Sum of product of all possible subsets of array

Question

I have written a code to find Sum of product of all possible subsets of array. I\'m getting the expected output but I\'m not able to make it fast enough to clear time related te

Answer 1

You need to use a little math. Lets say you have 3 values, like your example, but lets call them A, B, and C.

To get sum of products, you need to calculate:

Result3 = A + B + C + A*B + A*C + B*C + A*B*C
        = A + B + A*B + (1 + A + B + A*B) * C

Now, if we calculate A + B + A*B first, calling it Result2, then you get:

Result2 = A + B + A*B
Result3 = Result2 + (1 + Result2) * C

And we repeat that, so

Result2 = A + (1 + A) * B
Result1 = A
Result2 = Result1 + (1 + Result1) * B

Can you see the pattern? Let's use that with 4 values:

Result4 = A + B + C + D + A*B + A*C + A*D + B*C + B*D + C*D
        + A*B*C + A*B*D + A*C*D + B*C*D + A*B*C*D
        =      A + B + C + A*B + A*C + B*C + A*B*C
        + (1 + A + B + C + A*B + A*C + B*C + A*B*C) * D
        = Result3 + (1 + Result3) * D

Summary:

Result1 = A
Result2 = Result1 + (1 + Result1) * B
Result3 = Result2 + (1 + Result2) * C
Result4 = Result3 + (1 + Result3) * D

As code, this is:

private static long sumProduct(int... input) {
    long result = 0;
    for (int value : input)
        result += (result + 1) * value;
    return result;
}

Only one iteration, so O(n).

Test

System.out.println(sumProduct(2, 3));
System.out.println(sumProduct(2, 3, 5));
System.out.println(sumProduct(2, 3, 5, 7));

Output

11
71
575

UPDATE

Code can also be done using Java 8 Streams with a Lambda expression, using either IntStream.of(int...) or Arrays.stream(int[]) (they do the same).

// Using IntStream with result as int
private static int sumProduct(int... input) {
    return IntStream.of(input).reduce((a, b) -> a + (1 + a) * b).getAsInt();
}

// Using Arrays with result as long
private static long sumProduct(int... input) {
    return Arrays.stream(input)
                 .asLongStream()
                 .reduce((a, b) -> a + (1 + a) * b)
                 .getAsLong();
}