Solved

largest of given three integers

Posted on 2014-10-06
20
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Last Modified: 2014-10-09
Hi,

I am working on below challenge

http://codingbat.com/prob/p101887

I wrote as below

public int intMax(int a, int b, int c) {
if(a>b)
{
int minValue=Math.min(a,b);
return a;
}

if(a>c)
{
int minValue=Math.min(a,c);
return a;
}


if(b>c)
{
int minValue=Math.min(b,c);
return b;
}
return c;
  
}

Open in new window


I have two test cases failing. Please advise on how to fix and effective way to write this code.


Expected      Run            
intMax(1, 2, 3) → 3      3      OK         
intMax(1, 3, 2) → 3      3      OK         
intMax(3, 2, 1) → 3      3      OK         
intMax(9, 3, 3) → 9      9      OK         
intMax(3, 9, 3) → 9      9      OK         
intMax(3, 3, 9) → 9      9      OK         
intMax(8, 2, 3) → 8      8      OK         
intMax(-3, -1, -2) → -1      -1      OK         
intMax(6, 2, 5) → 6      6      OK         
intMax(5, 6, 2) → 6      5      X         
intMax(5, 2, 6) → 6      5      X
0
Comment
Question by:gudii9
  • 8
  • 4
  • 3
  • +3
20 Comments
 
LVL 4

Expert Comment

by:AnthonyHamon
Comment Utility
The logic you need to apply is:

if a > b and a > c then return a
if b > c then return b else return c
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
No need of covering below scenario?

if c> a then return c else return a
0
 
LVL 4

Expert Comment

by:AnthonyHamon
Comment Utility
That scenario is covered.  Try it with your test data to validate.
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
public int intMax(int a, int b, int c) {

  
if((a>b)&(a>c))
		{
		//int minValue=Math.min(a,b);
		return a;
		}

		if(b>c)
		{
		//int minValue=Math.min(a,c);
		return b;
		}

return 0;
		/*if(b>c)
		{
		int minValue=Math.min(b,c);
		return b;
		}
		return c;
		  
		}*/

}

Open in new window


I tried as above


still failing for few cases

Expected      Run            
intMax(1, 2, 3) → 3      0      X         
intMax(1, 3, 2) → 3      3      OK         
intMax(3, 2, 1) → 3      3      OK         
intMax(9, 3, 3) → 9      9      OK         
intMax(3, 9, 3) → 9      9      OK         
intMax(3, 3, 9) → 9      0      X         
intMax(8, 2, 3) → 8      8      OK         
intMax(-3, -1, -2) → -1      -1      OK         
intMax(6, 2, 5) → 6      6      OK         
intMax(5, 6, 2) → 6      6      OK         
intMax(5, 2, 6) → 6      0      X         
Correct for more than half the tests

complete code is

public class Test4 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int largeValue=intMax(5,6,2);
		System.out.println("largeValue is--->"+largeValue);

	}
	
	
	public static int intMax(int a, int b, int c) {
		if((a>b)&(a>c))
		{
		//int minValue=Math.min(a,b);
		return a;
		}

		if(b>c)
		{
		//int minValue=Math.min(a,c);
		return b;
		}

return 0;
		/*if(b>c)
		{
		int minValue=Math.min(b,c);
		return b;
		}
		return c;
		  
		}*/

}
}

Open in new window

0
 
LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 100 total points
Comment Utility
You want &&, not & (although it will do the right thing in this case, it's not the same thing in all cases).

You should not return 0 there. If a>b && a>c was false and b > c was also false, you return 0. What should you return there instead?
0
 
LVL 4

Assisted Solution

by:AnthonyHamon
AnthonyHamon earned 100 total points
Comment Utility
You have a 'return 0;' statement in your code which should be 'return c;'  The final 'if' block is remarked out - you can remove this completely if you wish.
0
 
LVL 20

Expert Comment

by:Amitkumar Panchal
Comment Utility
Below is the simplest one with the help of Math.max().

public int intMax(int a, int b, int c) {
  return Math.max(Math.max(a,b),c);
}

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0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
Let us say if i have 10 given int numbers (instead of 3 in this case) can i find the largest of the 10 using Math.max function. Please advise
0
 
LVL 84

Expert Comment

by:ozo
Comment Utility
You can do it with nine applications of the Math.max function.
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
You can do it with nine applications of the Math.max function.

you mean nine separate java apllications. Like nine different java main programs. Please advise
0
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LVL 20

Assisted Solution

by:Amitkumar Panchal
Amitkumar Panchal earned 100 total points
Comment Utility
Let us say if i have 10 given int numbers (instead of 3 in this case) can i find the largest of the 10 using Math.max function. Please advise

try the below out:

int[] values = {12,45,23,67,45,86,43,67,98,5,45,345,67,89,34,56};
int maxval = values[0];
for (int ind=1; ind < values.length; ind++) {
    maxval=Math.max(maxval, values[ind]);
}
System.out.println("Max Value : " + maxval);
0
 
LVL 37

Accepted Solution

by:
zzynx earned 100 total points
Comment Utility
>> you mean nine separate java apllications. Like nine different java main programs. Please advise
No, of course not!

What was meant was this:

public int intMax(int p1, int p2, int p3, int p4, int p5, int p6, int p7, int p8, int p9, int p10) {
  return Math.max(Math.max(Math.max(Math.max(Math.max(Math.max(Math.max(Math.max(Math.max(p1,p2),p3), p4),p5),p6),p7),p8),p9),p10);
}

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Which looks (and is) ridiculous, but works. ;-)
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
public class Test6 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] values = {12,45,23,67,45,86,43,67,98,5,45,345,67,89,34,56}; 
		int maxval = values[0];
		for (int ind=1; ind < values.length; ind++) {
		    maxval=Math.max(maxval, values[ind]);
		}
		System.out.println("Max Value : " + maxval);

	}
	
	
	

}

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above code returned 345 which is correct
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
public class Test7 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int largeValue=intMax(5,6,8);
		System.out.println("largeValue is--->"+largeValue);

	}
	
	
	public static int intMax(int a, int b, int c) {
		if((a>b)&&(a>c))
		{
		//int minValue=Math.min(a,b);
		return a;
		}

		if(b>c)
		{
		//int minValue=Math.min(a,c);
		return b;
		}

return c;
		

}
}

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above also returned 8
0
 
LVL 84

Expert Comment

by:ozo
Comment Utility
Those results are not unexpected.
Did you have reason to think it would do something different?
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
That scenario is covered.  Try it with your test data to validate.

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i validated with test data and works fine. But i did not understand how that scenario is covered. Please advise
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
Those results are not unexpected.
Did you have reason to think it would do something different?

I agree. The results are correct and expected only.
0
 
LVL 84

Expert Comment

by:ozo
Comment Utility
if you are referring to the scenario  c> a,
it was covered only when a<=b, otherwise your initial if(a>b)
would have immediately returned a
but instead of returning a, you could have returned Math.max(a,c);
0
 
LVL 37

Expert Comment

by:zzynx
Comment Utility
>> i did not understand how that scenario is covered. Please advise
Which scenario?

If you read your code - which I commented - you simply see that it covers all scenario's:

public static int intMax(int a, int b, int c) {
   if ((a>b)&&(a>c)) { // a is bigger than b and bigger than c...
     return a; // ...so return it
  }
  // If you're here, you already know that a is not the biggest one, so let's focus on b and c
  if (b>c) { // b is bigger than c...
    return b; // ... so return b
  }
  // if you're here, you know that a is not the biggest, b is not the biggest, so...
  return c; // c must be the biggest. Well, return it.
}

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Maybe it's even more clear when you write it with some more else's like this:

public static int intMax(int a, int b, int c) {
  if ((a>b)&&(a>c)) {
     return a;
  } else  if (b>c) {
    return b;
  } else {
    return c;
  }
}

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0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 100 total points
Comment Utility
I agree if you are referring to the code in http:#a40364357, but the question of http:#a40364184 was posted prior to that code, so wouldn't it have been referring to the code in http:#questionViewInlineCode10-28532206-1 ?
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