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int in the range of 30 to 50 challenge

Hi,

I was working on the int in the range of 30 to 50 challenge


http://codingbat.com/prob/p132134


I wrote code like below
public boolean in3050(int a, int b) {

if((Math.abs(a-30)>=10 &&Math.abs(b-30)>=10)  || (Math.abs(a-40)>=10&& Math.abs(b-40)>=10))
{
return true;
}

if((Math.abs(a-40)>=10 && Math.abs(b-40)>=10) || (Math.abs(a-50)>=10 && Math.abs(b-50)>=10))
{
return true;
}
  return false;
}

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I have couple of test cases failing as below

Expected      Run            
in3050(30, 31) → true      true      OK         
in3050(30, 41) → false      false      OK         
in3050(40, 50) → true      true      OK         
in3050(40, 51) → false      true      X         
in3050(39, 50) → false      false      OK         
in3050(50, 39) → false      false      OK         
in3050(40, 39) → true      true      OK         
in3050(49, 48) → true      true      OK         
in3050(50, 40) → true      true      OK         
in3050(50, 51) → false      true      X         
in3050(35, 36) → true      true      OK         
in3050(35, 45) → false      false      OK         
Correct for more than half the tests


How to fix and improve my program. Please advise
ASKER CERTIFIED SOLUTION
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ozo
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ASKER

public boolean in3050(int a, int b) {

if((Math.abs(a-30)>=10 &&Math.abs(b-30)>=10)  || (Math.abs(a-40)>=10&& Math.abs(b-40)>=10))
{
return true;
}

if((Math.abs(a-40)>=10 && Math.abs(b-40)>=10) || (Math.abs(a-50)>=10 && Math.abs(b-50)>=10))
{
return true;
}

if((a-50)>=1 && (b-50)>=1)
{
return false;
}
return false;

}

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I think i improved my code bit still 2 failing

Expected      Run            
in3050(30, 31) → true      true      OK         
in3050(30, 41) → false      false      OK         
in3050(40, 50) → true      true      OK         
in3050(40, 51) → false      true      X         
in3050(39, 50) → false      false      OK         
in3050(50, 39) → false      false      OK         
in3050(40, 39) → true      true      OK         
in3050(49, 48) → true      true      OK         
in3050(50, 40) → true      true      OK         
in3050(50, 51) → false      true      X         
in3050(35, 36) → true      true      OK         
in3050(35, 45) → false      false      OK
Think more carefully about what in the range 30..40 inclusive means, and what Math.abs(a-30)>=10 means
SOLUTION
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ASKER

thinking. Why thinking is takes time. why it is not spontaneous yet with java language unlike english language?
Avatar of gudii9

ASKER

in the mind i have answer but to teach computer to understand is bit hard for me
Avatar of dpearson
dpearson

Don't worry that it takes time.  It will all get faster with practice.

Everyone is slow when they start out.

Doug
Avatar of gudii9

ASKER

public class Test9 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		in3050(40,51);
	}

	
	public static boolean in3050(int a, int b) {

		if((Math.abs(a-30)<=10 &&Math.abs(b-30)<=10)  || (Math.abs(a-40)<=10&& Math.abs(b-40)<=10))
		{
		return true;
		}

		if((Math.abs(a-40)<=10 && Math.abs(b-40)<=10) || (Math.abs(a-50)<=10 && Math.abs(b-50)<=10))
		{
		return true;
		}

		if((a-50)>0 && (b-50)>0)
		{
		return false;
		}
		return false;

		}
}

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tried as above but still not working. WIll modify more
Think more carefully about what in the range 30..40 inclusive means, and what Math.abs(a-40)<=10 means
Think about what happens when a is 29, 30, 31, 39, 40, 41

Another hint:  How would you test for in the range -10..10?
>> why it is not spontaneous yet with java language unlike english language?
I think that's because you use the method Math.abs()
I wonder why you use it for this challenge?

Let's do it in English:
Checking if an int x is in the range 30...40 is just as simple as
1) checking that x is 30 or more
but ALSO
2) checking that x is no more than 40, thus 40 or less

In java
1) x >= 30
AND
2) x <= 40

Hence, checking if an int x is in the range 30...40 is just as simple as:

if ( x>=30 && x<=40) {
}

You have to check if both a and b are in the range 30..40, so that just doubling what you already had above:
if ( (a>=30 && a<=40) && (b>=30 && b<=40) ) {
}

And you have to check it for another range: 40...50, so that's replacing 30 by 40 and 40 by 50:
if ( (a>=40 && a<=50) && (b>=40 && b<=50) ) {
}

So the eventual result is:

public boolean in3050(int a, int b) {
      if ( (a>=30 && a<=40) && (b>=30 && b<=40) ) { // a and b in the range 30...40?
         return true; // Yes!
      }

      if ( (a>=40 && a<=50) && (b>=40 && b<=50) ) { // a and b in the range 40...50?
         return true; // Yes!
      }

      return false; // no succes (otherwhise, we shouldn't have come till here)
}

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Avatar of gudii9

ASKER

Above solution works great.. in my mind i was struck with one direction of Math.abs for this challenge. How to keep mind working  thinking on all different directions and possibilities.
For what range of numbers  is Math.abs(a-40)<=10 true?
Avatar of gudii9

ASKER

both sides of 40 ie 30-50 since -10 also considered as 10 with abs
Yes.  So can you see how to modify 30-50 to 30-40?
Avatar of gudii9

ASKER

public class Test8 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
    boolean b=in3050(30, 51);
    System.out.println("b is-->"+b);
	}

	
	public static boolean in3050(int a, int b) {
	      if ( a>=30 && a<=40 && b>=30 && b<=40 ) { // a and b in the range 30...40?
	         return true; // Yes!
	      }

	      if ( a>=40 && a<=50 && b>=40 && b<=50 ) { // a and b in the range 40...50?
	         return true; // Yes!
	      }

	      return false; // no succes (otherwhise, we shouldn't have come till here)
	}
}

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Even if i remove the inner brackets I still get same results.

Instead of

  if ( (a>=30 && a<=40) &&( b>=30 && b<=40) )
changed as below

  if ( a>=30 && a<=40 && b>=30 && b<=40 )

So no need of inner extra brackets right?
Please advise
In java, comparisons have precedence over boolean operators,

but I was asking whether you could figure out if
30-50 corresponds to Math.abs(a-40)<=10
then
30-40 corresponds to what?
Avatar of gudii9

ASKER

(a-35)<=5
which adds 5 on both sides of 35
Correct, Math.abs(a-35)<=5 would test for the range 30..40
>> Above solution works great..
Then why didn't you give it any points?