CODEWITHSUNDEEP
javajava-8java-stream
Artem Kuzmin
Artem Kuzmin

Reputation: 23

Sum of subtractions using Java Stream API

everyone.

I've faced a task, that looked pretty simple to me, but I spend like 4 hours and left with nothing. So, let's assume we have simple list of integers like

[1, 2, 3, 4, 5]

I want to get sum of subtractions for each element - for this list, it would be equal to 4 -> (2 - 1) + (3 - 2)... and so on.

To avoid answers with range, list may look like this:

[5, 12, 16, 25, 25, 48]

It's always growing, but not with the same range between.

Using simple for it's not hard to get something like: 

    int result = 0;
    for (int i = 1; i < lst.size(); i++) {
        result += lst.get(i) - lst.get(i-1);
    }

But I'm feeling myself utterly stupid when I'm trying to solve this case using streams.

Is this actually possible?

Upvotes: 2

Views: 110

Answers (2)

Jeremy Grand
Jeremy Grand

Reputation: 2370

Assuming you have a list of Integers like this : [ n0, n1, n2, ... , nM, nN ], what you're doing is :

  1. Create list [(n1 - n0), (n2 - n1), (n3 - n2), ..., (nN - nM)].
  2. Sum that list : X = (n1 - n0) + (n2 - n1) + (n3 - n2) + ... + (nN - nM). Which is rigourously the same (just swapped the inside of parenthesises as they don't matter) as X = (-n0) + (n1 - n1) + (n2 - n2) + ... + (nM - nM) + nN ].

Which is (as others pointed out) X = nN - n0. No matter what is inside your list (except null).

Thus, lst.get(lst.size()-1) - lst.get(0) is, by far, the best answer.

Upvotes: 4

Lino
Lino

Reputation: 19926

Try using this:

int result = IntStream.range(1, lst.size())  // creates a stream of indexes from 1 to lst.size()
        .map(i -> lst.get(i) - lst.get(i-1)) // maps the new value into a new stream
        .sum();                              // sums it up

Upvotes: 2

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