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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>
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