The challenge
You will be given a certain array of length n, such that n > 4, having positive and negative integers but there will be no zeroes and all the elements will occur once in it.
We may obtain an amount of n sub-arrays of length n - 1, removing one element at a time (from left to right).
For each subarray, let’s calculate the product and sum of its elements with the corresponding absolute value of the quotient, q = SubProduct/SubSum (if it is possible, SubSum cannot be 0). Then we select the array with the lowest value of |q|(absolute value)
e.g.: we have the array, arr = [1, 23, 2, -8, 5]
Sub Arrays SubSum SubProduct |q|
[23, 2, -8, 5] 22 -1840 83.636363
[1, 2, -8, 5] 0 -80 No value
[1, 23, -8, 5] 21 -920 43.809524
[1, 23, 2, 5] 31 230 7.419355 <--- selected array
[1, 23, 2, -8] 18 368 20.444444
Let’s compare the given array with the selected subarray:
[1, 23, 2, -8, 5]
[1, 23, 2, 5]
The difference between them is at the index 3 for the given array, with element -8, so we put both things for a result [3, -8].
That means that to obtain the selected subarray we have to take out the value -8 at index 3. We need a function that receives an array as an argument and outputs the pair [index, arr[index]] that generates the subarray with the lowest value of |q|.
select_subarray([1, 23, 2, -8, 5]) == [3, -8]
Another case:
select_subarray([1, 3, 23, 4, 2, -8, 5, 18]) == [2, 23]
In Javascript the function will be selectSubarray().
We may have some special arrays that may have more than one solution as the one that follows below.
select_subarray([10, 20, -30, 100, 200]) == [[3, 100], [4, 200]]
If there is more than one result the function should output a 2Darray sorted by the index of the element removed from the array.
Thanks to Unnamed for detecting the special cases when we have multiple solutions.
Features of the random tests:
Number of tests = 200
length of the array, l, such that 20 <= l <= 100
The solution in Java code
Option 1:
import static java.util.Arrays.stream;
import static java.util.stream.Collectors.toList;
import static java.util.stream.IntStream.*;
interface Solution {
static int[][] selectSubarray(int[] arr) {
var qs = rangeClosed(0, arr.length).mapToDouble(i -> Double.MAX_VALUE).toArray();
return range(0, arr.length)
.peek(i -> qs[i] = q(stream(arr).filter(e -> e != arr[i]).toArray()))
.peek(i -> qs[qs.length - 1] = Math.min(qs[i], qs[qs.length - 1]))
.boxed().collect(toList()).stream()
.filter(i -> qs[i] == qs[qs.length - 1])
.map(i -> new int[]{i, arr[i]})
.toArray(int[][]::new);
}
private static double q(int[] sub) {
double sum = of(sub).sum();
double prod = of(sub).mapToDouble(i -> i).reduce(1, (a, b) -> a * b);
return Math.abs(prod / (sum != 0 ? sum : 1));
}
}
Option 2:
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.stream.Collectors;
import static java.math.BigInteger.valueOf;
public class Solution {
public static int[][] selectSubarray(final int[] arr) {
final int ss = Arrays.stream(arr).sum();
final BigInteger pp = Arrays.stream(arr).mapToObj(BigInteger::valueOf).reduce((a, b) -> a.multiply(b)).get();
ArrayList<int[]> lt = new ArrayList<>();
BigInteger q = pp.abs();
BigInteger t;
for (int i = 0; i < arr.length; i++) {
if(ss == arr[i]) continue;
t = pp.divide(valueOf(arr[i])).divide(valueOf(ss - arr[i])).abs();
if (q.compareTo(t) >= 0) {
if(q.compareTo(t) > 0) lt.clear();
q = t;
lt.add(new int[] { i, arr[i] });
}
}
return lt.stream().toArray(a -> new int[a][]);
}
}
Option 3:
import java.util.stream.*;
public class Solution {
private static double quotient(final int[] arr, final int exclude) {
double sum = IntStream.of(arr).mapToDouble(x -> (double)x)
.sum() - arr[exclude],
product = IntStream.of(arr).mapToDouble(x -> (double)x)
.reduce(1.0, (a,b) -> a * b) / arr[exclude];
return sum != 0 ? Math.abs(product / sum) : Double.MAX_VALUE;
}
public static int[][] selectSubarray(final int[] arr) {
double minQ = IntStream.range(0, arr.length)
.mapToDouble(i -> quotient(arr, i))
.min().getAsDouble();
return IntStream.range(0, arr.length)
.filter(i -> quotient(arr, i) <= minQ)
.mapToObj(i -> new int[]{i, arr[i]})
.toArray(int[][]::new);
}
}
Test cases to validate our solution
import org.junit.Test;
import java.util.*;
import java.util.stream.*;
public class SolutionTest {
@Test
public void basicTests() {
Tester.doTest(new int[]{1, 23, 2, -8, 5}, new int[][]{{3, -8}});
Tester.doTest(new int[]{1, 3, 23, 4, 2, -8, 5, 18}, new int[][]{{2, 23}});
Tester.doTest(new int[]{10, 20, -30, 100, 200}, new int[][]{{3, 100}, {4, 200}});
}
final Random rand = new Random();
@Test
public void randomTests() {
final List<Integer> values = IntStream.concat(IntStream.rangeClosed(-200, -1),
IntStream.rangeClosed(1, 200))
.boxed().collect(Collectors.toList());
for (int trial = 1; trial <= 200; trial++) {
Collections.shuffle(values);
final int arr[] = values.stream().limit(rand.nextInt(81) + 20)
.mapToInt(x -> x).toArray();
Tester.doTest(arr, solution(arr));
}
}
public static int[][] solution(final int[] arr) {
double totalSum = 0.0, totalProduct = 1.0;
for (int x : arr) {
totalSum += x;
totalProduct *= x;
}
double qrr[] = new double[arr.length], qmin = Double.MAX_VALUE;
for (int i = 0; i < arr.length; i++)
if ( totalSum != arr[i] ) {
qrr[i] = Math.abs(totalProduct / arr[i] / (totalSum - arr[i]));
qmin = Math.min(qmin, qrr[i]);
} else
qrr[i] = Double.MAX_VALUE;
List<int[]> indices = new ArrayList<>();
for (int i = 0; i < arr.length; i++)
if ( qrr[i] <= qmin )
indices.add(new int[]{i, arr[i]});
return indices.toArray(int[][]::new);
}
}