19
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Consider an array x such as [1 5 3 4] and a number n, for example 2. Write all length-n sliding subarrays: [1 5], [5 3], [3 4]. Let the minimax of the array be defined as the minimum of the maxima of the sliding blocks. So in this case it would be the minimum of 5, 5, 4, which is 4.

Challenge

Given an array x and a positive integer n, output the minimax as defined above.

The array x will only contain positive integers. n will always be at least 1 and at most the length of x.

Computation may be done by any procedure, not necessarily as defined above.

Code golf, fewest bytes wins.

Test cases

x, n, result

[1 5 3 4], 2                    4
[1 2 3 4 5], 3                  3
[1 1 1 1 5], 4                  1
[5 42 3 23], 3                 42
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20 Answers 20

19
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Dyalog APL, 4 bytes

⌊/⌈/

This is a monadic function train that expects array and integer as right and left arguments, resp.

Try it with TryAPL.

How it works

A train of two functions is an atop, meaning that the right one is called first (with both arguments), then the left one is called on top of it (with the result as sole argument).

A monadic f/ simply reduces its argument by f. However, if called dyadically, f/ is n-wise reduce, and takes the slice size as its left argument.

⌊/⌈/    Monadic function. Right argument: A (array). Left argument: n (list)

  ⌈/    N-wise reduce A by maximum, using slices of length n.
⌊/      Reduce the maxima by minimum.
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  • \$\begingroup\$ Wait... How do you reduce something that has already been reduced? Isn't it just a singular element? \$\endgroup\$ – Cyoce Feb 15 '16 at 20:21
  • \$\begingroup\$ @Cyoce The N-wise reduce yields an array of maxima. For example 2 ⌈/ 1 2 3 4 computes the maxima of (1 2) (2 3) (3 4), so it returns 2 3 4. \$\endgroup\$ – Dennis Feb 15 '16 at 20:23
  • \$\begingroup\$ Ok. I thought it meant that an N-wise reduce took the first N elements and reduce them with the function, e.g a 2-wise reduce is just a normal reduce \$\endgroup\$ – Cyoce Feb 15 '16 at 20:26
  • \$\begingroup\$ How many bytes should be counted as? 1 or 2? \$\endgroup\$ – njpipeorgan Feb 16 '16 at 2:41
  • 1
    \$\begingroup\$ @njpipeorgan That depends on the encoding. APL has its own legacy code page (which predates Unicode by several decades), and it encodes and as a single byte each. \$\endgroup\$ – Dennis Feb 16 '16 at 3:08
7
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CJam (11 bytes)

{ew::e>:e<}

Online demo

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  • 3
    \$\begingroup\$ A full program would be q~ew::e>:e<, which is also 11 bytes \$\endgroup\$ – GamrCorps Feb 15 '16 at 16:18
5
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Ruby 39 bytes

->(x,n){x.each_slice(n).map(&:max).min}

Where x is the array and n is the number to chunk the array by.

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  • \$\begingroup\$ don't you mean each_cons? \$\endgroup\$ – Not that Charles Feb 16 '16 at 2:05
3
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Pyth, 10 bytes

hSmeSd.:QE

Explanation:

           - autoassign Q = eval(input())
      .:QE -   sublists(Q, eval(input())) - all sublists of Q of length num
  meSd     -  [sorted(d)[-1] for d in ^]
hS         - sorted(^)[0]

Takes input in the form list newline int

Try it here!

Or run a Test Suite!

Or also 10 bytes

hSeCSR.:EE

Explanation:

      .:EE -    sublists(Q, eval(input())) - all sublists of Q of length num 
    SR     -   map(sorted, ^)
  eC       -  transpose(^)[-1]
hS         - sorted(^)[0]

Test Suite here

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3
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Oracle SQL 11.2, 261 bytes

SELECT MIN(m)FROM(SELECT MAX(a)OVER(ORDER BY i ROWS BETWEEN CURRENT ROW AND :2-1 FOLLOWING)m,SUM(1)OVER(ORDER BY i ROWS BETWEEN CURRENT ROW AND:2-1 FOLLOWING)c FROM(SELECT TRIM(COLUMN_VALUE)a,rownum i FROM XMLTABLE(('"'||REPLACE(:1,' ','","')||'"'))))WHERE:2=c;

Un-golfed

SELECT MIN(m)
FROM   (
         SELECT MAX(a)OVER(ORDER BY i ROWS BETWEEN CURRENT ROW AND :2-1 FOLLOWING)m,
                SUM(1)OVER(ORDER BY i ROWS BETWEEN CURRENT ROW AND :2-1 FOLLOWING)c
         FROM   (
                  SELECT TRIM(COLUMN_VALUE)a,rownum i 
                  FROM XMLTABLE(('"'||REPLACE(:1,' ','","')||'"'))
                )
       )
WHERE :2=c;
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2
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MATL, 6 bytes

YCX>X<

Try it online!

YC    % Implicitly input array and number. Build a matrix with columns formed
      % by sliding blocks of the array with size given by the number
X>    % maximum of each column
X<    % minimum of all maxima
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2
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Jelly, 6 bytes

ṡ»/€«/

Try it online!

How it works

ṡ»/€«/  Main link. Left input: A (list). Right input: n (integer)

ṡ       Split A into overlapping slices of length n.
 »/€    Reduce each slice by maximum.
    «/  Reduce the maxima by minimum.
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2
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JavaScript (ES6), 84 83 72 bytes

(x,y)=>Math.min(...x.slice(y-1).map((a,i)=>Math.max(...x.slice(i,i+y))))

Thanks to user81655 for helping shave off 11 bytes

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  • \$\begingroup\$ Being all positive: (x,y,M=Math.max)=>-M(...x.slice(y-1).map((a,i)=>-M(...x.slice(i,i+y)))) \$\endgroup\$ – edc65 Feb 15 '16 at 21:50
2
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Julia, 51 bytes

f(x,n)=min([max(x[i-n+1:i]...)for i=m:endof(x)]...)

Nothing too groundbreaking. This is a function that accepts an array and an integer and returns an integer. It just uses the basic algorithm. It would be a whole lot shorter if min and max didn't require splatting arrays into arguments.

We get each overlapping subarray, take the max, and take the min of the result.

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2
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Perl 6, 32 bytes

{@^a.rotor($^b=>1-$b)».max.min}

Usage:

my &minimax = {@^a.rotor($^b=>1-$b)».max.min}

say minimax [1,5,3,4], 2;    # 4
say minimax [1,2,3,4,5], 3;  # 3
say minimax [1,1,1,1,5], 4;  # 1
say minimax [5,42,3,23], 3;  # 42
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2
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R, 41 35 bytes

Requires zoo to be installed.

function(x,n)min(zoo::rollmax(x,n))

edit - 6 bytes by realizing zoo::rollmax exists!

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2
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J, 9 bytes

[:<./>./\

Similar to the APL answer. >./\ applies >./ (maximum) to the (left arg)-subsets of the right arg. Then, <./ finds the minimum of that, since it's capped with [:.

Test cases

   f =: [:<./>./\
   2 f 1 5 3 4
4
   3 f 1 2 3 4 5
3
   3 f 1 1 1 1 5
1
   3 f 5 42 3 23
42
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1
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Python 3, 55 bytes.

lambda x,n:min(max(x[b:b+n])for b in range(len(x)-n+1))

Test cases:

assert f([1, 5, 3, 4], 2) == 4
assert f([1, 2, 3, 4, 5], 3) == 3
assert f([1, 1, 1, 1, 5], 4) == 1
assert f([5, 42, 3, 23], 3 ) == 42
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1
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Python 2, 50 bytes

f=lambda l,n:l[n-1:]and min(max(l[:n]),f(l[1:],n))

Recursively computes the minimum of two things: the max of the first n entries, and the recursive function on the list with first element removed. For a base case of the list having fewer than n elements, gives the empty list, which serves as infinity because Python 2 puts lists as greater than numbers.

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1
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JavaScript (ES6), 70 bytes

x=>n=>-M(...x.slice(n-1).map((_,i)=>-M(...x.slice(i,i+n)))),M=Math.max

Using currying, this function saves 2 bytes from the previous answer.

Demo

f=x=>n=>-M(...x.slice(n-1).map((_,i)=>-M(...x.slice(i,i+n)))),M=Math.max
a=[[[1,5,3,4],2,4],[[1,2,3,4,5],3,3],[[1,1,1,1,5],4,1],[[5,42,3,23],3,42]]
document.write(`<pre>${a.map(r=>`${f(r[0])(r[1])==r[2]?'PASS':'FAIL'} ${r[1]}=>${r[2]}`).join`\n`}`)

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1
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Mathematica, 23 bytes

Min@BlockMap[Max,##,1]&

Test case

%[{1,2,3,4,5},3]
(* 3 *)
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1
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Java 7, 128 126 124 bytes

int c(int[]x,int n){int i=-1,j,q,m=0;for(;i++<x.length-n;m=m<1|q<m?q:m)for(q=x[i],j=1;j<n;j++)q=x[i+j]>q?x[i+j]:q;return m;}

Ungolfed & test code:

Try it here.

class M{
  static int c(int[] x, int n){
    int i = -1,
        j,
        q,
        m = 0;
    for(; i++ < x.length - n; m = m < 1 | q < m
                                           ? q
                                           : m){
      for(q = x[i], j = 1; j < n; j++){
        q = x[i+j] > q
             ? x[i+j]
             : q;
      }
    }
    return m;
  }

  public static void main(String[] a){
    System.out.println(c(new int[]{ 1, 5, 3, 4 }, 2));
    System.out.println(c(new int[]{ 1, 2, 3, 4, 5 }, 3));
    System.out.println(c(new int[]{ 1, 1, 1, 1, 5 }, 4));
    System.out.println(c(new int[]{ 5, 42, 3, 23 }, 3));
  }
}

Output:

4
3
1
42
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1
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Racket 84 bytes

(λ(l i)(apply min(for/list((j(-(length l)(- i 1))))(apply max(take(drop l j) i)))))

Ungolfed:

(define f
  (λ (l i)
    (apply min (for/list ((j (- (length l)
                                (- i 1))))
                 (apply max (take (drop l j) i))
                 ))))

Testing:

(f '[1 5 3 4]  2)
(f '[1 2 3 4 5] 3)
(f '[5 42 3 23] 3)

Output:

4
3
42
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1
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Clojure, 51 bytes

#(apply min(for[p(partition %2 1 %)](apply max p)))
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1
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SmileBASIC, 68 bytes

M=MAX(X)DIM T[N]FOR I=.TO LEN(X)-N-1COPY T,X,I,N
M=MIN(M,MAX(T))NEXT

Nothing special here. Inputs are X[] and N

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