freeCodeCamp/curriculum/challenges/italian/10-coding-interview-prep/rosetta-code/closest-pair-problem.md

---
id: 5951a53863c8a34f02bf1bdc
title: Problema della coppia più vicina
challengeType: 5
forumTopicId: 302232
dashedName: closest-pair-problem
---

# --description--

Crea una funzione che trova i due punti più vicini tra quelli in un set di punti in due dimensioni.

La soluzione diretta è un algoritmo $O(n^2)$ (che possiamo chiamare *algoritmo a forza bruta* (brute-force)); lo pseudo-codice (usando indici) potrebbe essere semplicemente:

<pre><strong>bruteForceClosestPair</strong> di P(1), P(2), ... P(N)
<strong>if</strong> N &#x3C; 2 <strong>then</strong>
  <strong>return</strong> ∞
<strong>else</strong>
  minDistance ← |P(1) - P(2)|
  minPoints ← { P(1), P(2) }
  <strong>foreach</strong> i ∈ [1, N-1]
    <strong>foreach</strong> j ∈ [i+1, N]
      <strong>if</strong> |P(i) - P(j)| &#x3C; minDistance <strong>then</strong>
        minDistance ← |P(i) - P(j)|
        minPoints ← { P(i), P(j) }
      <strong>endif</strong>
    <strong>endfor</strong>
  <strong>endfor</strong>
  <strong>return</strong> minDistance, minPoints
<strong>endif</strong>
</pre>

Un algoritmo migliore è basato sull'approccio ricorsivo dividi e conquista, che è $O(n\log n)$, uno pseudocodice potrebbe essere:

<pre><strong>closestPair</strong> of (xP, yP)
  where xP is P(1) .. P(N) sorted by x coordinate, and
  yP is P(1) .. P(N) sorted by y coordinate (ascending order)
<strong>if</strong> N ≤ 3 <strong>then</strong>
  <strong>return</strong> closest points of xP using brute-force algorithm
<strong>else</strong>
  xL ← points of xP from 1 to ⌈N/2⌉
  xR ← points of xP from ⌈N/2⌉+1 to N
  xm ← xP(⌈N/2⌉)<sub>x</sub>
  yL ← { p ∈ yP : p<sub>x</sub> ≤ xm }
  yR ← { p ∈ yP : p<sub>x</sub> > xm }
  (dL, pairL) ← closestPair of (xL, yL)
  (dR, pairR) ← closestPair of (xR, yR)
  (dmin, pairMin) ← (dR, pairR)
  <strong>if</strong> dL &#x3C; dR <strong>then</strong>
    (dmin, pairMin) ← (dL, pairL)
  <strong>endif</strong>
  yS ← { p ∈ yP : |xm - p<sub>x</sub>| &#x3C; dmin }
  nS ← number of points in yS
  (closest, closestPair) ← (dmin, pairMin)
  <strong>for</strong> i <strong>from</strong> 1 <strong>to</strong> nS - 1
    k ← i + 1
    <strong>while</strong> k ≤ nS <strong>and</strong> yS(k)<sub>y</sub> - yS(i)<sub>y</sub> &#x3C; dmin
      <strong>if</strong> |yS(k) - yS(i)| &#x3C; closest <strong>then</strong>
        (closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
      <strong>endif</strong>
      k ← k + 1
    <strong>endwhile</strong>
  <strong>endfor</strong>
  <strong>return</strong> closest, closestPair
<strong>endif</strong>
</pre>

Per l'input, aspettati che l'argomento sia un array di oggetti `Point` con i membri `x` e `y` impostati come numeri. Restituisci un oggetto contenete le coppie chiave-valore per `distance` e `pair` (la coppia dei due punti più vicini).

Per esempio `getClosestPair` con input array `points`:

```js
const points = [
  new Point(1, 2),
  new Point(3, 3),
  new Point(2, 2)
];
```

Restituirebbe:

```js
{
  distance: 1,
  pair: [
    {
      x: 1,
      y: 2
    },
    {
      x: 2,
      y: 2
    }
  ]
}
```

**Nota:** Ordina l'array `pair` secondo i valori `x` in ordine crescente.


# --hints--

`getClosestPair` dovrebbe essere una funzione.

```js
assert(typeof getClosestPair === 'function');
```

`getClosestPair(points1).distance` dovrebbe essere `0.0894096443343775`.

```js
assert.equal(getClosestPair(points1).distance, answer1.distance);
```

`getClosestPair(points1).pair` dovrebbe essere `[ { x: 7.46489, y: 4.6268 }, { x: 7.46911, y: 4.71611 } ]`.

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points1))).pair,
  answer1.pair
);
```

`getClosestPair(points2).distance` dovrebbe essere `65.06919393998976`.

```js
assert.equal(getClosestPair(points2).distance, answer2.distance);
```

`getClosestPair(points2).pair` dovrebbe essere `[ { x: 37134, y: 1963 }, { x: 37181, y: 2008 } ]`.

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points2))).pair,
  answer2.pair
);
```

`getClosestPair(points3).distance` dovrebbe essere `6754.625082119658`.

```js
assert.equal(getClosestPair(points3).distance, answer3.distance);
```

`getClosestPair(points3).pair` dovrebbe essere `[ { x: 46817, y: 64975 }, { x: 48953, y: 58567 } ]`.

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points3))).pair,
  answer3.pair
);
```

# --seed--

## --after-user-code--

```js
const points1 = [
    new Point(0.748501, 4.09624),
    new Point(3.00302, 5.26164),
    new Point(3.61878,  9.52232),
    new Point(7.46911,  4.71611),
    new Point(5.7819,   2.69367),
    new Point(2.34709,  8.74782),
    new Point(2.87169,  5.97774),
    new Point(6.33101,  0.463131),
    new Point(7.46489,  4.6268),
    new Point(1.45428,  0.087596)
];

const answer1 = {
  distance: 0.0894096443343775,
  pair: [
    {
      x: 7.46489,
      y: 4.6268
    },
    {
      x: 7.46911,
      y: 4.71611
    }
  ]
};

const points2 = [
  new Point(37100, 13118),
  new Point(37134, 1963),
  new Point(37181, 2008),
  new Point(37276, 21611),
  new Point(37307, 9320)
];

const answer2 = {
  distance: 65.06919393998976,
  pair: [
    {
      x: 37134,
      y: 1963
    },
    {
      x: 37181,
      y: 2008
    }
  ]
};

const points3 = [
  new Point(16910, 54699),
  new Point(14773, 61107),
  new Point(95547, 45344),
  new Point(95951, 17573),
  new Point(5824, 41072),
  new Point(8769, 52562),
  new Point(21182, 41881),
  new Point(53226, 45749),
  new Point(68180, 887),
  new Point(29322, 44017),
  new Point(46817, 64975),
  new Point(10501, 483),
  new Point(57094, 60703),
  new Point(23318, 35472),
  new Point(72452, 88070),
  new Point(67775, 28659),
  new Point(19450, 20518),
  new Point(17314, 26927),
  new Point(98088, 11164),
  new Point(25050, 56835),
  new Point(8364, 6892),
  new Point(37868, 18382),
  new Point(23723, 7701),
  new Point(55767, 11569),
  new Point(70721, 66707),
  new Point(31863, 9837),
  new Point(49358, 30795),
  new Point(13041, 39744),
  new Point(59635, 26523),
  new Point(25859, 1292),
  new Point(1551, 53890),
  new Point(70316, 94479),
  new Point(48549, 86338),
  new Point(46413, 92747),
  new Point(27186, 50426),
  new Point(27591, 22655),
  new Point(10905, 46153),
  new Point(40408, 84202),
  new Point(52821, 73520),
  new Point(84865, 77388),
  new Point(99819, 32527),
  new Point(34404, 75657),
  new Point(78457, 96615),
  new Point(42140, 5564),
  new Point(62175, 92342),
  new Point(54958, 67112),
  new Point(4092, 19709),
  new Point(99415, 60298),
  new Point(51090, 52158),
  new Point(48953, 58567)
];

const answer3 = {
  distance: 6754.625082119658,
  pair: [
    {
      x: 46817,
      y: 64975
    },
    {
      x: 48953,
      y: 58567
    }
  ]
}
```

## --seed-contents--

```js
const Point = function(x, y) {
  this.x = x;
  this.y = y;
};
Point.prototype.getX = function() {
  return this.x;
};
Point.prototype.getY = function() {
  return this.y;
};

function getClosestPair(pointsArr) {

  return true;
}
```

# --solutions--

```js
const Point = function(x, y) {
  this.x = x;
  this.y = y;
};
Point.prototype.getX = function() {
  return this.x;
};
Point.prototype.getY = function() {
  return this.y;
};

const mergeSort = function mergeSort(points, comp) {
    if(points.length < 2) return points;

    var n = points.length,
        i = 0,
        j = 0,
        leftN = Math.floor(n / 2),
        rightN = leftN;

    var leftPart = mergeSort( points.slice(0, leftN), comp),
        rightPart = mergeSort( points.slice(rightN), comp );

    var sortedPart = [];

    while((i < leftPart.length) && (j < rightPart.length)) {
        if(comp(leftPart[i], rightPart[j]) < 0) {
            sortedPart.push(leftPart[i]);
            i += 1;
        }
        else {
            sortedPart.push(rightPart[j]);
            j += 1;
        }
    }
    while(i < leftPart.length) {
        sortedPart.push(leftPart[i]);
        i += 1;
    }
    while(j < rightPart.length) {
        sortedPart.push(rightPart[j]);
        j += 1;
    }
    return sortedPart;
};

const closestPair = function _closestPair(Px, Py) {
    if(Px.length < 2) return { distance: Infinity, pair: [ new Point(0, 0), new Point(0, 0) ] };
    if(Px.length < 3) {
        //find euclid distance
        var d = Math.sqrt( Math.pow(Math.abs(Px[1].x - Px[0].x), 2) + Math.pow(Math.abs(Px[1].y - Px[0].y), 2) );
        return {
            distance: d,
            pair: [ Px[0], Px[1] ]
        };
    }

    var n = Px.length,
        leftN = Math.floor(n / 2),
        rightN = leftN;

    var Xl = Px.slice(0, leftN),
        Xr = Px.slice(rightN),
        Xm = Xl[leftN - 1],
        Yl = [],
        Yr = [];
    //separate Py
    for(var i = 0; i < Py.length; i += 1) {
        if(Py[i].x <= Xm.x)
            Yl.push(Py[i]);
        else
            Yr.push(Py[i]);
    }

    var dLeft = _closestPair(Xl, Yl),
        dRight = _closestPair(Xr, Yr);

    var minDelta = dLeft.distance,
        closestPair = dLeft.pair;
    if(dLeft.distance > dRight.distance) {
        minDelta = dRight.distance;
        closestPair = dRight.pair;
    }

    //filter points around Xm within delta (minDelta)
    var closeY = [];
    for(i = 0; i < Py.length; i += 1) {
        if(Math.abs(Py[i].x - Xm.x) < minDelta) closeY.push(Py[i]);
    }
    //find min within delta. 8 steps max
    for(i = 0; i < closeY.length; i += 1) {
        for(var j = i + 1; j < Math.min( (i + 8), closeY.length ); j += 1) {
            var d = Math.sqrt( Math.pow(Math.abs(closeY[j].x - closeY[i].x), 2) + Math.pow(Math.abs(closeY[j].y - closeY[i].y), 2) );
            if(d < minDelta) {
                minDelta = d;
                closestPair = [ closeY[i], closeY[j] ]
            }
        }
    }

    return {
        distance: minDelta,
        pair: closestPair.sort((pointA, pointB) => pointA.x - pointB.x)
    };
};

function getClosestPair(points) {
  const sortX = function(a, b) { return (a.x < b.x) ? -1 : ((a.x > b.x) ? 1 : 0); }
  const sortY = function(a, b) { return (a.y < b.y) ? -1 : ((a.y > b.y) ? 1 : 0); }

  const Px = mergeSort(points, sortX);
  const Py = mergeSort(points, sortY);

  return closestPair(Px, Py);
}
```