I came up with

        (%~ */) 4 3 2 8
    48 64 96 24

but I think your solution is more readable.

                        ... peter

On 20140327, 1:10 PM, Dan Bron wrote:
Oh, I like this game!

           (*/ % ]) 4 3 2 8
        48 64 96 24

           (*/@#~1-=@i.) 4 3 2 8
        48 64 96 24

-Dan

----- Original Message ---------------

Subject: Re: [Jprogramming] J as an interviewing tool
    From: Roger Hui <[email protected]>
    Date: Thu, 27 Mar 2014 12:44:59 -0700
      To: Programming forum <[email protected]>

Suppose you did not have the outfix operator, but still have array ops.

    x=: 4 3 2 8
    I=: =i.#x
    I
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
    (-.I)*/ .(^~) x
48 64 96 24





On Thu, Mar 27, 2014 at 12:37 PM, Dan Bron <[email protected]> wrote:

A question on StackExchange [1]:

"
    I was asked the following interview question over the phone:
    Given an array of integers, produce an array whose values
    are the product of every other integer excluding the current index.


    Example:
    [4, 3, 2, 8] -> [3*2*8, 4*2*8, 4*3*8, 4*3*2] -> [48, 64, 96, 24]

"

His answer is given in the postscript. It spans 3 kilobytes over 86 lines.
The community's responses aren't much of an improvement (by that metric;
by other metrics - Java-specific ones, they decent advancements).

The J answer?

    1 */\. 4 3 2 8
48 64 96 24

Could've answered the interviewer in a sentence, right over the phone.

-Dan


[1] Code Review StackExchange question
     "Array whose values are the product of every other integer":


http://codereview.stackexchange.com/questions/45498/array-whose-values-are-the-product-of-every-other-integer



import java.math.BigInteger;
import java.util.Arrays;

public class ProductOfAnArray {

     public static void main(String[] args) {

         try {
             System.out.println(Arrays.toString(ProductOfAnArray
                     .calcArray(new int[] { 4, 3, 2, 8 })));
             System.out.println(Arrays.toString(ProductOfAnArray
                     .calcArray(new int[] { 4, 0, 2, 8 })));
             System.out.println(Arrays.toString(ProductOfAnArray
                     .calcArray(new int[] { 4, 0, 2, 0 })));
             System.out.println(Arrays.toString(ProductOfAnArray
                     .calcArray(new int[] {})));
             System.out
                     .println(Arrays.toString(ProductOfAnArray
                             .calcArray(new int[] { 4, 3, 2, 8, 3, 2, 4, 6,
7,
                                     3, 2, 4 })));
             System.out
                     .println(Arrays.toString(ProductOfAnArray
                             .calcArray(new int[] { 4, 3, 2, 8, 3, 2, 4, 6,
7,
                                     3, 2, 4 })));
             System.out.println(Arrays.toString(ProductOfAnArray
                     .calcArray(new int[] { 4432432, 23423423, 34234,
23423428,
                             4324243, 24232, 2342344, 64234234, 4324247,
                             4234233, 234422, 234244 })));
         } catch (Exception e) {
             // debug exception here and log.
         }
     }

     /*
      * Problem: Given an array of integers, produce an array whose values
are
      * the product of every other integer excluding the current index.
      *
      * Assumptions. Input array cannot be modified. input is an integer
array
      * "produce an array" - type not specified for output array
      *
      * Logic explanation:
      *
      * Assume we have array [a,b,c,d] Let multiple be multiple of each
element
      * in array. Hence multiple = 0 if one of the element is 0; To produce
the
      * output. Ans at i -> multiple divided by the value at i. if 2 numbers
are
      * 0 then entire output will be 0 because atleast one 0 will be a
multiple
      * if 1 number is 0 then every thing else will be 0 except that index
whole
      * value is to be determined
      *
      */
     public static BigInteger[] calcArray(final int[] inp) throws Exception
{
         if (inp == null)
             throw new Exception("input is null");

         int cnt = 0;
         BigInteger multiple = new BigInteger("1");
         boolean foundZero = false;

         for (int i : inp) {
             if (i == 0) {
                 cnt++;
                 foundZero = true;
                 if (cnt < 2)
                     continue;
                 else
                     break;
             }
             multiple = multiple.multiply(BigInteger.valueOf(i));
         }

         BigInteger ans[] = new BigInteger[inp.length];

         for (int i = 0; i < inp.length; i++) {
             if (foundZero) {
                 if (cnt < 2) {
                     ans[i] = (inp[i] == 0) ? multiple : new
BigInteger("0");
                 } else {
                     ans[i] = new BigInteger("0");
                 }
             } else {
                 ans[i] = multiple.divide(BigInteger.valueOf(inp[i]));
             }
         }
         return ans;
     }

}
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