Question

I’m currently writing an interview question for a java expert profile. Here it is:

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Considering this code :

listing 1

package com.example;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

public class Searching {
    public static void main(String[] args) {
        int input = Integer.valueOf(args[0]);
        String[] strings = {"1", "2", "4", "8", "16", "32", "64", "128"};
        List<Integer> integers = new ArrayList<Integer>();
        for (String s : strings) {
            integers.add(Integer.valueOf(s));
        }
        System.out.println("index of "+input+" is:"+Collections.binarySearch(integers, input, cmp));
    }

    static Comparator<Integer> cmp = new Comparator<Integer>() {
        public int compare(Integer i, Integer j) {
            return i < j ? -1 : (i == j ? 0 : 1);
        }
    };
}

This code is then compiled with this cmd line

listing 2

javac com/example/Searching.java

and run with this command line

listing 3

java com/example/Searching 128

QUESTION A:

Executing listing 3 produce:

index of 128 is:-8

Can you explain this output ?

QUESTION B:

Considering this

 java com/example/Searching 32

output is

index of 32 is:5

Can you explain this output ?

QUESTION C:

Assuming you have a JRE 1.6, a shell, and a text editor. What would you change to listing 1 and/or listing 2 and/or listing 3 to produce this output:

index of 128 is:7

remark: the less you change, the better it is.

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My questions are :

• what would be your answer to those questions ?
• how to improve it ?

Answer 1

As an interview question, I would make the problem simpler. I have found that in interviews it can be much harder to solve these sort of problems without allot of hints. After a couple of questions they can’t answer, interviewees can give up which isn’t always productive.

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Can you explain this output ?

There is a bug in the code with i == j which impact A & B differently. In one case the sort assumes the value is less than 128 and in the second it matches 32 because this is cached.

If you try something like -XX:+AggressiveOpts` or another option to increase the Integer cache size it would match in each case.

What would you change to listing 1

Change i == j ? 0 : -1 to i > j ? -1 : 0

Of course using Integer.compare() would some the problem 😉

how to improve it

Depending on the purpose of the program, I would use

int count = 0;
for(int n = Integer.parseInt(args[0]); n != 0; n >>>= 1)
  count++;
System.out.println(count);