Bxdty

I was reading through a problem and was trying to solve this problem.

You've invited N people over for dinner. Let's say 4.

You have a circular dinner table and you wish to seat everyone around
it. Unfortunately, not all of your friends are friends with each
other, but you'd like to seat everyone optimally so that as many
people as possible are seated next to people they consider friends and
not enemies.

You've charted everyone's friendships and hatreds in a matrix of size
NxN and represented friendships with the integer 1, hatreds with -1,
and sheer indifference with 0.

[[ 0, 1, 1, 1, 1],    ← yes you like all your friends

 [-1, 0, 1,-1, 0],

 [-1, 1, 0, 1, 0],

 [ 1, 1, 1, 0,-1],

 [ 1, 0, 0,-1, 0]]

Question:

-> Write a Javascript method that computes an optimal seating arrangement as an Array, e.g. [0,4,2,1,3], for a given input matrix. (assuming indexes 0
and N-1 are seated adjacently). What is the time complexity for the
solution? Add thoughts on possible optimizations.

I've tried solving this manually however, I didn't understand the question's example [0,4,2,1,3] for the given input matrix.

Can someone Enlighten me?

How did he/she come up with [0,4,2,1,3]?

Thanks and very much appreciate your time.

How did he/she come up with [0,4,2,1,3]?

That permutation certainly isn't the right answer for the example input (see reasoning below), so I think that Emma's comment above is spot-on: the problem statement is just demonstrating what a "seating arrangement as an Array" should look like in general, not specifically demonstrating the optimal seating arrangement for the example input.

As for why I say that [0,4,2,1,3] certainly isn't the right answer for the example you've given . . . I don't completely understand how we decide whether one permutation is better than another, but it's clear that [0,4,1,2,3] is better by any measure. For both [0,4,2,1,3] and [0,4,1,2,3], the first person (0) likes both neighbors; the second person (4) is neutral toward both neighbors; and the third and fifth people (2 and 3 in the former, 1 and 3 in the latter) each like one neighbor and are neutral toward the other. The only difference between the two permutations is that in [0,4,2,1,3], the fourth person (1) is neutral toward one neighbor and dislikes the other, whereas in [0,4,1,2,3], the fourth person (2) is neutral toward one neighbor and likes the other. So the latter is obviously superior, no matter whether we consider it more important to increase likes or to decrease dislikes.

Checking all possible orders is a classical permutation task even if there might be a more efficient algorithm for this specific problem.

One optimization can be done by reducing the permutation to array length-1 since in circular orders e.g. 0,1,2,3,4 and 4,0,1,2,3 (and all further rotations) are the same. You can view the order from you own seat always starting at position 0.

(function ()

{

  'use strict';



  let popularity =

  [

    [ 0, 1, 1, 1, 1],   // ← yes you like all your friends

    [-1, 0, 1,-1, 0],

    [-1, 1, 0, 1, 0],

    [ 1, 1, 1, 0,-1],

    [ 1, 0, 0,-1, 0],

  ];



  function permutation(arr)

  {

    let

      l = arr.length,

      perms = 

    ;



    if(l<2)

      return [arr];



    for(let i=0; i<l; i++)

    {

      let

        cpy    = Array.from(arr),

        [perm] = cpy.splice(i, 1)

      ;

      perms.push(...permutation(cpy).map(v => [perm, ...v]));

    }



    return perms;

  }





  let

    keys = Array.from(popularity.keys()).slice(1),

    permutations = permutation(keys),

    rating = permutations.map(v =>

    {

      let

        last = v.length -1,



        // start with our own relationships to the left and right neighbour

        // (each: we like him, he likes us)

        rate =

            popularity [0]       [v[0]]

          + popularity [v[0]]    [0]

          + popularity [0]       [v[last]]

          + popularity [v[last]] [0]

      ;



      for(let i = 0; i<last; i++)

        rate += popularity[v[i]][v[i+1]] + popularity[v[i+1]][v[i]];



      return [rate, [0, ...v]];

    }

  ).sort( (v1, v2) => ( v1[0] === v2[0] ? 0 : (v1[0] > v2[0] ? -1 : 1))  );



  console.log(rating);



})();

output:

[ [ 8, [ 0, 4, 1, 2, 3 ] ],

  [ 8, [ 0, 3, 2, 1, 4 ] ],

  [ 6, [ 0, 3, 1, 2, 4 ] ],

  [ 6, [ 0, 4, 2, 1, 3 ] ],

  [ 4, [ 0, 1, 4, 2, 3 ] ],

  [ 4, [ 0, 1, 2, 3, 4 ] ],

  [ 4, [ 0, 4, 1, 3, 2 ] ],

  [ 4, [ 0, 1, 3, 2, 4 ] ],

  [ 4, [ 0, 2, 3, 1, 4 ] ],

  [ 4, [ 0, 3, 2, 4, 1 ] ],

  [ 4, [ 0, 4, 2, 3, 1 ] ],

  [ 4, [ 0, 4, 3, 2, 1 ] ],

  [ 2, [ 0, 3, 4, 2, 1 ] ],

  [ 2, [ 0, 3, 1, 4, 2 ] ],

  [ 2, [ 0, 2, 4, 1, 3 ] ],

  [ 2, [ 0, 4, 3, 1, 2 ] ],

  [ 2, [ 0, 3, 4, 1, 2 ] ],

  [ 2, [ 0, 1, 2, 4, 3 ] ],

  [ 2, [ 0, 2, 1, 4, 3 ] ],

  [ 2, [ 0, 2, 1, 3, 4 ] ],

  [ 0, [ 0, 1, 4, 3, 2 ] ],

  [ 0, [ 0, 2, 3, 4, 1 ] ],

  [ -2, [ 0, 1, 3, 4, 2 ] ],

  [ -2, [ 0, 2, 4, 3, 1 ] ] ]

As we can see, there still are reversed permutations combined with yourself (0) having the same rating of course. Eleminating mirrored orders, i.e. reversed permutations, would be another optimization.

I did this for demonstration in single steps to have a more readable code facing single problems step by step. You could refactor the rating calculation directly into the permutation algorithm.

A permutation has an O(N!) complexity. Since we ignore our own seat, the permutation part is reduced to O((N-1)!).