How to print array if the output is 0?

Question

I solve this question and it's working good but a small problem that if the ray contain only one number or same number it return 0 but i want that number. My code:

public class RaySmallest
{
    public static int go(int[] ray)
    {
          int smallest = 0;
      for(int i =0;i<ray.length;i++) 
      {
            if(smallest > ray[i]) 
            {
                smallest = ray[i];
            }
      } 
      return smallest;
 }
}

The Runner for code

class Main
{
  public static void main(String[] args)
  {
    RaySmallest rt = new RaySmallest();

    System.out.println( rt.go( new int[]{-99,1,2,3,4,5,6,7,8,9,10,12345} ) );
    System.out.println( rt.go( new int[]{10,9,8,7,6,5,4,3,2,1,-99} ) );
    System.out.println( rt.go( new int[]{10,20,30,40,50,-11818,40,30,20,10} ) );
    System.out.println( rt.go( new int[]{32767} ) );
    System.out.println( rt.go( new int[]{255,255} ) );
    System.out.println( rt.go( new int[]{9,10,-88,100,-555,1000} ) );
    System.out.println( rt.go( new int[]{10,10,10,11,456} ) );
    System.out.println( rt.go( new int[]{-111,1,2,3,9,11,20,30} ) );
    System.out.println( rt.go( new int[]{9,8,7,6,5,4,3,2,0,-2,-989} ) );
    System.out.println( rt.go( new int[]{12,15,18,21,23,1000} ) );
    System.out.println( rt.go( new int[]{250,19,17,15,13,11,10,9,6,3,2,1,-455} ) );
    System.out.println( rt.go( new int[]{9,10,-8,10000,-5000,1000} ) );

  }
}

Like in line four my code give me 0 but i need there output as 32767.

I need this answers.

Thank you

Maxdola · Accepted Answer · 2020-01-12 02:39:14Z

1

If the smallest number is greater than zero the number isn't actually set and this is why you aren't getting the smallest number if you have only one, which is probably below zero.

public static int go(int[] ray) {
    int smallest = ray.length > 0 ? ray[0] : -1;
    for (int value : ray) {
        if (smallest > value) {
            smallest = value;
        }
    }
    return smallest;
}

edited Jan 12, 2020 at 2:39

answered Jan 12, 2020 at 2:35

Maxdola

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5 Comments

Turing85 Over a year ago

even simpler: initialize smallest with Integer.MAX_VALUE or with ray[0] (after checking that ray has a lenght > 0).

Maxdola Over a year ago

Yes that is actually better.

Faheem Over a year ago

Can any one explain why he use [? ray[0] : -1;]

Maxdola Over a year ago

this checks if the array is at least 1 element long and if it is then it uses the firest value for comparison and otherwise it uses -1, which is then returned if the element has the size of 0.

Maxdola Over a year ago

No Problem at all :)

Nate · Accepted Answer · 2020-01-12 03:03:53Z

Ok, here is part of your code:

 int smallest = 0;
  for(int i =0;i<ray.length;i++) 
  {
        if(smallest > ray[i]) 
        {
            smallest = ray[i];
        }
  } 
  return smallest;

Essentially this tells me, smallest is 0. You compare if 0 > ray[i], which on line 4 is 32767. 0 is not greater than 32767, so smallest is still equal to 0 and returned as 0.

int smallest = 0;
int[] numArray = new int[ray.length];
 for(int i=0;i<ray.length;i++)
 { 
       if(ray.length == 1)
             return ray[i];
       if(smallest > ray[i])
             smallest = ray[i];
       if(ray[i] == ray[ray.length - 1] && i-ray.length > 0 && i==ray.length-2){
             int x = 0;
             for(int x=0;x<ray.length-1;x++){
                  if(ray[i] != ray[x])
                    break;
             }
             if(ray[i] == ray[x])
                  return ray[i];
       }

}
return smallest;

This solution should take care of having a single value in the array, or if all values in the array are equal.

This is not really and answer, this is just a deeper description to the problem, not a solution.

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