if(tmp[j+1] == 'O'){
tmp[j+1] = '.';
}
}
if(tm3[j] == 'O'){
tm3[j] = '.';
}
if(tm2[j] == 'O'){
tm2[j] = '.';
}
}
}
else if(flag == 1 && flag1 == 0){
//flag = 0;
if(tmp[j] == '1'){
tmp[j]='.';
if(j!=0){
tmp[j-1] = '.';
}
if(j!=tmp.length-1){
if(tmp[j+1] == 'O'){
tmp[j+1] = '.';
}
}
if(tm3[j] == 'O'){
tm3[j] = '.';
}
}
}
else if(flag1 == 1 && flag == 0){
//flag1 = 0;
if(tmp[j] == '1'){
tmp[j]='.';
if(j!=0){
tmp[j-1] = '.';
}
if(j!=tmp.length-1){
if(tmp[j+1] == 'O'){
tmp[j+1] = '.';
}
}
if(tm2[j] == 'O'){
tm2[j] = '.';
}
}
}
else{
if(tmp[j] == '1'){
tmp[j]='.';
if(j!=0){
tmp[j-1] = '.';
}
if(j!=tmp.length-1){
if(tmp[j+1] == 'O'){
tmp[j+1] = '.';
}
}
}
}
}
String a3 = new String(tmp);
grid[i] = a3;
if(flag == 0 && flag1 == 0){
String a1 = new String(tm2);
String a2 = new String(tm3);
grid[i-1] = a1;
grid[i+1] = a2;
}
else if(flag == 1 && flag1 == 0){
//String a1 = new String(tm2);
String a2 = new String(tm3);
grid[i+1] = a2;
flag = 0;
}
else if(flag == 0 && flag1 == 1){
String a1 = new String(tm2);
//String a2 = new String(tm3);
grid[i-1] = a1;
flag1 = 0;
}
}
count++;
ans = grid;
if((n+1)%4==0){
//grd = grid;
break;
}
}
}
}
return(ans);
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
String[] rcn = scanner.nextLine().split(" ");
int r = Integer.parseInt(rcn[0]);
int c = Integer.parseInt(rcn[1]);
int n = Integer.parseInt(rcn[2]);
String[] grid = new String[r];
for (int i = 0; i < r; i++) {
String gridItem = scanner.nextLine();
grid[i] = gridItem;
}
String[] result = bomberMan(n, grid);
for (int i = 0; i < result.length; i++) {
bufferedWriter.write(result[i]);
if (i != result.length - 1) {
bufferedWriter.write("\n");
}
}
bufferedWriter.newLine();
bufferedWriter.close();
scanner.close();
}
}
Abhiskek
The pattern is like if n=1 ans = grid
if n%2 == 0 than all are 0's
now the headache is when it is other than above cases
I. For all cases the solution for n = 3,7,11,15....(i.e. the configration of grid at n=3) is same while for n = 5,9,13.... it is same as that of initial i.e. at t=1.
II. While for the second case it is like, for n = 3,7,11,15.... the solution is still same as that of the n=3 while for n = 5,9,13.... it is different from that of t=1 or initital. But for all n E (5,9,13,17...) it is same. So for worst case you need to follow the pattern mentioned(i.e. {i,j} = 0 will affect {i+-1 and j+-1}) in question for 2 times i.e. first time to find the grid for n=3,7,11... and second time for n=5,9,13,17... !!
I am not able to figure out for what cases of n=5,9,13,17..... we have to use I and for what cases II.
The Bomberman Game
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import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;
public class Solution {
}
Abhiskek The pattern is like if n=1 ans = grid if n%2 == 0 than all are 0's now the headache is when it is other than above cases I. For all cases the solution for n = 3,7,11,15....(i.e. the configration of grid at n=3) is same while for n = 5,9,13.... it is same as that of initial i.e. at t=1.
II. While for the second case it is like, for n = 3,7,11,15.... the solution is still same as that of the n=3 while for n = 5,9,13.... it is different from that of t=1 or initital. But for all n E (5,9,13,17...) it is same. So for worst case you need to follow the pattern mentioned(i.e. {i,j} = 0 will affect {i+-1 and j+-1}) in question for 2 times i.e. first time to find the grid for n=3,7,11... and second time for n=5,9,13,17... !!
I am not able to figure out for what cases of n=5,9,13,17..... we have to use I and for what cases II.
Hope it will be helpful for u.
You should have use modularization. Create a function process and return, use it how many times you want.