Solved

countX

Posted on 2016-09-09
22
191 Views
Last Modified: 2016-09-14
Hi,

I am working on below challenge
http://codingbat.com/prob/p170371

Given a string, compute recursively (no loops) the number of lowercase 'x' chars in the string.

countX("xxhixx") → 4
countX("xhixhix") → 3
countX("hi") → 0
i was not clear on how to find the number of x using recursion? please advise
0
Comment
Question by:gudii9
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 14
  • 6
  • 2
22 Comments
 
LVL 12

Expert Comment

by:tel2
ID: 41792066
Do you know what recursion is?
If so, tell us what you know about recursion, and we might be able to make sure your understanding is correct.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41792181
http://introcs.cs.princeton.edu/java/23recursion/

The idea of calling one function from another immediately suggests the possibility of a function calling itself. The function-call mechanism in Java supports this possibility, which is known as recursion.
Your first recursive program. The "Hello, World" for recursion is the factorial function, which is defined for positive integers n by the equation
n!=n×(n−1)×(n−2)×…×2×1
n!=n×(n−1)×(n−2)×…×2×1
The quantity n! is easy to compute with a for loop, but an even easier method in Factorial.java is to use the following recursive function:
public static long factorial(int n) {
    if (n == 1) return 1;
    return n * factorial(n-1);
}
We can trace this computation in precisely the same way that we trace any sequence of function calls.
factorial(5)
   factorial(4)
      factorial(3)
         factorial(2)
            factorial(1)
               return 1
            return 2*1 = 2
         return 3*2 = 6
      return 4*6 = 24
   return 5*24 = 120

reading from above link as above.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41792183
how to get base case for this challenge?
how to draw diagram like below all the way till base case

factorial(5)
   factorial(4)
      factorial(3)
         factorial(2)
            factorial(1)
               return 1
            return 2*1 = 2
         return 3*2 = 6
      return 4*6 = 24
   return 5*24 = 120
0
Congratulations! You’re Certified – Now What?

Starting a new career can be overwhelming. Becoming certified in your field of expertise is a great start, but where do you go from here?  Here are some tips to help you on your career journey.

 
LVL 27

Expert Comment

by:rrz
ID: 41793087
This challenge is the same programming  problem as
https://www.experts-exchange.com/questions/28968901/countHi-challenge.html   
how to get base case for this challenge?
The base case is
When you found all the "x"s  return 0.
You can use a String method to search for x.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41793539
Return 1  or return 0?
0
 
LVL 12

Assisted Solution

by:tel2
tel2 earned 250 total points
ID: 41793567
In the case of factorials, you'll return 1 (or alternatively return n when n == 1).  If you returned 0, then 0 would be multiplied by the other numbers which would always result in 0, but that's not how factorials work, so use 1.  When the other numbers are multiplied by 1 the result will not be changed, which is good.

In the case of counting, return 0 (or alternatively return n when n == 0).  0 will be added to the other numbers, which will not change the result, which is good.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41793633
sure. let me try
0
 
LVL 27

Expert Comment

by:rrz
ID: 41793768
how to draw diagram like below all the way till base case
I couldn't do it that way. Here is what I wrote.
public class Fact{
	public static void main(String[] args){
	    System.out.println(" factorial(5) equals " + factorial(5));
	}
	public static int factorial(int n) {
	    System.out.println(" factorial(" + n + ")");
		if(n==1){
			System.out.println(" return 1");
			return 1;
		}
		System.out.println(" return " + n + " * factorial(" + (n-1) + ")");
        return n * factorial(n-1);
    }
}

Open in new window

My output:
 factorial(5)
 return 5 * factorial(4)
 factorial(4)
 return 4 * factorial(3)
 factorial(3)
 return 3 * factorial(2)
 factorial(2)
 return 2 * factorial(1)
 factorial(1)
 return 1
 factorial(5) equals 120
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796709
package com.solution;

import java.io.*;

public class CountX {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		System.out.print("Return Value :");
		System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println(countX("value is" + countX("xxhixx")));
	}

	public static int countX(String str) {
	    if (str.length() == 0){ 
	    	return 0;
	    		};
	    if (str.charAt(0) == 'x') {
	    	return 1 + countX(str.substring(1));
	    	};
	    	else {
		   return countX(str.substring(1));}
	
}
}

Open in new window


above gives below error at line 25

Syntax error on token "else", delete this token

please advise
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796867
package com.solution;

import java.io.*;

public class Count8 {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		System.out.print("Return Value :");
		System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println("value is" + count8(8818));
	}

	public static int count8(int n) {
		//public int count8(int n) {
			  if (n < 1) return 0;
			  if (n % 10 == 8 && (n / 10) % 10 == 8) return 2 + count8(n/10);
			  else if (n % 10 == 8) return 1 + count8(n/10);
			  else return count8(n/10);
			}
	}
		 /* else {
		  return count8(n/10);
		  }*/
		   /* if(n % 10 == 8){
				if(next digit is 8)return 2 + count8(n / 10);
				else return 1 + count8(n / 10);*/
		    

Open in new window

public int countX(String str) {
   
	    if (str.length() == 0){ 
	    	return 0;
	    		};
	    if (str.charAt(0) == 'x') {
	    	return 1 + countX(str.substring(1));
	    	};
	    	else {
		   return countX(str.substring(1));
	    	  
	    	}
	    	
	    	return 1;
	
}

Open in new window

i see i made a stupid mistake of putting ; next to } of if block above passed all tests
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796877
public int countX(String str) {
   
	    if (str.length() == 0){ 
	    	return 0;
	    		};
	    if (str.charAt(0) == 'x') {
	    	return 1 + countX(str.substring(1));
	    	}
	    	else {
		   return countX(str.substring(1));
	    	  
	    	}
	    	
	    //	return 1;
	
}

Open in new window


i wonder why same code in eclipse gives wrong result

package com.solution;

import java.io.*;

public class CountX {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		System.out.print("Return Value :");
		System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println(countX("value is" + countX("xxhixx")));
	}

	public static int countX(String str) {
		if (str.length() == 0) {
			return 0;
		}
		if (str.charAt(0) == 'x') {
			return 1 + countX(str.substring(1));
		} else {
			//else {
				return countX(str.substring(1));
			//}
		}

		// return 1;

		/*
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); }; else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 * 
		 * return 1;
		 */

		/*
		 * public int countX(String str) {
		 * 
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); } else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 */

		// return 1;

	}

}

Open in new window

Return Value :xhixx
0
0
 
LVL 27

Expert Comment

by:rrz
ID: 41796927
Your logic in your method was fine.  It works like the following.
import java.io.*;
public class X {
	public static void main(String args[]) {
		String inputStr = new String("xxhixx");
		System.out.println(" value is " + countX(inputStr));
	}
	public static int countX(String str) {
	    if (str.length() == 0){ 
	    	return 0;
	    		}
	    if (str.charAt(0) == 'x') {
	    	return 1 + countX(str.substring(1));
	    	}
	    	else {
		   return countX(str.substring(1));}
	
	}
}

Open in new window

0
 
LVL 27

Accepted Solution

by:
rrz earned 250 total points
ID: 41796932
My solution was a little different.
public int countX(String str) {
  int index = str.indexOf("x");
  if(index == -1) return 0;
  else return 1 + countX(str.substring(index + 1));
}

Open in new window

0
 
LVL 7

Author Comment

by:gudii9
ID: 41796945
package com.solution;

import java.io.*;

public class CountX {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		// System.out.print("Return Value :");
		// System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println(countX("value is" + countX("xxhixx")));
	}

	public static int countX(String str) {
		if (str.length() == 0) {
			return 0;
		}
		if (str.charAt(0) == 'x') {
			return 1 + countX(str.substring(1));
		} else {
			// else {
			return countX(str.substring(1));
			// }
		}

		// return 1;

		/*
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); }; else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 * 
		 * return 1;
		 */

		/*
		 * public int countX(String str) {
		 * 
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); } else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 */

		// return 1;

	}

}

Open in new window


what is difference between above not working code and below working coe
package com.solution;

import java.io.*;

public class CountX2 {
	public static void main(String args[]) {
		String inputStr = new String("xxhixx");
		System.out.println(" value is " + countX(inputStr));
	}

	public static int countX(String str) {
		if (str.length() == 0) {
			return 0;
		}
		if (str.charAt(0) == 'x') {
			return 1 + countX(str.substring(1));
		}

		else {
			return countX(str.substring(1));
		}

	}
}

Open in new window


eclipse also aligned it differently with else block. I used beyondcompare for text comparison but could not notice any difference to my eyes?
compare.png
0
 
LVL 27

Expert Comment

by:rrz
ID: 41796964
		System.out.println(countX("value is" + countX("xxhixx")));

Open in new window

That is wrong. It should be
System.out.println(" value is " + countX(inputStr));

Open in new window

0
 
LVL 7

Author Comment

by:gudii9
ID: 41798247
           System.out.println(countX("value is" + countX("xxhixx")));

Select all
 
Open in new window
That is wrong. It should be
System.out.println(" value is " + countX(inputStr));

i could not catch even after using beyondcompare. how to make my eyes, brain as sharp as yours?
what is the difference of passing inputStr or passign directly value of inputStr which is xxhixx

To me both seems one and the same?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41798270
package com.solution;

import java.io.*;

public class CountX {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		// System.out.print("Return Value :");
		// System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println(countX("value is" + countX(Str)));
	}

	public static int countX(String str) {
		if (str.length() == 0) {
			return 0;
		}
		if (str.charAt(0) == 'x') {
			return 1 + countX(str.substring(1));
		} else {
			// else {
			return countX(str.substring(1));
			// }
		}

		// return 1;

		/*
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); }; else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 * 
		 * return 1;
		 */

		/*
		 * public int countX(String str) {
		 * 
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); } else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 */

		// return 1;

	}

}

Open in new window


above also gives 0?
0
 
LVL 27

Expert Comment

by:rrz
ID: 41798364
how to make my eyes, brain as sharp as yours?
We(coders) have all been there. Sometimes we can't see something right in front on our faces. One time I spent 4 hours looking for a error that was just a simple capitalization mistake. I can't explain it. It is just a occupational hazard.  
what is the difference of passing inputStr or passign directly value of inputStr which is xxhixx

 To me both seems one and the same?
It is the same.
above also gives 0?
You still have
System.out.println(countX("value is" + countX(Str)));

Open in new window

Where you should have
System.out.println("value is" + countX(Str));

Open in new window

0
 
LVL 7

Author Comment

by:gudii9
ID: 41798378
You still have
System.out.println(countX("value is" + countX(Str)));

Select all
 
Open in new window
Where you should have
System.out.println("value is" + countX(Str));

i used countX two times
0
 
LVL 7

Author Comment

by:gudii9
ID: 41798380
occupational hazard.  
like this word
0
 
LVL 7

Author Comment

by:gudii9
ID: 41798382
package com.solution;

import java.io.*;

public class CountX {
	public static void main(String args[]) {
		String Str = new String("xxhixx");

		// System.out.print("Return Value :");
		// System.out.println(Str.substring(1));
		/*
		 * System.out.print("Return Value :" );
		 * System.out.println(Str.substring(10, 15) );
		 */
		System.out.println("value is" + countX(Str));
	}

	public static int countX(String str) {
		if (str.length() == 0) {
			return 0;
		}
		if (str.charAt(0) == 'x') {
			return 1 + countX(str.substring(1));
		} else {
			// else {
			return countX(str.substring(1));
			// }
		}

		// return 1;

		/*
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); }; else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 * 
		 * return 1;
		 */

		/*
		 * public int countX(String str) {
		 * 
		 * if (str.length() == 0){ return 0; }; if (str.charAt(0) == 'x') {
		 * return 1 + countX(str.substring(1)); } else { return
		 * countX(str.substring(1));
		 * 
		 * }
		 */

		// return 1;

	}

}

Open in new window


above finally works
0
 
LVL 7

Author Comment

by:gudii9
ID: 41798392
gives below output
value is4
0

Featured Post

What does it mean to be "Always On"?

Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This is about my first experience with programming Arduino.
Make the most of your online learning experience.
The viewer will learn additional member functions of the vector class. Specifically, the capacity and swap member functions will be introduced.
Progress
Suggested Courses

615 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question