String challenge ( piyush and magical park )

#1

sir my three of the test case is getting wrong ,plz clarify…

package chall_Strings;

import java.util.*;

public class piyushAndMagicalPark {
static Scanner s=new Scanner(System.in);
public static void main(String[] args) {
int row=s.nextInt();
int col=s.nextInt();
int min=s.nextInt();
int str=s.nextInt();
char[][] arr=new char[row][col];
for(int i=0 ;i<row ;i++) {
for(int j=0 ;j<col ;j++) {
arr[i][j] =s.next().charAt(0);
}
}
for(int i=0 ;i <row ;i++) {
for(int j=0 ;j <col ;j++) {
char ch=arr[i][j];
if( j == (col -1)) {
if(ch == ‘*’) {
str =str+5;
}else if(ch == ‘.’) {
str =str-2;
}else {
break;
}
}else {

	if(ch == '*'){
		str = str-1;
		str=str + 5;
	}else if( ch == '.') {
str=str-1;
		str = str-2;
	}else {
		
		break;
	}
	}
}

}
if(str > min) {
System.out.print(“Yes”+"\n" +str);
}else {
System.out.print(“No”);
}
}
}

this is my code

#2

@sameeksha,
You will not print “No” in any case.

  • Start from the 0-0 index of the 2 - D matrix( No step would be count hence, strength remain intact).
  • Changing row won’t take any strength.
  • If at any time, the Strength become smaller than K ( threshold for piyush), then no need to traverse rest of the array, you can say no nothing else.

Algorithm

  • Take input N, M, K, S.
  • Take input 2 - D matrix of size N x M.
  • Put a loop on the array starting from 0 - 0 index to (N - 1) - (M - 1).
  • check if the Strength is lower than the threshold viz, K, print “No” and return.
  • otherwise,

    1. if character is ‘*’, add 5 to the strength.

    2. else if, character is ‘.’ , subtract 2 from the strength.

    3. else, if character is ‘#’, break.

  • If you are not in the last column, decrement strength by 1.
  • After the loop, print ‘Yes’ and strength separated by a new line.
#3

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