In this blog post we will Provide Coding Ninjas Basics of Java with Data Structures and Algorithms Solution. This will have solutions to all the problems that are included in Coding Ninjas Java Course. Coding Ninjas question – Given a NxM matrix containing Uppercase question with solution..
NxM matrix containing Uppercase – Coding Ninjas question
Input Format :
Given a NxM matrix containing Uppercase English Alphabets only. Your task is to tell if there is a path in the given matrix which makes the sentence “CODINGNINJA” .
There is a path from any cell to all its neighbouring cells. For a particular cell, neighbouring cells are those cells that share an edge or a corner with the cell.
Input Format :
The first line of input contains two space separated integers N and M, where N is number of rows and M is the number of columns in the matrix.
Each of the following N lines contain M characters. Please note that characters are not space separated.
Output Format :
Print 1 if there is a path which makes the sentence “CODINGNINJA” else print 0.
Constraints :
1 <= N <= 1000
1 <= M <= 1000
Time Limit: 1 second
Sample Input 1:
2 11
CXDXNXNXNXA
XOXIXGXIXJX
Sample Output 1:
1
Coding ninjas solutions
Java Language [Solutions ]
public class Solution {
int solve(String[] Graph , int N, int M)
{
String searchingString = "CODINGNINJA";
boolean[][] visited = new boolean[Graph.length][];
for(int i=0;i<Graph.length;i++) { //Graph.length gives the number of rows, Graph[i].length give sthe number of columns per row
visited[i] = new boolean[Graph[i].length()];
}
for(int i=0;i<Graph.length;i++) {
for(int j=0;j<Graph[i].length();j++) {
if(Graph[i].charAt(j) == 'C' ) {
boolean ans = dfs( Graph,visited,searchingString.substring(1),i,j);
if(ans) {
return 1;
}
}
}
}
return 0;
}
public static boolean dfs(String[] graph, boolean[][] visited, String searchString,int i,int j) {
if(searchString.length() == 0) {
visited[i][j] = true;
return true;
}
visited[i][j] = true;
// To traverse in 8 different directions
int[] X_Dir = {-1,1,0,0,1,-1,1,-1 };
int[] Y_Dir = {0,0,-1,1,1,-1,-1,1 };
for(int k=0;k<X_Dir.length;k++) {
int x = i+ X_Dir[k];
int y = j+ Y_Dir[k];
if( x>=0 && y>=0 && x< graph.length &&
y < graph[x].length() &&
graph[x].charAt(y) ==searchString.charAt(0) && !visited[x][y] ) {
boolean smallAns = dfs(graph, visited, searchString.substring(1), x, y);
if(smallAns) {
return smallAns;
}
}
}
visited[i][j] = false;
return false;
}
}Conclusion
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