# Borderless table

In this challenge you are going to place letters from the alphabet in a Cartesian plane and output the result as a text.

Your input will consist in a list of list with 3 parameters:

• X coordinate
• Y coordinate
• String

How?

We know that a Cartesian plane contain 2 axes $$\(X, Y)\$$ and 4 quadrants where the signs of the $$\(X,Y)\$$ coordinates are $$\(+,+)\$$, $$\(−,+)\$$, $$\(−,−)\$$, and $$\(+,−)\$$. For example

Consider the following 3 by 3 matrix as a Cartesian plane

$$\begin{matrix} (-1,1) & (0,1) & (1,1) \\ (-1,0) & (0,0) & (1,0) \\ (-1,-1) & (0,-1) & (1,-1) \end{matrix}$$

If we are given in the input something like [[-1,1,L],[0,1,F]] our matrix will look something similar to

$$\begin{matrix} L & F & (1,1) \\ (-1,0) & (0,0) & (1,0) \\ (-1,-1) & (0,-1) & (1,-1) \end{matrix}$$

And the final output LF

In addition to that there are some points we need to follow in order to get the correct output:

• When a X,Y coord is repeated, you will need to concatenate the strings. Example: assume in (-1,1) the string F is placed and you need to place the string a in the same point. You concatenate both strings resulting in Fa and that is the value that will go in (-1,1).
• Your output need to be consistent to the matrix. Example imagine this as your final result:

$$\begin{matrix} Ma & r & ie \\ i & s & (1,0) \\ cute & (0,-1) & (1,-1) \end{matrix}$$

You must output

Ma  rie
i   s
cute


Why?

You can view this as a table where the columns are the values of the x-axis and the rows the y-axis.

        Column 1    |   Column 2    |   Column 3
----------------------------------------
Row 1   |  "Ma"     |      "r"      |     "ie"
Row 2   |  "i"      |      "s"      |
Row 3   |  "cute"   |               |


All columns values must have the same length

        Column 1    |   Column 2    |   Column 3
----------------------------------------
Row 1   |  "Ma  "   |      "r"      |     "ie"
Row 2   |  "i   "   |      "s"      |
Row 3   |  "cute"   |               |


Finnaly we output the result

Ma  rie
i   s
cute


Test Cases

Input
------------
[[3, 3, "c"]
[4, 1, "un"]
[5, 3, "e"]
[4, 3, "od"]
[4, 2, "lf"]
[1, 2, "go"]
[2, 1, "i"]
[2, 1, "s f"]]

Output
--------------
code
go     lf
is f un


Input
--------------
[[0, 0, 's'],
[-1,1, 'M'],
[0, 1, 'r'],
[-1,1, 'a'],
[1, 1, 'i'],
[-1, 0, 'i'],
[1, 1, 'e'],
[-1,- 1, 'c'],
[-1,- 1, 'u'],
[-1, -1, 'te']]

Output.
----------------
Ma  rie
i   s
cute


Notes

• This is supposed to be
• You can wrap the coordinates in a single list e.g [[3, 3], "c"]
• You can take the input in any reasonable format
• You can assume there wont be any number or empty spaces only in the input. e.g. There can be something like a a but never 1 or " " or 1a or 1 1
• sandbox Feb 5 '19 at 14:53
• @KevinCruijssen You can assume there wont be any number or empty spaces only in the input. There can be something like a a but never 1 or   or 1a or 1 1 Feb 5 '19 at 17:52
• @LuisfelipeDejesusMunoz Thanks. Oh, and one more question I'm sure more people here would want to know: who is Marie? ;p Feb 5 '19 at 18:02
• @KevinCruijssen My crush 5 years ago :c Feb 5 '19 at 18:02
• Can we take input as a list of named tuples? Something like this: (int a,int b,string c)? Feb 5 '19 at 19:23

# JavaScript (ES8),  186 180  179 bytes

Saved 1 byte thanks to @Shaggy

a=>(g=d=>a.some(([x,y,s])=>(w[x+=d]>(l=((r=o[y=d-y]=o[y]||[])[x]=[r[x]]+s).length)||(w[x]=l),x|y)<0,w=[o=[]])?g(-~d):o.map(r=>w.map((w,x)=>(r[x]||'').padEnd(w)).join).join
)


Try it online!

### Negative indices in JS (or the lack of them)

Given an array A[], it's perfectly legal in JS to do something like A[-1] = 5. However, this will not save the value in the array itself. Instead, it will implicitly coerce this negative index to a string ("-1") and set the corresponding property in the surrounding object of the array.

The bad news is that properties are not iterable with methods such as map():

a = [];
a = 3;
a[-1] = 5;
a.map((v, i) => console.log(v + ' is stored at index ' + i))


Try it online!

The above code will only display 3 is stored at index 1.

A possible workaround would be:

a = [];
a = 3;
a[-1] = 5;
Object.keys(a).map(k => console.log(a[k] + ' is stored with key ' + k))


Try it online!

But:

• This is not very golf-friendly.
• The keys are not sorted in numerical order.

### What we do here

We definitely want to work with positive values of both $$\x\$$ and $$\y\$$ in order to avoid the problems described above.

We could do a first pass on the data, looking for the minimum value of $$\x\$$ and the minimum value of $$\y\$$. But that would be quite lengthy.

• we start with $$\d=0\$$
• we process an iteration where $$\x\$$ is replaced with $$\x+d\$$ and $$\y\$$ is replaced with $$\d-y\$$
• if we have either $$\x<0\$$ or $$\y<0\$$ for any entry, we abort and recursively start another attempt with $$\d+1\$$
• I think you can save a byte by declaring o within w: w=[o=[]]. Feb 5 '19 at 17:56
• @Shaggy I think that's safe indeed. Thanks. :) Feb 5 '19 at 18:23

# APL (Dyalog Unicode), 39 bytesSBCS

Anonymous infix lambda taking* lists of coordinates and strings as left and right arguments.

{⊃,/↑¨↓⌽m⊣m[c],←⍵⊣m←(⊃⌈/c←1+⍺-⌊/⍺)⍴⊂''}


Try it online!

{} "dfn"; left (coordinates) and right (strings) arguments are ⍺ and ⍵:

⊂'' enclosed empty string, so use as fill for an array

()⍴ cyclically reshape into an array of the following dimensions:

⌊/⍺ the lowest value along each axis of the coordinates

⍺- subtract that from all the coordinates

1+ add that to one (since we want the inclusive range)

c← store in c (for coordinates)

⌈/ the highest value along each axis of those

⊃ unpack to use as dimensions

m← store in m (for matrix)

⍵⊣ discard that in favour of the strings

m[c],← append those to m at the coordinates c

m⊣ discard those in favour of the amended m

⌽ mirror

↓ split into list of lists of strings

↑¨ mix each list of strings into a character matrix, padding with spaces

,/ reduce by horizontal concatenation

⊃ unpack (since reduction reduces rank from 1 to 0)

* If taking a single argument of interwoven coordinates and strings is required, it will be 5 bytes longer.

# Python 2, 188185 181 bytes

s=sorted;d={};k={}
for x,y,c in input():d[x]=d.get(x,{});k[~y]=d[x][y]=d[x].get(y,'')+c
for y in s(k):print''.join('%*s'%(-max(map(len,d[x].values())),d[x].get(~y,''))for x in s(d))


Try it online!

# 05AB1E, 45 44 bytes

WsàŸãεUõIvyнXQiyθ«]IZsß->ôεíDéθgjí}øRJʒðKĀ}»


Takes the input-coordinates as an inner list.

Explanation:

Wsà           # Get the minimum and maximum of the (implicit) input-list
Ÿ          # Create a list in the range [min, max]
ã         # Create each possible pair by taking the cartesian product with itself
ε             # Map each coordinate to:
U            #  Pop and store the coordinate in variable X
õ            #  Push an empty string ""
Iv          #  Loop y over the input-items:
yн        #   Get the coordinates of item y
XQi     #   If it's equal to variable X:
yθ   #    Get the string of item y
«  #    Concat it to the (non-)empty string
]             # Close the if-statement, loop, and map
IZsß         # Get the maximum and minimum of the input-list
-        # Subtract them from each other
>       # And increase it by 1
ô      # Split the list into parts of this size
ε             # Map each part to:
í            #  Reverse each inner string
Déθg        #  Get the length of the longest inner string
j       #  Prepend spaces to each item up to that length
í      #  And reverse every item back again
}ø            # After the map: zip/transpose; swapping rows/columns
R           # Reverse the entire list
J          # Join each inner list together to a single string
ʒðKĀ}         # Remove all strings consisting only of spaces
»        # Join the strings by newlines (and output implicitly)


# Charcoal, 60 bytes

≔Ｅθ§ι¹η≔Ｅθ§ι⁰ζＦ…·⌊ζ⌈ζ«≔Ｅ…·⌊η⌈η⭆θ⎇∧⁼ι§μ⁰⁼κ§μ¹§μ²ωε⮌εＭ⌈ＥεＬκ±Ｌε


Try it online! Link is to verbose version of code. Explanation:

≔Ｅθ§ι¹η≔Ｅθ§ι⁰ζ


Extract the coordinates from the input.

Ｆ…·⌊ζ⌈ζ«


Loop over the x-coordinates.

≔Ｅ…·⌊η⌈η⭆θ⎇∧⁼ι§μ⁰⁼κ§μ¹§μ²ωε


Loop over the y-coordinates, extracting and concatenating all of the strings at the given coordinates.

⮌ε


Print the strings in reverse order as the y-coordinates are reversed compared to Charcoal's coordinate system.

Ｍ⌈ＥεＬκ±Ｌε


Move to the start of the next column.

# Perl 5 -p00-MList::Util=max, 148 bytes

s/(\S+) (\S+) (.*)
/$a{$1}=max$a{$1},length($h{$2}{$1}.=$3);''/ge;for$y(sort{$b-$a}keys%h){map{printf"%-$a{$_}s",$h{$y}{$_}}sort{$a-$b}keys%a;say""}


TIO

How

• s/(\S+) (\S+) (.*) / ... ;''/ge;, substitution flags /g loop /e eval, replacement evaluates to empty clearing line input/default variable
• $a{$1}=max$a{$1},length($h{$2}{$1}.=$3), autovivifies a map %h of map whose first level keys y second level x and concatenate string $3 to the value, get the length and autovivifies a second map %a whose keys x and value the max of length over the column (x) • for$y(sort{$b-$a}keys%h){ ... ;say""}, for row indices $y in keys of %h sorted numerically reverse, say"" at the end to print a newline • map{ ... }sort{$a-$b}keys%a, for column index $_ in keys %a sorted numerically
• printf"%-$a{$_}s",$h{$y}{_}, print string aligned to the left with column width # Clean, 212 206 bytes import StdEnv,StdLib ? =minList m=maxList c=flattena#(x,y,z)=unzip3 a
=flatlines(map c(transpose[[ljustify(m(map length l))k\\l<-[reverse[c[s\\(u,v,s)<-a|u==i&&v==j]\\j<-[?y..m y]]],k<-l]\\i<-[?x..m x]]))


Try it online!