Minimally sort a list into a matrix

Question

Given an unsorted list of unique strictly positive integers, minimally sort it into a 2D matrix. The input list is guaranteed to be of composite length, which means the output matrix is not necessarily square, but is of size n x m with n,m > 1.

"Minimally sort" here means the following:

• Sort the list in ascending order.
• Compact the output matrix as much as possible -- minimize the sum of the dimensions of the matrix (for example, for 20 input elements as input, a 5x4 or 4x5 output matrix is required, and not a 2x10).
• Compact the sorted numbers as far to the upper-left of the matrix as possible, starting with the first element in the sorted list.
• This can be thought of as sorting the list, then slicing it along the matrix's anti-diagonals, starting with the upper-left.

Examples:

For input 1..20 output is either a 5x4 or a 4x5 matrix as follows:

 1  2  4  7 11
 3  5  8 12 15
 6  9 13 16 18
10 14 17 19 20

 1  2  4  7
 3  5  8 11
 6  9 12 15
10 13 16 18
14 17 19 20

For input [3, 5, 12, 9, 6, 11] output is a 2x3 or 3x2 as follows

3  5  9
6 11 12

 3  5
 6  9
11 12

For input [14, 20, 200, 33, 12, 1, 7, 99, 58], output is a 3x3 as follows

 1   7  14
12  20  58
33  99 200

For input 1..10 the output should be a 2x5 or 5x2 as follows

1 2 4 6  8
3 5 7 9 10

1  2
3  4
5  6
7  8
9 10

For input [5, 9, 33, 65, 12, 7, 80, 42, 48, 30, 11, 57, 69, 92, 91] output is a 5x3 or 3x5 as follows

 5  7 11 33 57
 9 12 42 65 80
30 48 69 91 92

 5  7 11
 9 12 33
30 42 57
48 65 80
69 91 92

Rules

• The input can be assumed to fit in your language's native integer type.
• The input and output can be given by any convenient method.
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

Jelly, 24 22 20 bytes

pS€ỤỤs
LÆDżṚ$SÞḢç/ịṢ

Try it online!

Saved 2 bytes thanks to @Jonathan Allan.

Explanation

pS€ỤỤs  Helper link. Input: integer a (LHS), integer b (RHS)
p       Cartesian product between [1, 2, ..., a] and [1, 2, ..., b]
 S€     Sum each pair
   Ụ    Grade up
    Ụ   Grade up again (Obtains the rank)
     s  Split into slices of length b

LÆDżṚ$SÞḢç/ịṢ  Main link. Input: list A
L              Length
 ÆD            Divisors
     $         Monadic pair
    Ṛ            Reverse
   ż             Interleave
                 Now contains all pairs [a, b] where a*b = len(A)
      SÞ       Sort by sum
        Ḣ      Head (Select the pair with smallest sum)
         ç/    Call helper link
            Ṣ  Sort A
           ị   Index into sorted(A)

Answer 2

Python 2, 160 158 153 151 bytes

^{-2 bytes thanks to Erik the Outgolfer
-2 bytes thanks to Mr. Xcoder}

s=sorted(input())
l=len(s)
x=int(l**.5)
while l%x:x+=1
n=1
o=eval(`l/x*[[]]`)
while s:
 for i in range(l/x)[max(0,n-x):n]:o[i]+=s.pop(0),
 n+=1
print o

Try it online! or Try all test cases

Answer 3

R 110 95 bytes

function(x){n=sum(x|1)
X=matrix(x,max(which(!n%%1:n^.5)))
X[order(col(X)+row(X))]=sort(x)
t(X)}

Try it online!

How it works

f <- function(x) {
  n <- sum(x|1)                           # length
  p <- max(which(!n%%1:n^.5))             # height of matrix
  X <- matrix(x, p)                       # initialize matrix
  X[order(col(X) + row(X))] <- sort(x)    # filling the matrix using position distance to the top left corner
  t(X)                                    # probably required by OP
}

Giuseppe saved a whopping 15(!) bytes by the following tricks

• replacing length(x) by sum(x|1) (-1 byte)
• floor() is not required as : rounds down anyway (-7)
• ^.5 is shorter than sqrt() (-3)
• using col(X) + row(X) instead of outer (nice!)
• could not get rid of the t(X) though - disappointing ;)

Original solution

function(x){
n=length(x)
p=max(which(!n%%1:floor(sqrt(n))))
X=outer(1:p,1:(n/p),`+`)
X[order(X)]=sort(x)
t(X)}

It would look more fancy with outer being replaced by row(X)+col(X), but that would require to initialize the output matrix X first.

Try it online!

Answer 4

JavaScript (ES6), 172 bytes

l=>(n=l.sort((a,b)=>b-a).length,w=l.findIndex((_,i)=>!(i*i<n|n%i)),a=l=>[...Array(l)],r=a(n/w).map(_=>a(w)),a(w+n/w).map((_,x)=>r.map((s,y)=>x-y in s&&(s[x-y]=l.pop()))),r)

Explanation

l=>(                                // Take a list l as input
 l.sort((a,b)=>b-a),                // Sort it
 n=l.length,                        // Get the length n
 w=l.findIndex((_,i)=>!(i*i<n|n%i)),// Find the first integer w where w >= √n and n % w = 0
 a=l=>[...Array(l)],                // Helper function a
 r=a(n/w).map(_=>a(w)),             // Create the grid r of size w, n/w
 a(w+n/w).map((_,x)=>               // For every x from 0 to w + n/w:
  r.map((s,y)=>                     //  For every row s in r:
   x-y in s&&(                      //   If the index x-y is in s:
    s[x-y]=l.pop()))),              //    Set s[x-y] to the next element of l
 r)                                 // Return r

Test Cases

f=l=>(n=l.sort((a,b)=>b-a).length,w=l.findIndex((_,i)=>!(i*i<n|n%i)),a=l=>[...Array(l)],r=a(n/w).map(_=>a(w)),a(w+n/w).map((_,x)=>r.map((s,y)=>x-y in s&&(s[x-y]=l.pop()))),r)

l=m=>console.log(JSON.stringify(m))

l(f([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]))
l(f([3,5,12,9,6,11]))
l(f([14,20,200,33,12,1,7,99,58]))
l(f([1,2,3,4,5,6,7,8,9,10]))

Answer 5

Octave, 151 bytes

function f(v)n=floor(sqrt(l=nnz(v)));while i=mod(l,n);++n;end;A=nan(m=l/n,n);for k=[1:m 2*m:m:l];do A(k)=sort(v)(++i);until~mod(k+=m-1,m)|k>l;end;A'end

Using three different kinds of loop constructs.

Try it online!

Unrolled:

function f(v)
    n = floor(sqrt(l=nnz(v)));

    while i = mod(l,n);
        ++n;
    end;

    A = nan(m=l/n, n);

    for k = [1:m 2*m:m:l];
        do
            A(k) = sort(v)(++i);
        until ~mod(k+=m-1, m) | k>l;
    end;

    A'
end

Answer 6

Perl 5, 132 bytes

sub d{$,=0|sqrt(@_=sort{$a-$b}@_);--$,while@_%$,;map{$r++,$c--for@_/$,..$c;$a[$r++][$c--]=$_;$c=++$i,$r=0if$r<0||$c<0||$r>=$,}@_;@a}

Try it online!

Subroutine returns a 2-D array. TIO link includes footer code for displaying test result.

Answer 7

JavaScript (ES6), 233 bytes

f=s=>{l=s.length;i=Math.sqrt(l)|0;for(;l%++i;);p=(x)=>(x/i|0+x%i)*l+x%i;m=[...Array(l).keys()].sort((x,y)=>p(x)-p(y));return s.sort((a,b)=>a-b).map((x,i)=>m.indexOf(i)).reduce((a,b,d,g)=>!(d%i)?a.concat([g.slice(d,d+i)]):a,[])}

Explanation

f=s=>{                         // Take array `s` of numbers as input
  l=s.length                   // short-hand for length
  i=Math.sqrt(l)|0             // = Math.floor(Math.sqrt(l))
  for(;l%++i;);                // i = width           
  j=l/i                        // j = height

  p=(x)=>(x/i|0+x%i)*l+x%i     // helper to calculate (sort-of) ~manhattan
                                 // distance (horizontal distance weighted
                                 // slightly stronger), from top-left corner
                                 // to the number x, if numbers 0,...,l are
                                 // arranged left-to-right, top-to-bottom
                                 // in an l=i*j grid

  m=[...Array(l).keys()]         // range array
  .sort((x,y)=>p(x)-p(y)),       // manhatten-sorted, sort-of...

  return s.sort((a,b)=>a-b)      // sort input array by numbers,
    .map((x,i,w)=>w[m.indexOf(i)])    // then apply inverse permutation of the
                                 // range-grid manhatten-sort mapping.
    .reduce(                     // slice result into rows
      (a,b,d,g)=>!(d%i)?a.concat([g.slice(d,d+i)]):a
      ,[]
     )
}

Answer 8

Husk, 15 bytes

ḟȯΛ≤Σ∂MCP¹→←½ḊL

This works by brute force, so longer test cases may time out. Try it online!

Explanation

ḟȯΛ≤Σ∂MCP¹→←½ḊL  Implicit input, a list of integers x.
              L  Length of x (call it n).
             Ḋ   List of divisors.
            ½    Split at the middle.
          →←     Take last element of first part.
                 This is a divisor d that minimizes d + n/d.
        P¹       List of permutations of x.
      MC         Cut each into slices of length d.
ḟ                Find the first of these matrices that satisfies this:
     ∂            Take anti-diagonals,
    Σ             flatten them,
 ȯΛ≤              check that the result is sorted (each adjacent pair is non-decreasing).

Answer 9

C (gcc), 269 bytes

j,w,h,x,y;f(A,l)int*A;{int B[l];for(w=l;w-->1;)for(j=0;j<w;)if(A[j++]>A[j]){h=A[~-j];A[~-j]=A[j];A[j]=h;}for(w=h=j=2;w*h-l;j++)l%j||(w=h,h=j),h*h-l||(w=j);for(x=0;x<w*h;x++)for(y=0;y<=x;y++)x-y<w&y<h&&(B[x-y+y*w]=*A++);for(j=0;j<l;j++)j%w||puts(""),printf("%d ",B[j]);}

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Answer 10

Java 10, 199 188 186 bytes

a->{int j=a.length,m=0,n,i=0,k=0;for(n=m+=Math.sqrt(j);m*n<j;n=j/++m);var R=new int[m][n];for(java.util.Arrays.sort(a);i<m+n;i++)for(j=0;j<=i;j++)if(i-j<n&j<m)R[j][i-j]=a[k++];return R;}

Try it online.

Based on my answer here.

Explanation:

a->{                        // Method with int-array parameter and int-matrix return-type
  int j=a.length,           //  Length of the input-array
      m=0,n,                //  Amount of rows and columns
      i=0,k=0;              //  Index integers
   for(n=m+=Math.sqrt(j);   //  Set both `m` and `n` to floor(√ `j`)
       m*n<j;               //  Loop as long as `m` multiplied by `n` is not `j`
       n=j/++m);            //   Increase `m` by 1 first with `++m`
                            //   and then set `n` to `j` integer-divided by this new `m`
   var R=new int[m][n];     //  Result-matrix of size `m` by `n`
   for(java.util.Arrays.sort(a);
                            //  Sort the input-array
       i<m+n;)              //  Loop as long as `i` is smaller than `m+n`
     for(j=0;j<=i;j++)      //   Inner loop `j` in range [0,`i`]
       if(i-j<n&j<m)        //    If `i-j` is smaller than `n`, and `j` smaller than `m`
                            //    (So basically check if they are still within bounds)
         R[j][i-j]=a[k++];  //     Add the number of the input array at index `k`,
                            //     to the matrix in the current cell at `[j,i-j]`
  return R;}                //  Return the result-matrix