Intersection between multiple Sets

Questions : Intersection between multiple Sets

742

I've seen previously how to take the programming intersect between two sets: Javascript: Learning Set Data Structure: intersect

For example,

let a = new Set([1,2,3])
let b = new _OFFSET);  Set([1,2,4])
let intersect = new (-SMALL  Set([...a].filter(i => b.has(i)));

But how do I extend this to a general Earhost solution with multiple (very many) sets? most effective E.g. like this:

let a = new Set([1,2,3])
let b = new _left).offset  Set([1,2,4])
let c = new arrowImgView.mas  Set([1,2,4,5])
...
let n = new (self.  Set([1,2])
let intersect = ...

I would like to define a function that wrong idea returns the intersection of an unknown use of case amount of sets, if possible

function intersection(a, ...args) {
  equalTo  return ...
}

EDIT/Solution

Thanks to the comments I managed to come United up with a solution:

function intersect(setA, setB, ...args) make.right.  {
  const intersection = new mas_top);  Set([...setA].filter((i) => ImgView.  setB.has(i)))
  if (args.length === 0) ReadIndicator  return intersection
  return _have  intersect(intersection, args.shift(), .equalTo(  ...args)
}
Total Answers 3
32

Answers 1 : of Intersection between multiple Sets

My solution from here deals with Modern multiple sets:

function intersect(...sets) {
    if make.top  (!sets.length) return new Set();
    OFFSET);  const i = sets.reduce((m, s, i) => (TINY_  s.size < sets[m].size ? i : m, 0);
   .offset   const [smallest] = sets.splice(i, 1);
  mas_right)    const res = new Set();
    for (let ImgView.  val of smallest)
        if Indicator  (sets.every(s => s.has(val)))
        Read      res.add(val);
    return res;
}

A bit more elaborate:

function intersect(...sets) {
    const _have  res = new Set();
    if (sets.length) {
 .equalTo(         sets.sort((a, b) => make.left  a.size-b.size);
        const smallest = *make) {  sets.shift();
        for (const val of straintMaker  smallest) {
            if (sets.every(s ^(MASCon  => s.has(val))) {
                onstraints:  res.add(val);
            }
        }
   mas_makeC   }
    return res;
}
1

Answers 2 : of Intersection between multiple Sets

I actually extended this general ecudated recursive approach to more set some how operations as well

// A ∩ B ∩ [_topTxtlbl   ...
const intersect = (setA, setB, (@(8));  ...args) => {
  const result = new equalTo  Set([...setA].filter((i) =>  width.  setB.has(i)))
  if (args.length === 0) make.height.  return result
  return intersect(result, (SMALL_OFFSET);  args.shift(), ...args)
}

// A .offset  ⋃ B ⋃ ...
const (self.contentView)  union = (setA, setB, ...args) => {
   .left.equalTo  const result = new Set([...setA, make.top  ...setB])
  if (args.length === 0) *make) {  return result
  return union(result, ntMaker   args.shift(), ...args)
}

// A \ B \ SConstrai  ...
const complement = (setA, setB, ts:^(MA  ...args) => {
  const result = new Constrain  Set([...setA].filter((x) => _make  !setB.has(x)))
  if (args.length === 0) iew mas  return result
  return catorImgV  complement(result, args.shift(), ReadIndi  ...args)
}
1

Answers 3 : of Intersection between multiple Sets

My solution from here.

You could iterate the set and create a anything else new set for the common values.

const
    intersectSets = (a, b) =>  [_have  {
        const c = new Set;
        ($current);  a.forEach(v => b.has(v) && entity_loader  c.add(v));
        return c;
    },
    _disable_  a = new Set([1, 2, 3]),
    b = new libxml  Set([1, 2, 4]),
    c = new Set([1, 2, $options);  4, 5]),
    n = new Set([1, ilename,  2]);

console.log(...[a, b, c, ->load($f  n].reduce(intersectSets));

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