Let's consider a list \$L\$ (initially empty) and a pointer \$p\$ into this list (initialized to \$0\$).

Given a pair of integers \$(m,n)\$, with \$m\ge 0\$ and \$n>0\$:

  1. We set all uninitialized values in \$L\$ up to \$p+m+n\$ (excluded) to \$0\$.
  2. We advance the pointer by adding \$m\$ to \$p\$.
  3. We create a vector \$[1,2,...,n]\$ and 'add' it to \$L\$ at the position \$p\$ updated above. More formally: \$L_{p+k} \gets L_{p+k}+k+1\$ for each \$k\$ in \$[0,..,n-1]\$.

We repeat this process with the next pair \$(m,n)\$, if any.

Your task is to take a list of pairs \$(m,n)\$ as input and to print or return the final state of \$L\$.


Input: [[0,3],[1,4],[5,2]]

  • initialization:

    p = 0, L = []
  • after [0,3]:

    p = 0, L = [0,0,0]
             + [1,2,3]
             = [1,2,3]
  • after [1,4]:

    p = 1, L = [1,2,3,0,0]
             +   [1,2,3,4]
             = [1,3,5,3,4]
  • after [5,2]:

    p = 6, L = [1,3,5,3,4,0,0,0]
             +             [1,2]
             = [1,3,5,3,4,0,1,2]


  • Instead of a list of pairs, you may take the input as a flat list \$(m_0,n_0,m_1,n_1,...)\$ or as two separated lists \$(m_0,m_1,...)\$ and \$(n_0,n_1,...)\$.

  • You may assume that the input is non-empty.

  • The output must not contain any trailing \$0\$'s. However, all intermediate or leading \$0\$'s must be included (if any).

  • This is .

Test cases




  • 2
    \$\begingroup\$ @KevinCruijssen Given that I usually try to avoid edge cases and given that another answer is already doing something special with the empty list, supporting it is now optional. \$\endgroup\$ – Arnauld Oct 25 '19 at 7:39
  • \$\begingroup\$ Thanks, that saves me a bit of trouble fixing it, since my program was outputting 0 due to the sum at the end. \$\endgroup\$ – Kevin Cruijssen Oct 25 '19 at 7:40

18 Answers 18


J, 27 bytes


Try it online!

Takes two separate lists for m an n - the list for n is the left argument of the function, the list for m - the right one.

K (ngn/k), 42 bytes


Try it online!

Takes two separate lists for m an n

It's too long currently, I'll try to golf it.

|improve this answer|||||
  • 1
    \$\begingroup\$ I tried solving this independently and came up with something very similar, but 25 bytes. Kind of fun 3 nested dyadic hooks +/@(((,~#&0)~1+i.)"0~+/\): Try it online! \$\endgroup\$ – Jonah Oct 26 '19 at 3:42
  • \$\begingroup\$ @Jonah Great! I don't mind if you post it separately. \$\endgroup\$ – Galen Ivanov Oct 26 '19 at 4:08

Jelly,  10  9 bytes


A dyadic Link accepting a list of integers on each side, \$M\$ on the left \$N\$ on the right, which yields a list of integers.

Try it online! Or see the test-suite.


Ä0ẋżR}F€S - Link: list of integers, M; list of integers, N
Ä         - cumulative sums (M) = [m1, m1+m2, m1+m2+m3, ...]
 0ẋ       - zero repeated = [[0]*m1,[0]*(m1+m2),[0]*(m1+m2+m3), ...]
    R}    - range (right=N) = [[[1,2,3,...,n1]],[[1,2,3,...n2]],[[1,2,3,...,n3]], ...]
   ż      - zip together = [[[0]*m1,[[1,2,3,...,n1]]],[[0]*(m1+m2),[[1,2,3,...,n2]]],[[0]*(m1+m2+m3),[[1,2,3,...,n3]]], ...]
      F€  - flatten each = [[0,0,...,0,1,2,3,...,n1],[0,0,...,0,1,2,3,...,n2],[0,0,...,0,1,2,3,...,n3], ...]
        S - sum
|improve this answer|||||
  • \$\begingroup\$ Nice. I’d missed the fact that two lists were acceptable input. \$\endgroup\$ – Nick Kennedy Oct 25 '19 at 18:28

Jelly, 12 11 bytes


Try it online!

A full program that takes a list of lists of integers as its argument and returns a list of integers. Can be adapted to work as a link by resetting the register to zero after each call (as implemented in the footer on TIO).

Saved a byte now list can be non-empty.

|improve this answer|||||

Java 8, 148 147 bytes

N->M->{int l=M.length,s=M[l-1],p=0,L[],i=0,j;for(int n:N)s+=n;for(L=new int[s];i<l;i++)for(p+=N[i],j=s;j-->0;)L[j]-=j<p|j>=p+M[i]?0:~j+p;return L;}

-1 byte thanks to @ceilingcat.

Takes both integer-arrays as separated inputs.

Try it online.


N->M->{            // Method with integer-array as two parameters as well as return-type
  int l=M.length,  //  Length of the input-arrays
      s=           //  Length of the output-array,
        M[l-1],    //  starting at the last pointer-item of array `M`
      p=0,         //  Position `p` as specified in the challenge, starting at 0
      L[],         //  Output list, starting uninitialized
      i=0,j;       //  Index integers
  for(int n:N)     //  Loop over the value-list `N`:
    s+=n;          //   And add all of them to the output-length
  for(L=new int[s];//  Now initialize the output-list of size `s`, filled with 0s by default
      i<l;i++)     //  Loop `i` in the range [0, `l`):
    for(p+=N[i],   //   Increase position `p` by the `i`'th pointer of `N`
        j=s;j-->0;)//   Inner loop `j` in the range (`s`,0]:
     L[j]-=        //    Decrease the `j`'th item of the output-list by:
       j<p         //     If `j` is smaller than position `p`
       |j>=        //     or `j` is larger than or equal to
        p+M[i]?    //     pointer `p` and the `i`'th value of `M` combined
         0         //      Leave the `j`'th item of output-array the same by adding 0
       :           //     Else:
        ~j+p;      //      Decrease the `j`'th item of the output-array by `-j-1+p`
                   //      (increase it by the difference between `j` and `p`, plus 1)
  return L;}       //  And after the nested loops: return the resulting array
|improve this answer|||||

Zsh, 59 56 54 bytes

-3 bytes by changing to my Bash strategy, -2 bytes by switching back to my original strategy, now that the rules specify the list is non-empty.

for m n;for ((i=0,p+=m;i<n;a[p+i]+=++i)):

Try it online: [Original strat handling empty case] [Adapted Bash strat] [Current]

Setting a[5]=1 causes a[1] through a[4] to be initialized empty. The final parameter expansion replaces all empty elements with 0.

for m n                                 # implicit 'in "$@"'
    for ((i=0, p+=m; i<n; a[p+i]+=++i)) # increment p by m, add to a[p,p+n-1]
        :                               # ':' is a no-op builtin
<<<${a/#%/0}                            # the glob #% matches empty elements
|improve this answer|||||

Haskell, 90 87 bytes

(foldl(%)[].).zipWith(\o p->(0<$[1..o])++[1..p]).scanl1(+)

Takes the input as two separate lists [m0,m1,...] and [n0,n1,...].

Try it online!

A variant (function % same as above), also 87 bytes:

|improve this answer|||||

APL (Dyalog Unicode), 19 bytes


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Uses the algorithm described in my J answer.

How it works

1⊥0⌈(↑(⍳+)¨-⊢)∘(+\)  left = n's, right = m's
              ∘(+\)  m2 = cumulative sum of m's
      (⍳+)¨          Nested array of 1..(each element of n+m2)
           -⊢        Subtract each elem of m2 from each row
    (↑       )       Promote nested array to matrix
  0⌈                 Take max with 0 (change negatives to 0)
1⊥                   Sum in column direction

APL (Dyalog Unicode), 20 bytes


Try it online!

A dyadic dfn whose left argument is the ns and right argument is ms.

How it works

{                  }  ⍺ = n's, ⍵ = m's
                +\⍵   Cumulative sum of offsets (M)
    ⍺{        }¨      Zip with n's
      (⍵⍴0),⍳⍺        ... to create M zeros followed by 1..n
   ↑                  Promote to matrix, filling zeros where necessary
 1⊥                   Sum column-wise
|improve this answer|||||
  • \$\begingroup\$ Hey Bubbler can you improve this J solution +/@(((,~#&0)~1+i.)"0~+/\)? Try it online! \$\endgroup\$ – Jonah Oct 26 '19 at 3:45
  • \$\begingroup\$ @Jonah I couldn't shorten that one specifically, but I could come up with a shorter solution using a different approach. \$\endgroup\$ – Bubbler Oct 30 '19 at 7:46
  • \$\begingroup\$ full program: 1⊥0⌈↑a-⍨⍳¨⎕+a←+\⎕ \$\endgroup\$ – ngn Oct 31 '19 at 23:05
  • \$\begingroup\$ 1⊥↑⎕{+\⍸⍵⍺}¨+\⎕ in some future version of dyalog, with ⎕io←0 \$\endgroup\$ – ngn Oct 31 '19 at 23:08

05AB1E, 13 bytes


Takes two separated lists. The list of positions \$(m_0, m_1, m_2, ...)\$ as first input and list of values \$(n_0, n_1, n_2, ...)\$ as second input.

Try it online or verify all test cases.


η        # Get all prefixes of the (implicit) input-list of positions
         #  i.e. [0,1,5] → [[0],[0,1],[0,1,5]]
 O       # Sum each inner list
         #  → [0,1,6]
  Å0     # Convert each of the inner values into a list of 0s of that length
         #  → [[],[0],[0,0,0,0,0,0]]
s        # Get the second (implicit) input-list of values
 L       # And create a list in the range [1,n] of each value
         #  i.e. [3,4,2] → [[1,2,3],[1,2,3,4],[1,2]]
‚        # Pair the two lists together
         #  → [[[],[0],[0,0,0,0,0,0]],[[1,2,3],[1,2,3,4],[1,2]]]
 ø       # Zip/transpose, swapping rows/columns
         #  → [[[],[1,2,3]],[[0],[1,2,3,4]],[[0,0,0,0,0,0],[1,2]]]
  €˜     # Flatten each inner pair to a single inner list
         #  → [[1,2,3],[0,1,2,3,4],[0,0,0,0,0,0,1,2]]
    0ζ   # Zip/transpose, swapping rows/column, with 0 as trailing filler
         #  → [[1,0,0],[2,1,0],[3,2,0],[0,3,0],[0,4,0],[0,0,0],[0,0,1],[0,0,2]]
      O  # Take the sum of each inner list
         #  → [1,3,5,3,4,0,1,2]
         # (after which the result is output implicitly)
|improve this answer|||||

PHP, 93 bytes


Try it online!

Input is a flat list of m0,n0,m1,n1,... passed by command arguments ($argv) and output is string representation of L.

|improve this answer|||||

Python 3, 144 119 118 111 110 106 bytes

def f(m,n):
 for i,j in zip(n,m):p+=j;l+=[0]*(p+i-len(l));exec("l[~k]+=i-k;k+=1;"*i)
 return l

Try it online!

Thanks to:
-@mypetlion for saving me 25 bytes

|improve this answer|||||

Wolfram Language (Mathematica), 62 60 56 53 50 bytes


Try it online!

-3 with guarantee that input is non-empty.

Takes a list of pairs as the argument.

|improve this answer|||||

Ruby, 92 88 84 bytes


Input is a list of pairs, in the form [[m1, n1], [m2, m2], ...].

The approach is pretty much the algorithm as described.

Golfy tricks I've used include:

  • l[i]||=0 setting the value for L to 0 if it isn't already set
  • That's pretty much it.

Thanks to Value Ink for

Try it online!

Ruby can be pretty terse. When the version with numbered block parameters comes out, this can be reduced by a few bytes to this (note the @1 ins:

|improve this answer|||||
  • \$\begingroup\$ Why are you ending your map block with returning l and then taking [-1], when you can just do ;l after the map? Like so. \$\endgroup\$ – Value Ink Oct 26 '19 at 23:46
  • \$\begingroup\$ Two more things I thought up: n,*l=0 will initiate l to the empty array for -2 bytes, and you can wrap the n+=a into your upto call like so: 0.upto(b-1+n+=a) \$\endgroup\$ – Value Ink Oct 26 '19 at 23:51

J, 20 bytes


Try it online!

I guess this is quite different from both Galen Ivanov's and Jonah's answers. A dyadic train that takes ns as left argument and ms as right.

One trick was to avoid (...)"0 (apply to each item) in favor of ...@+. The conjunction u@v has the effect of u@:v"v. The rank-forcing effect is usually not desirable when v is an arithmetic verb, but it works perfectly here.

How it works

+/@(0>.>:@i.@+-])+/\  Left(n): range generators ex) 3 4 2
                      Right(m): offsets ex) 0 1 5
                 +/\  cm: Cumulative sum of offsets ex) 0 1 6
             +        Add n and cm element-wise ex) 3 5 8
          i.@         Generate 0..x-1 for each x above
                      ex) 0 1 2;0 1 2 3 4;0 1 2 3 4 5 6 7
       >:@            Increment each value
                      ex) 1 2 3;1 2 3 4 5;1 2 3 4 5 6 7 8
                      (Implicit) Form a 2D array, padding with 0s where needed
              -]      Subtract each item of cm from each row of above
                      ex)  1  2  3  0  0  0  0  0
                           0  1  2  3  4 -1 -1 -1
                          -5 -4 -3 -2 -1  0  1  2
    0>.               Max with 0; Change negative numbers to 0s
+/@(            )     Take sum in column direction
|improve this answer|||||
  • \$\begingroup\$ very nice @Bubbler \$\endgroup\$ – Jonah Oct 30 '19 at 11:51

Bash, 77 bytes

for N;{
for((i=0,p+=b;i<p+b[1];a[i]+=++i>p?i-p:0)){ :;}
echo ${a[@]}

Try it online!

I couldn't port my first Zsh answer directly, since ${a[@]/#%/0} doesn't work in Bash. So instead of fixing the empty elements at the end, I set all the elements with a[i]+=0 along the way. In the end, this strategy works out better for Zsh anyway!

|improve this answer|||||

C, 204 196 195 192 185 bytes

*c(m,n,q,x,p,z)int*m,*n;{int i,j,*a=malloc(4);for(i=p=z=0;m[i]+1;i++){q=n[i]+(p+=m[i]);a=realloc(a,8*((x=z)>q?z:(z=q)));bzero(a+x,(z-x)*8);for(j=p;j<q;j++)a[j]+=j-p+1;}a[z]--;return a;}

De-golfed version:

* count (m, n, size_of_array, pointer_into_array, previous_size_of_array, final_element_counted_to) //all but m and n are implicitly int - return type of function is implicitly int pointer
     int* m, * n;
  pointer_into_array = size_of_array = 0;
  int* array = malloc(sizeof(int)); //get a pointer so we can realloc later
  for (int i = 0; m[i] + 1; i++) { //iterate through m and n, stopping at -1 sentinel value
    final_element_counted_to = n[i] + (pointer_into_array += m[i]); //update pointer, get final element counted to on this pair
    array = realloc(array, sizeof(int) * 2 * ((previous_size_of_array = size_of_array) > final_element_counted_to ? size_of_array : (size_of_array = final_element_counted_to) + 1)); //reallocate array to size of max(size_of_array, final_element_counted_to+1) * sizeof(int) * 2 [multiplying by 2 for the final statement of a[z] = -1]
    bzero(array + previous_size_of_array, (size_of_array - previous_size_of_array) * sizeof(int) * 2); //initialises new memory to zero
    for(int j = pointer_into_array; j < final_element_counted_to; j++) //do the actual counting + adding
      array[j] += j - pointer_into_array + 1;
  array[size_of_array]--; //make sure it's terminated by -1
  return array;

Try it online!

Function c takes two arguments (rest are for free declaration) m and n, both -1 terminated arrays corresponding respectively to the two separate lists for m and n. It returns a -1 terminated array.

Please note that this requires sizeof(int) to be 4, and also requires the non-standard and deprecated but widely-implemented function bzero.

|improve this answer|||||
  • \$\begingroup\$ Welcome to Code Golf! You can use the header and footer sections of TIO to store boilerplate code: like this. They do not count towards your byte score. \$\endgroup\$ – Arnauld Oct 26 '19 at 17:39
  • \$\begingroup\$ And it seems like you current score is actually 204. \$\endgroup\$ – Arnauld Oct 26 '19 at 17:39
  • \$\begingroup\$ 171 bytes \$\endgroup\$ – ceilingcat Oct 30 '19 at 8:00

K (ngn/k), 30 bytes


Try it online!

|improve this answer|||||
  • 1
    \$\begingroup\$ i tried to beat this with "amend" but i couldn't. my best is: {@[&0|/y+j;i+j:+\x;+;1+i:!'y]} \$\endgroup\$ – ngn Oct 31 '19 at 23:39

Clojure, 170 Bytes

(fn [a](reduce #(mapv +(concat %1(repeat 0))%2)(mapv (fn [[x y]](concat(repeat x 0)(range 1(+ 1 y))))(reduce #(conj %1(mapv +[(nth(last %1)0)0]%2))[(first a)](rest a)))))

Try it online!

|improve this answer|||||
  • \$\begingroup\$ Hello and welcome to PPCG! \$\endgroup\$ – Jonathan Frech Oct 30 '19 at 19:20

Java, 642 Bytes

This I think is kind of awful, but doing it with Java who knows what I was expecting. If someone could help me make this better I would really appreciate that, a bunch of bytes are taken up dealing with the fact that combiner can get two different length lists and need to add them, and also with the fact that there is a type needed for the BiConsumer lambda. I think there are a few ways, but it's getting late...

Input is n, expected to be an array of arrays.

    IntStream.range(0, n.length).mapToObj((x) -> new int[] { n[x][0] + (x > 0 ? n[x - 1][0] : 0), n[x][1]})
            .map((v) -> IntStream.range(0, v[0] + v[1]).map((x) -> x >= v[0] ? 1 + x - v[0] : 0).toArray())
                () -> new ArrayList<>(),
                (BiConsumer<List<Integer>, int[]>) (l, v) -> IntStream.range(0, Math.max(l.size(), v.length)).forEach((i) -> {
                    if (i >= l.size()) l.add(0);
                    if (i < v.length) l.set(i, l.get(i) + v[i]);
                (a, b) -> {
                    while (a.size() > b.size()) b.add(0);
                    while (b.size() > a.size()) a.add(0);
                    return IntStream.range(0, a.size()).mapToObj((i) -> a.get(i) + b.get(i)).collect(Collectors.toList());

Improved and golf'd by Kevin Cruijssen:

n->IntStream.range(0,n.length).mapToObj(x->new int[]{n[x][0]+(x>0?n[x-1][0]:0),n[x][1]}).map(v->IntStream.range(0,v[0]+v[1]).map(x->x<v[0]?0:1+x-v[0]).toArray()).collect(Collector.of(()->new Stack<>(),(java.util.function.BiConsumer<List<Integer>,int[]>)(l,v)->IntStream.range(0,Math.max(l.size(),v.length)).forEach(i->{if(i>=l.size())l.add(0);if(i<v.length)l.set(i,l.get(i)+v[i]);}),(a,b)->{while(a.size()>b.size())b.add(0);while(b.size()>a.size())a.add(0);return IntStream.range(0,a.size()).mapToObj(i->a.get(i)+b.get(i)).collect(Collectors.toList());},Collector.Characteristics.IDENTITY_FINISH))
|improve this answer|||||
  • \$\begingroup\$ How does this receive input? And surely you can golf away pretty much all the whitespace at least \$\endgroup\$ – Jo King Oct 25 '19 at 5:32
  • \$\begingroup\$ I'm not sure how you ended up with 581 bytes, because your current code above is 842 bytes and still missing required imports and n-> as input. Removing all whitespaces; parenthesis around the (L)-> everywhere; changing ArrayList to Stack; reversing some checks so we could use a<b?0:1 instead of a>=b?1:0; and some other basic golfs, I end up at 642 bytes. I'm sure this can be a lot shorter without the stream builtins, though. \$\endgroup\$ – Kevin Cruijssen Oct 25 '19 at 9:56
  • 2
    \$\begingroup\$ Anyway, welcome to CGCC! If you haven't seen them yet tips for golfing in Java and tips for golfing in <all languages> might both be interesting to read through. Enjoy your stay! :) \$\endgroup\$ – Kevin Cruijssen Oct 25 '19 at 9:57
  • \$\begingroup\$ To get 581 bytes I minified with codebeautify.org/javaviewer and pasted into mothereff.in/byte-counter . Should I be pasting the minified version into the code area when posting here? \$\endgroup\$ – Disco Mike Oct 25 '19 at 16:42
  • \$\begingroup\$ I realize now that I excluded (n) -> though \$\endgroup\$ – Disco Mike Oct 25 '19 at 16:43

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