# Sum the array times n, except the last

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

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

• Can we take the list in reverse?
– user92069
Commented Jul 11, 2020 at 8:20
• @Third-party'Chef' No. Commented Jul 11, 2020 at 8:26
• I wonder if your J solution used mixed base conversion
– xnor
Commented Jul 11, 2020 at 9:37
• Can we take the input as numbers instead of a single array? Commented Jul 11, 2020 at 9:54
• @HighlyRadioactive Yes, that's fine. Commented Jul 11, 2020 at 12:19

# J, 3 bytes

That was fun to find.

&+/


Try it online!

### 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.

• Perfect, you nailed it! Commented Jul 11, 2020 at 12:17

# JavaScript (ES6),  28  23 bytes

Saved 3 bytes thanks to @Mukundan314

Expects (A)(n).

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


Try it online!

### 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.

• I'm simultaneously amazed and disgusted. Good work!
– Jhal
Commented Jul 11, 2020 at 22:02

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


Try it online!

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.

# Python 3, 27 bytes

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


Port of my Japt solution to python

Try it online!

• Snap! Just wrote the exact same code without looking at any answers! Commented Jul 11, 2020 at 10:31
• +1 for 2 min earlier Commented Jul 11, 2020 at 11:29

# Wolfram Language (Mathematica), 19 bytes

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


Try it online!

# Python 3, 27 bytes

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


Try it online!

## Clojure 41 bytes

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

Unfortunately, + does have to be applyed.

Try It Online

• Welcome to the community! Why not add a Try-It-Online link for your code so users can try it out :)
– mkst
Commented Jul 14, 2020 at 8:45
• You can save 2 bytes by writing just % instead of %1, it is always recognized as the first argument, even when there are more. Commented Jul 15, 2020 at 12:42
• And another one by swapping the order of operands: tio.run/… Commented Jul 15, 2020 at 12:49

# 05AB1E, 5 bytes

-2 bytes thanks to @KevinCruijssen.

*²÷O


Try it online!

## 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


Try it online!

## 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

• Commented Jul 13, 2020 at 7:05

# APL (Dyalog Extended), 9 bytes (SBCS)

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

⊢/+.×+×∘~


Try it online!

×∘~$$\A×(1-n)\$$

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

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

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

• @Bubbler Ugh, thanks, that'll be much harder to explain. Also, it isn't really related to my solution at all. You should self-answer with that, I guess.
Commented Jul 13, 2020 at 7:55

# APL (Dyalog Unicode), 5 bytes

+⍣⎕/⎕


Try it online!

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
5


# APL (Dyalog Extended), 8 bytes

1¨⍛,⊥0,⊣


Try it online!

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

• Another interesting method could be (⎕⊥,)/⎕, but I don't think you can get rid of the brackets
– Jo King
Commented Jul 16, 2020 at 1:25

# R, 3736 35 bytes

-2 bytes with help from Giuseppe

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


Try it online!

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.

• If the list were guaranteed to be at least length 2, this would be a byte shorter, but as it is, it's longer by one instead. Commented Jul 13, 2020 at 18:09
• @Giuseppe Good old seq_along! I parlayed it into a 36-byter. Thanks! Commented Jul 14, 2020 at 8:36
• @Giuseppe seq(!l) works and is equivalent to seq(a=l), even if l is of length 1! Commented Jul 15, 2020 at 10:59
• Wow, it's been a while since I saw such a neat and applicable golfing trick! That's probably more due to my lack of participation here than anything else. Commented Jul 15, 2020 at 18:15

# Pyramid Scheme, 407 bytes

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


Try it online!

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.

# K (oK), 14 13 bytes

Solution:

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


Try it online!

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!
• You can trim a byte by doing {*|x+/y*-1_x} Commented Nov 19, 2020 at 14:04

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

$_=pop(@F)+<>*sum@F  Try it online! • Did you mean -MList::Util=sum? Commented Jul 14, 2020 at 12:12 • Interesting, I didn't know = would work there. I've always used + for places where spaces can't be used without quotes or to help precedence (e.g. /\d+\D/,say+($1)x$&). Looks like a bunch of things work though. Good to know, thanks! :) Commented Jul 14, 2020 at 12:56 # Japt, 7 bytes o +V*Ux  Try it online! ## Explanation o +V*Ux o // Pop and return last element of first input + // plus V* // second input times Ux // Sum of first input  # 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. # Pyth, 7 bytes +*sPQEe  Try it online! ## 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  # 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. # Golfscript, 13 bytes ~:i;-1%{i*+}*  Try it online! 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 • 11 bytes, if you take input in a different format – user92069 Commented Jul 29, 2020 at 13:47 # 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.) • You forgot to add the last back in. Commented Jul 15, 2020 at 16:06 • My bad, unfortunately the answer is now much longer. Commented Jul 16, 2020 at 0:55 # Pyramid Scheme, 7747366819851732 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  • Some golfing for 1042 bytes. There's more you can do though (especially with that first do loop) – Jo King Commented Aug 4, 2020 at 8:46 # 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  • 19 bytes, as an anonymous function – user96495 Commented Aug 2, 2020 at 14:11 • I should use anonymous functions more. Thanks! Commented Aug 4, 2020 at 3:43 ## 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! • The new program works now. Only took 6 days! Commented Aug 31, 2020 at 4:16 • +1 for persistence! Commented Aug 31, 2020 at 4:28 # Jelly, 4 bytes Ṫṭ×S  Try it online! 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  Try it online!  / 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. # PHP, 41 bytes fn($a,$n)=>array_pop($a)+array_sum($a)*$n


Try it online!

Just trying to use all the built-ins!

# Raku, 20 bytes

{@^a.pop+$^b*@a.sum}  By using twigils, @^a matches the first arg (the array), and $^b the second (the multiplier).

Try it online!

# 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.

Try it online!

ṪṭSƊḅ - Link: list of numbers, A; number n
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


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

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



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

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

+--+--+--+--+-+
|30|10|40|10|5|
+--+--+--+--+-+

95


# 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.

# Julia 1.0, 21 19 bytes

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


Try it online!