# Sum the array times n, except the last

Code Golf Asked on October 27, 2021

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


# TI-BASIC, 30 bytes (on-calc) / 37 bytes (as text)

Assumes list variable L₁ contains the array and variable N contains n.

dim(L₁
Nsum(L₁,1,Ans-1)+L₁(Ans


## Explanation

• dim(L₁: Get the size of L₁, to be used in the next line as Ans.
• Nsum(L₁,1,Ans-1)+L₁(Ans: The sum( function has optional arguments to specify the beginning and end of a region of the list to be summed. The sum is implicitly multiplied by N, then added to the last element of the list, and implicitly printed.

Closing parentheses can be left out at the end of lines in TI-BASIC.

Answered by orangey on October 27, 2021

# Risky, 9 bytes

-_{1+0+00_?+0+?*?


Try it online!

Answered by xigoi on October 27, 2021

# Vyxals, 5 bytes

*ṫ⁰/J


Try it Online!

Answered by wasif on October 27, 2021

# Factor + math.unicode, 23 bytes

[ dup pop -rot Σ * + ]


Try it online!

Answered by chunes on October 27, 2021

# Stax, 6 4 bytes

╡÷E@


Run and debug it

Answered by Razetime on October 27, 2021

# 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.

Answered by Unrelated String on October 27, 2021

# PowerShell Core, 29 bytes

param($a,$b)$a-join"*$b+"|iex


Try it online!

^ inspired by Arnauld's solution

My original one

# PowerShell Core, 75 bytes

param($a,$b)(($a[($l=$a.Length-1)]+$b*($a[0..($l-1)]-join'+'|iex)),$a)[!$l]


Try it online!

Answered by Julian on October 27, 2021

# Julia 1.0, 21 19 bytes

A$n=pop!(A)+sum(A)n  Try it online! Answered by MarcMush on October 27, 2021 # Japt-h, 6 bytes Ç+Ux*V  Try it Answered by Shaggy on October 27, 2021 ## Pip, 12 bytes b*$+@<a+@RVa


### Explanation

b*$+@<a+@RVa b* Second argument(b) times$+         Sum of elements of
@<a      First argument list(a) without last element
+@RVa Plus first element of the reverse of a


Try it online!

Answered by Razetime on October 27, 2021

## 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!

Answered by Razetime on October 27, 2021

# 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.

Answered by Jo King on October 27, 2021

# Perl 6, 29 bytes

->$a,$b {$a.pop+([+]$a)*$b};  Try it online! Answered by sugarfi on October 27, 2021 # 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  Answered by Razetime on October 27, 2021 # Rust, 45 bytes |a,n|a.pop().unwrap()+a.iter().sum::<i64>()*n  Try it online The expected argument types are a: &mut Vec<i64> and n: i64. Answered by TehPers on October 27, 2021 # C (gcc), 53 bytes Saved 2 bytes thanks to ceilingcat s;f(a,n)int*a,n;{for(s=0;a[1];s+=*a++);return*a+s*n;}  Try it online! Answered by kaadmy on October 27, 2021 # MATL, 9 bytes *1G0)0Z(s  Try it online! ## Explanation *1G0)0Z(s * % Multiply the inputs and push it onto the stack 1G % Push first input onto the stack 0) % Replace top stack element with its last element 0Z( % Set last element of multiplied inputs to top element in stack s % Replace top stack element with its sum  Answered by Mukundan314 on October 27, 2021 # 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  Answered by MarcinKonowalczyk on October 27, 2021 # Scala, 22 bytes a=>a.last+_*a.init.sum  This returns a List[Int] => Int => Int Try it online Answered by user on October 27, 2021 # Lua, 58 bytes r,t=0,{...}for i=2,#t-1 do r=r+t[i]end print(r*t[1]+t[#t])  Try it online! Having almost no array functions, Lua doesn't do too well in this challenge. Answered by val is still with Monica on October 27, 2021 ## excel, 25 bytes Enter A followed by n in one column, followed by this formula =SUM(R1:R[-3])*R[-1]+R[-2] Answered by Conor James Thomas Warford Hen on October 27, 2021 # Raku, 21 bytes {*.join("*$_+").EVAL}


Try it online!

This works the same as the Perl 5 answer and also works with non-mutable lists, unlike the existing Raku answer, which is only one byte smaller than this.

Note: My previous answer of the same size does not work for singleton lists, which I've lodged an issue about.

Answered by Jo King on October 27, 2021

# 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

Answered by psIQos on October 27, 2021

## dc, 21 bytes

ss0[+z2<o]dso0rx*ls+p

Examples:

dc -e '10 3 1 4 1 5 ss0[+z2<o]dso0rx*ls+p'
95
dc -e '1 3 1 4 1 5 ss0[+z2<o]dso0rx*ls+p'
14
dc -e '999 1 ss0[+z2<o]dso0rx*ls+p'
1


Answered by Sandra on October 27, 2021

# 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


Answered by M L on October 27, 2021

# Perl 5 + -pi, 19 bytes

s/,/*$^I+/g;$_=eval


For example, for multiplicand 10, perl -pi10 -e's/,/*$^I+/g;$_=eval

Try it online. Idea stolen from this J answer. Accuracy improved by Dom Hastings in comments on this answer (many thanks!).

Answered by msh210 on October 27, 2021

## Clojure 41 bytes

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

Unfortunately, + does have to be applyed.

Try It Online

Answered by M Smith on October 27, 2021

# 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!

Answered by mkst on October 27, 2021

# R, 3736 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.

Answered by Robin Ryder on October 27, 2021

# 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


Answered by Jonathan Allan on October 27, 2021

# SimpleTemplate, 81 bytes

It's fairly chunky :/
Mostly due to limitations in my language...

{@eachargv.0}{@ifloop.last is equal0}{@set*_ _,argv.1}{@/}{@incby_ R}{@/}{@echoR}


Takes a string of numbers or an array as the first argument, and the n as the 2nd argument.

Ungolfed:

The code above is a little weird to read...

{@set result 0}
{@each argv.0 as number}
{@if loop.last is equal to false}
{@set* number number, argv.1}
{@/}
{@inc by number result}
{@/}
{@echo result}


Notes:

• {@set result 0} is optional, as {@inc} will create the variable, if it doesn't exist
• {@each argv.0 as number} doesn't need need the as number, defaulting to the variable _ to store values
• {@if loop.last is equal to false} uses the special variable loop to know if it is the last item in the {@each} Also, false is the same as 0.
• {@set* number number, argv.1} multiplies number with argv.1 and stores it inside number. If number was an array, it would multiply with all the values.
• {@inc by number result} will increment the variable result by number (effectivelly, result = result + number)
• {@/} is usually optional, but, with this code, they are all required.

Answered by Ismael Miguel on October 27, 2021

# T-SQL, 40 bytes

I am using a table instead of an array, sql doesn't have arrays

The test uses a temporary table instead of a real table, because of lack of permissions to create a table.

SELECT sum(a*@-i/@@rowcount*a*~-@)FROM t


Try it online

Answered by t-clausen.dk on October 27, 2021

# 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


Answered by Bubbler on October 27, 2021

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

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

⊢/+.×+×∘~


Try it online!

×∘~$$A×(1-n)$$

+.×+$$big(sum_{i=1}^N A_i×nbig)+$$

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

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

# 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!

Answered by user0721090601 on October 27, 2021

# 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:
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
03 C3       ADD  AX, BX         ; AX = AX + BX


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.

Answered by 640KB on October 27, 2021

# Zsh-P, 24 bytes

a=(0 *<&0+$@) <<<$[a]


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Alternate solution using the -P flag, which enables RC_EXPAND_PARAM to do the same thing.

Answered by GammaFunction on October 27, 2021

# Wolfram Language (Mathematica), 16 bytes

a___~f~b_=+a#+b&


Try it online!

This one doesn't need a strange-looking input.

Takes input as f[A][n].

older, 17 bytes

#//.a_@b_:>a#2+b&


Try it online!

Takes a singly linked list as input: if X is a value and L is a list, X is the single-element list with element X, and X[L] is the list with element X, followed by the elements of L. For example, 3@1@4@1@5=3[1[4[1[5]]]] is the list with 3, 1, 4, 1, 5.

Per this discussion, we can take the first element with Head, and the tail with Last.

Answered by att on October 27, 2021

# 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


Answered by Mukundan314 on October 27, 2021

# 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.

Answered by Neil on October 27, 2021

# Charcoal, 10 9 bytes

Ｉ⁺⊟θ×η↨θ¹


Try it online! Link is to verbose version of code. Presumably Charcoal has the same problem as IO when it comes to summing empty lists. Edit: saved 1 byte by converting from base 1 instead. Explanation:

   θ        A
⊟         Remove last element
θ    Remaining elements of A
↨ ¹   Take the sum by converting from base 1
×       Multiplied by
η      n
Ｉ           Cast to string
Implicitly print


Answered by Neil on October 27, 2021

# Python 3, 27 bytes

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


Try it online!

Answered by Noodle9 on October 27, 2021

# Python 3, 27 bytes

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


Port of my Japt solution to python

Try it online!

Answered by Mukundan314 on October 27, 2021

# 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.

Answered by xash on October 27, 2021

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.

Answered by xnor on October 27, 2021

# 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.

Answered by Arnauld on October 27, 2021

# Wolfram Language (Mathematica), 19 bytes

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


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Answered by ZaMoC on October 27, 2021

# 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


Answered by Mukundan314 on October 27, 2021

# 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


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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


Answered by user92069 on October 27, 2021

# Io, 31 bytes

Saves a byte over the 32 byte solution. But it's boring.

method(a,N,a sum*N-a pop*(N-1))


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## Explanation

method(a, N,
a sum * N         // Sum the input, multiply by N
- a pop * (N - 1)   // Minus the last item, multiplied by N - 1
)


# Io, 32 bytes

Just as comparison

method(a,N,a push(a pop/N)sum*N)


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## Explanation

method(a, N,
a pop           // Extract the last item from the input list
/N         // Divide it by N
a push(       )        // And then put it back into the original list
sum     // Sum the list
*N   // Multiply the value by N
)


Answered by user92069 on October 27, 2021

# 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.

Answered by ovs on October 27, 2021

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

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


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Answered by Dom Hastings on October 27, 2021