Sum the array times n, except the last

Question

I've been posting relatively hard challenges recently, so here goes an easy one.

Task

Given an array \$A\$ and a number \$n\$, calculate the sum of all numbers of \$A\$ multiplied by \$n\$, except the last one. All numbers (the elements of \$A\$ and the value of \$n\$) are positive integers, and \$A\$ is non-empty. Shortest code in bytes wins.

I have a 3-byte J solution. Can you find it (or beat it in a different language)?

Test cases

A           N   Ans   Explanation
3 1 4 1 5   10  95    (3+1+4+1)*10+5
3 1 4 1 5   1   14    (3+1+4+1)*1+5
1           999 1     1

Answer 1

J, 3 bytes

That was fun to find.

&+/

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How it works

10 (&+/) 3 1 4 1 5 will bind 10 as an argument of + as 10&+, one verb that gets inserted between the elements of the list by /. So we have: 3 (10&+) 1 (10&+) 4 (10&+) 1 (10&+) 5. Now x n&v y means that y gets applied to n&v for x times. With J's right to left evaluation we get: to 5 add 1 times 10, add 4 times 10, add 1 times 10, add 3 times 10. A challenge made for J's stranger parts. :-) And because + is commutative, +&/ would also be a valid solution.

Answer 2

JavaScript (ES6),  28  23 bytes

Saved 3 bytes thanks to @Mukundan314

Expects (A)(n).

A=>n=>eval(A.join`*n+`)

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How?

We simply join the input array with "*n+", so that [1,2,3] is turned into "1*n+2*n+3" and evaluate the resulting string.

Answer 3

Haskell, 17 bytes

foldr1.((+).).(*)

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It turns out this this was close to a port of the intended J solution. The pointfree function ((+).).(*) takes the argument n to the map \a b->a*n+b, that is, to add n times the left value to the right value. This creates the same "verb" as J used, and the foldr1 does the same a J's automatic right to left evaluation. It starts with the rightmost value in the list, which never gets multiplied by n, and applies it right-to-left, effectively increasing the sum so far with n times to the new element.

Answer 4

Python 3, 27 bytes

lambda a,n:a.pop()+sum(a)*n

Port of my Japt solution to python

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Answer 5

Wolfram Language (Mathematica), 19 bytes

#2Tr@Most@#+Last@#&

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Answer 6

Python 3, 27 bytes

lambda a,n:a.pop()+sum(a)*n

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Answer 7

Clojure 41 bytes

#(+(last %1)(* %2(apply +(butlast %1))))

Unfortunately, + does have to be applyed.

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Answer 8

05AB1E, 5 bytes

-2 bytes thanks to @KevinCruijssen.

*`²÷O

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Explanation

*     Multiply list by second operand
 `    Dump
   ÷  Divide the last item by
  ²   the second operand
    O Sum the stack

05AB1E, 7 bytes

„²*ý.VO

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Explanation

„       2-char string
 ²*     (Which does when evaluated) Multiply by the second input
   ý    Join the input list by this
    .V  Evaluate
      O Sum the resulting stack

Answer 9

APL (Dyalog Extended), 9 bytes (SBCS)

Anonymous tacit infix function. Takes \$A\$ as left argument and \$n\$ as right argument.

⊢/+.×+×∘~

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×∘~ \$A×(1-n)\$

+.×+ \$\big(\sum_{i=1}^N A_i×n\big)+\$

⊢/ rightmost element (lit. right-argument reduction)

So this effectively implements: $$ \Bigg(\bigg(\sum_{i=1}^N A_i×n\bigg)+A×(1-n)\Bigg)_N\\ \bigg(\sum_{i=1}^N A_i×n\bigg)+A_N×(1-n)\\ \bigg(\sum_{i=1}^N A_i×n\bigg)+A_N-n×A_N\\ \bigg(\sum_{i=1}^{N-1} A_i×n\bigg)+A_N $$

Answer 10

APL (Dyalog Unicode), 5 bytes

+⍣⎕/⎕

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A full program, which pretty much works like the 3-byte J solution. Takes two lines of input, \$A\$ first and \$n\$ second.

How it works

+⍣⎕/⎕
    ⎕  ⍝ Take the input A
   /   ⍝ Reduce by...
+      ⍝   Add the left argument
 ⍣⎕    ⍝   n times

For n=10 and A = 3 1 4 1 5, this becomes:
+⍣10/3 1 4 1 5
3 (+⍣10) 1 (+⍣10) 4 (+⍣10) 1 (+⍣10) 5
3 added 10 times to
         1 added 10 times to
                  4 added 10 times to
                           1 added 10 times to
                                    5

---

APL (Dyalog Extended), 8 bytes

1¨⍛,⊥0,⊣

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A longer but more interesting one. A tacit dyadic function that takes \$A\$ on its left and \$n\$ on the right.

Uses mixed base conversion ⊥, which does the following:

Base:        1  1  1  ... 1    n
Digit value: n  n  n  ... n    1
Array value: 0  a1 a2 ... ax-1 ax
Total: a1n + a2n + ... + ax-1n + ax

How the code works

1¨⍛,⊥0,⊣  ⍝ Input: left=A, right=n
1¨        ⍝ An array of ones as long as A
  ⍛,      ⍝ Append n, which becomes the base
     0,⊣  ⍝ A prepended with single zero, which becomes the values
    ⊥     ⍝ Mixed base conversion as described above

Answer 11

R, 37 36 35 bytes

-2 bytes with help from Giuseppe

function(l,n)rev(l)%*%n^(seq(!l)>1)

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Reverse the vector, and perform dot product with the vector \$(n^0, n^1, n^1, \ldots,n^1) = (1, n, n,\ldots, n)\$.

I just discovered this behaviour of seq, which gains 1 byte on item 4 of this tip: seq(!l) is equivalent to seq(along.with = l) (giving the vector 1 2 3 ... length(l)) in all situations, even if l is of length 1. That is because !l is a logical, not an integer, and so we avoid the call to seq.int when l is a (length 1) integer.

Answer 12

Pyramid Scheme, 407 bytes

     ^      ^
    /l\    /+\
   /oop\  ^---^
  ^-----^ -  /x\
 / \   / \   ---
/arg\ /set\
-----^-----^
    /2\   /+\
    ---  ^---^
        ^-  /1\
       ^-   ---
      ^-
     /]\
    ^---^
   / \ /2\
  /set\---
 ^-----^
/x\   ^- 
---  /]\
    ^---^ 
   ^-  /#\
  / \  ---^
 /set\   / \
^-----^ /arg\
-    /+\-----^
    ^---^   /2\
   /*\  -   ---
  ^---^
 ^-  /#\
/x\ ^---
---/ \
  /arg\
 ^-----
/1\
---

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Takes input through command arguments, with n as the first argument. This basically implements the algorithm:

i = 2
x = 0
o = 0
while args[i]:
  o += x*args[1]
  x = args[i]
  i += 1

print(o + x)

But with more nesting and some shortcuts, like using the variable 2.

Answer 13

K (oK), 14 13 bytes

Solution:

{*|x+/y*-1_x}

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Explanation:

Couldn't figure out a smart way of solving this.

{*|x+/y*-1_x} / the solution
{           } / lambda taking implicity x, y
        -1_x  / drop (_) 1 element from end of x
      y*      / multiply by y
   x+/        / sum up with x as accumulator
 *|           / take last (reverse, first)

Notes:

• -1 byte thanks to coltim - thanks!

Answer 14

Perl 5 + -pa -MList::Util+sum, 19 bytes

$_=pop(@F)+<>*sum@F

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Answer 15

Japt, 7 bytes

o +V*Ux

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Explanation

o +V*Ux
o         // Pop and return last element of first input
  +       // plus
   V*     // second input times
     Ux   // Sum of first input

Answer 16

Retina 0.8.2, 31 bytes

\d+
$*
1(?=.*,1*;(1*)|1*$)
$1
1

Try it online! Link includes test cases. Explanation:

\d+
$*

Convert to unary.

1(?=.*,1*;(1*)|1*$)
$1

Multiply all but the last element of A by n and delete A.

1

Take the sum and convert to decimal.

Answer 17

Pyth, 7 bytes

+*sPQEe

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Explanation

+*sPQEe
    Q    # First input
   P     # Remove the last element
  s      # Sum elements
 *   E   # Multiply by the second input
+     e  # Add the last element of the first input

Answer 18

x86-16 machine code, 18 bytes

33 DB       XOR  BX, BX         ; clear running sum
49          DEC  CX             ; decrement array length
74 09       JZ   ADD_LAST       ; handle array length of 1 case
        LOOP_SUM:
AD          LODSW               ; load next value into AX
03 D8       ADD  BX, AX         ; BX = BX + AX
E2 FB       LOOP LOOP_SUM       ; keep looping
93          XCHG AX, BX         ; move sum into AX
F7 E2       MUL  DX             ; DX:AX = AX * DX
93          XCHG AX, BX         ; move result back to BX
        ADD_LAST:
AD          LODSW               ; load last value into AX
03 C3       ADD  AX, BX         ; AX = AX + BX
C3          RET                 ; return to caller

As a callable function: [SI] to input array, CX array length, DX = N. Output to AX.

Rather than make an elaborate test program, here's it being run using DOS DEBUG, entering the input array into memory and setting registers as they would be called:

Explanation of above:

Enter input array into memory address DS:200 as 16-bit, little-endian words:

-e 200 3 0 1 0 4 0 1 0 5 0

Point SI to this input array:

-r SI
:200

Set CX to array's length:

-r CX
:5

Set N to 10 (0xA in hex):

-r DX
:A

Execute and stop before last instruction (RET will "return to DOS" and clobber registers):

-g 111

Result is AX=005F or 95 in decimal.

Answer 19

Golfscript, 13 bytes

~:i;-1%{i*+}*

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Explanation: ~ to convert string input to array and integer on stack. :i; assigns \$n\$ to i and pops value. -1% reverses the array and {i*+}* folds the array with (a, b) -> a*n + b

Answer 20

Emacs Lisp with dash library: 38 51 bytes

(lambda(n A)(+(car(last A))(* n(-sum(butlast A)))))

(38 bytes was the function body' size only.)

Answer 21

Pyramid Scheme, 7747 3668 1985 1732 1619 bytes

Edits

• -4079 bytes by giving the variables shorter names... 😅
• -1638 bytes by doing some more manual optimisation
• -253 bytes by using more 0-height trees
• -113 bytes because out can output the final result directly

    ^  ^        ^       ^        ^  ^        ^        ^        ^
   / \ -^      / \     / \      / \ -^      / \      / \      / \
  /set\ -^    / do\   /set\    /   \ -^    /set\    /   \    /out\
 ^-----^ -^  ^-----^ ^-----^  / set \ -^  ^-----^  /     \   -----^
/l\   / \ -^/c\   ^-/n\   ^- ^-------^ -^/i\   ^- / loop  \      /+\
---  /arg\^----  ^- ---  ^- /N\     /#\^----  ^- ^---------^    ^---^
    ^------^    / \     /-\ ---    ^----^    ^- / \       / \  /s\ /#\
   / \    ^-   ^---^   ^---^      / \  ^-   ^- /<=>\     ^---^ --- ---^
  /99 \  ^-   / \ / \ /n\ /1\    /arg\^-   ^- ^-----^   / \ / \      / \
  ----- ^-   ^---/   \--- ---   ^------^  ^- /i\   /0\ ^---/set\    /arg\
       ^-   / \  -----^        /n\    ^- /-\ ---   ---/ \ ^-----^   -----^
      ^-   ^---      / \       ---   ^- ^---^        ^---/s\   / \      /-\
     ^-   / \       /set\           ^- /n\ /1\      / \  ---  / + \    ^---^
    ^-   /set\     ^-----^         / \ --- ---     ^---      ^-----^  /n\ /1\
   ^-   ^-----^   /c\   /!\       /set\           / \       /s\   /*\ --- ---
   -^  /n\   /+\  ---  ^---      ^-----^         /set\      ---  ^---^
   / \ ---  ^---^     /=\       /s\   /0\       ^-----^         /#\ /N\
  /set\    /n\ /1\   ^---^      ---   ---      /i\   /-\       ^--- ---
 ^-----^   --- ---  / \ /l\                    ---  ^---^     / \
/n\   /0\          /arg\---                        /i\ /1\   /arg\
---   ---         ^-----                           --- ---  ^-----
                 /n\                                       /i\
                 ---                                       ---

Try it online!

Explanation

(set nil (arg 99)) // Make nil

// Count the number of input arguments - n
(set nargin 0)
(do
    cond
    (
        (set nargin (+ nargin 1))
        (set cond (! (= (arg nargin) nil)))
    )
)
(set nargin (- nargin 1))

(set N (# (arg nargin))) // N - the number all but last of the array elements is getting multiplied by

// Add all but last elements of A
(set sum 0)
(set i (- nargin 1))
(loop
    (<=> i 0)
    (
        (set i (- i 1))
        (set sum (+ sum (* (# (arg i)) N))) // A[i] multiplied by N
    )
)
(set sum (+ sum (# (arg (- nargin 1))))) // Add the last element of A
(out sum) // Print

Answer 22

Ruby, 46 19 bytes

->a,n{eval a*"*n+"}

Courtesy of petStorm.

Old answer:

n,*A,l=gets.split(' ').map(&:to_i)
p A.sum*n+l

Answer 23

MAWP, 26 bytes

%@_2A<\:.>2M3A[1A~M~]%\WM:

Now it works properly on the testcases. Works on MAWP 1.1's integer input.

Try it!

Answer 24

Jelly, 4 bytes

Ṫṭ×S

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Ṫ       Pop the last element of the left argument,
 ṭ      append it to
  ×     the right argument times what's left of the left argument,
   S    and sum.

A more fun solution, which borrows Jonathan Allan's base conversion trick:

Jelly, 5 bytes

S,¥/ḅ

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   /     Reduce the left argument by
 ,       pair right with
S ¥      the sum of left,
    ḅ    and convert from base right.

Bonus: Ä-.ịḅ’} is a whole 7 bytes, and doesn't even work if the left argument only has one element, but it's just kind of funny.

Answer 25

PHP, 41 bytes

fn($a,$n)=>array_pop($a)+array_sum($a)*$n

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Just trying to use all the built-ins!

Answer 26

Raku, 20 bytes

{@^a.pop+$^b*@a.sum}

By using twigils, @^a matches the first arg (the array), and $^b the second (the multiplier).

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Answer 27

Jelly, 5 bytes

ṪṭSƊḅ

A dyadic Link accepting a list of numbers on the left and a number on the right which yields a number.

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ṪṭSƊḅ - Link: list of numbers, A; number n
   Ɗ  - last three links as a monad - f(A):
Ṫ     -   remove the tail (of A) and yield its value
  S   -   sum (the remaining elements in A)
 ṭ    -   tack -> [sum_of_remaining, tail]
    ḅ  - convert from base (n) -> n×sum_of_remaining+1×tail

Answer 28

Q'Nial, 33 bytes (20 bytes without operator definition)

s is op n a{+link[n*front,last]a}   %full operator definition

+link[n*front,last]a                %simple function

Explanation:

+                                   sum, reduce by +
 link                               list of the items of the argument
     [                              atlas (argument of link operation), point-free notation
      n*                            n *
        front                       all elements but the last of the argument
             ,
              last                  last element of the argument
                  ]                 end atlas
                   a                array a (argument of atlas)

Intermediate results, for n=10 and a=3 1 4 1 5

         10 s 3 1 4 1 5
or
         s 10 (3 1 4 1 5)

or
         +link[10*front,last]3 1 4 1 5

         [n*front,last] a
+-------------+-+
|+--+--+--+--+|5|
||30|10|40|10|| |
|+--+--+--+--+| |
+-------------+-+

         link[n*front,last] a
+--+--+--+--+-+
|30|10|40|10|5|
+--+--+--+--+-+

         +link[n*front,last]a
95

Answer 29

Befunge-98 (PyFunge), 29 27 bytes

j&10p#v&\10g*\4
_\.@  >+\:#

Try it online! Input is first N, then A. Note that there has to be a trailing space.

Animation of the code:

^{The pilcrow (¶) represents a newline (value 10) in the grid.}

Answer 30

Julia 1.0, 21 19 bytes

A$n=pop!(A)+sum(A)n

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