Find the String | Editorial by abhinavkeshav

We have been given a matrix of size **N** rows and **M** columns and a string **S**.
We need to find if we can form the string **S** using the characters in the matrix using these two rules :
1. Since there are **N** rows we need to choose **1st** character of S from **1st** row ,
** 2nd** from **2nd** row and so on until we reach **(N+1)th** character of S then we will again start from the **1st** row in cyclic order.
 
2.  We cannot pick the same occurrence twice from a row.

We will create a integer matrix which will store the occurred character of **S** in the character matrix.
Now the problem can be solved by looping through the string **S** and finding the occurrence of character at index i of **S** in row** i%N **of character matrix.
If we get the value as **-1** which means that the character is not present in the row we will print **"No"** , else we will keep checking in each row and updating the integer matrix.

private static String print(Scanner s) {
		String mat[] = s.nextLine().split(" ");
		int m=Integer.parseInt(mat[0]);
		int n=Integer.parseInt(mat[1]);
		String z[] = new String[m];
		int chk[][]=new int[m][n];
		for(int i=0;i<m;i++) {
				z[i]=s.nextLine();
		}
		String str = s.nextLine();
		int b;
		for(int k=0;k<str.length();k++) {
			b=z[k%m].indexOf(str.charAt(k));
			while(b>-1 && chk[k%m][b]==1) {
				b=z[k%m].indexOf(str.charAt(k), b+1);
			}
			if(b<0) {
				return "No";
			} else {
                chk[k%m][b]=1;
			}
		}
		return "Yes";
	}

Find the String | Editorial by abhinavkeshav

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