I read about this problem in TC forums.
How to efficiently search for a word in a 2D matrix of letters , given the conditions that , we can start at any position and for every position we can go to only the adjacent positions (vertically,horizontally and diagonally ,i.e in all 8 directions).
Example- searching topcoder in this matrix :
t o p t
g f c q
t y o d
d f r e
How to efficiently search for a word in a 2D matrix of letters , given the conditions that , we can start at any position and for every position we can go to only the adjacent positions (vertically,horizontally and diagonally ,i.e in all 8 directions).
Example- searching topcoder in this matrix :
t o p t
g f c q
t y o d
d f r e
My solution to this problem is a simple recursive DFS. Along with yes or no it highlights the word itself in the matrix by displaying the word and '-' at all other locations. My implementation is as shown below.
You can also find it here
import java.util.*; import java.io.*; public class Main{ public static int flag=0,m,n,dx[]={0,0,1,-1,1,-1,-1,1}; public static int dy[]={1,-1,0,0,1,-1,1,-1}; public static char wrd[]; static class S{char a;int p,x,y; S(){} S(char b,int pos,int c,int d){a=b;p=pos;x=c;y=d;} } public static LinkedList<S>vis=new LinkedList<S>(); public static LinkedList<S>init=new LinkedList<S>(); public static Stack<S>fc=new Stack<S>(); public static char mat[][]; public static void main(String args[]) throws Exception{ BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); m=Integer.parseInt(br.readLine()); n=Integer.parseInt(br.readLine()); int i,j; long a=System.currentTimeMillis(); mat=new char[m][n]; wrd=br.readLine().toCharArray(); for(i=0;i<m;i++) mat[i]=br.readLine().toCharArray(); for(i=0;i<m;i++) for(j=0;j<n;j++) if(mat[i][j]==wrd[0]){init.add(new S(wrd[0],0,i,j));} for(i=0;i<init.size();i++) {solve(init.get(i));if(flag==1) break;} if(flag==0) {System.out.println("NO");return;} for(i=0;i<n;i++) for(j=0;j<n;j++) mat[i][j]='-'; S tp=null; for(;!fc.isEmpty();) {tp=fc.pop();mat[tp.x][tp.y]=tp.a;} for(i=0;i<m;i++) { for(j=0;j<n;j++) System.out.print(mat[i][j]); System.out.println(); } System.out.println((System.currentTimeMillis()-a)+"ms"); } public static void solve(S s){ int i,j; if(s.p==wrd.length-1){System.out.println("YES");fc.push(s);flag=1;return;} for(i=0;i<8;i++) if((s.x+dx[i])>0&&(s.y+dy[i])>0&&(s.x+dx[i])<m&&(s.y+dy[i])<n&&mat[s.x+dx[i]][s.y+dy[i]]==wrd[s.p+1]) solve(new S(wrd[s.p+1],(s.p+1),(s.x+dx[i]),(s.y+dy[i]))); if(s.p>=0&&s.p<wrd.length&&flag==1){fc.push(s);} } }
21 21 topcoder ataabmtdvegybznbzqwtb odiurtrdutatureiortzn godytqoqdxjwvxbnxwaxm jhhxicpsqzfaxcctcefck lcjsoideqlcwevkovrfyj padervwrtjjthnjpbtuth qacjgkaektdewmhsnyyfg zbxbigjdgrythhgdmudef crufjakjerwwsdffliywd vqzbnkmfjtgrwkdgkodas naohuyhfgofgfjuhjrvaa mjgfdskuifyiumyjgleaw lrepwoiepodsobtbdjdcf hfueydfughyuicavsfosh iovhjuhodjitgzzcaxctg hjgsiugfyiusfgioqzpeb jbhsfgudfghdhfghgsoqv fgyhdutsfyirtyiytrtac gjihysoduftfuihfytadx gyrwydfiuzadytiudfufs jgiosjtisodfstotisryt YES --------------------- --------------------- --------------------- --------------------- --------------------- --------------------- --------------------- --------------------- --------------------- --------------------- -----------------r--- ------------------e-- ------------------d-- ------------------o-- ------------------c-- ------------------p-- ------------------o-- ------------------t-- --------------------- --------------------- --------------------- 125ms
0 comments:
Post a Comment