Multiple Array Intersection In Javascript

How to write in javascript(w/wth JQuery) to find values that intersect between arrays? It should be like var a = [1,2,3] var b = [2,4,5] var c = [2,3,6] and the intersect functio

Solution 1:

There are many ways to achieve this.

Since you are using jQuery I will suggest use grep function to filter the value that are present in all three array.

var a = [1, 2, 3]
var b = [2, 4, 5]
var c = [2, 3, 6]

var result = $.grep(a, function(value, index) {
  return b.indexOf(value) > -1 && c.indexOf(value) > -1;
})
console.log(result)

<scriptsrc="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Explanation: Loop over any array and filter out the values that are present in other array.

Update (for multimidensional array):

Concept - flatten the multidimensional array that is transform [[1,2],3,4] to [1,2,3,4] and then use the same logic used for single dimensional array.

Example:

var a = [
   [1, 4], 2, 3
 ]
 var b = [2, 4, 5]
 var c = [2, 3, 6, [4, 7]]

 //flatten the array's//[1,4,2,3]var aFlattened = $.map(a, function(n) {
   return n;
 })

 //[2,4,5]var bFlattened = $.map(b, function(n) {
   return n;
 })

 //[2,3,6,4,7]var cFlattened = $.map(c, function(n) {
   return n;
 })

 var result = $.grep(aFlattened, function(value) {
   return (bFlattened.indexOf(value) > -1 && cFlattened.indexOf(value) > -1);
 });

 console.log(result);

<scriptsrc="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Solution 2:

// First this is how you declare an arrayvar a = [1,2,3];
var b = [2,4,5];
var c = [2,3,6];


// Second, this function should handle undetermined number of parameters (so arguments should be used)functionintersect(){
  var args = arguments;
  // if no array is passed then return empty arrayif(args.length == 0) return [];
  
  // for optimisation lets find the smallest arrayvar imin = 0;
  for(var i = 1; i < args.length; i++)
    if(args[i].length < args[imin].length) imin = i;
  var smallest = Array.prototype.splice.call(args, imin, 1)[0];
  
  return smallest.reduce(function(a, e){
    for(var i = 0; i < args.length; i++)
      if(args[i].indexOf(e) == -1) return a;
    a.push(e);
    return a;
  }, []);
}

console.log(intersect(a, b, c));

Solution 3:

First of all '{}' means Object in JavaScript. Here is my suggestion.(this is another way of doing it)

// declarations  var a = [1,2,3];  
var b = [2,4,5];  
var c = [2,3,6];  

// filter property of array
a.filter(function(val) {
    if (b.indexOf(val) > -1  && c.indexOf(val) > -1)
        returnval;
});

what it does is it checks for each element in array 'a' and checks if that value is present in array 'b' and array 'c'. If it is true it returns the value. Simple!!!. The above code should work for String as well but it wouldn't work for IE < 9, so be careful.

Solution 4:

// Intersecting 2 ordered lists of length n and m is O(n+m)// This can be sped up by skipping elements// The stepsize is determined by the ratio of lengths of the lists// The skipped elements need to be checked after skipping some elements:// In the case of step size 2 : Check the previous element// In case step size>2 : Binary search the previously skipped range// This results in the best case complexity of O(n+n), if n<m// or the more propable complexity of O(n+n+n*log2(m/n)), if n<mfunctionbinarySearch(array, value, start = 0, end = array.length) {
    var j = start,
        length = end;
    while (j < length) {
        var i = (length + j - 1) >> 1; // move the pointer toif (value > array[i])
            j = i + 1;
        elseif (value < array[i])
            length = i;
        elsereturn i;
    }
    return -1;
}
functionintersect2OrderedSets(a, b) {
    var j = 0;
    var k = 0;
    var ratio = ~~(b.length / a.length) - 1 || 1;
    var result = [];
    var index;
    switch (ratio) {
        case1:
            while (j < a.length) {
                if (a[j] === b[k]) {
                    result.push(a[j]);
                    j++;
                    k++;
                } elseif (a[j] < b[k]) {
                    while (a[j] < b[k]) j++;
                } else {
                    while (b[k] < a[j]) k++;
                    if (k >= b.length) break;
                }
            }
            break;
        case2:
            while (j < a.length) {
                if (a[j] === b[k]) {
                    result.push(a[j]);
                    j++;
                    k++;
                } elseif (a[j] < b[k]) {
                    while (a[j] < b[k]) j++;
                } else {
                    while (b[k] < a[j]) k += 2;
                    if (k - 1 >= b.length) break;
                    if (a[j] <= b[k - 1]) k--;
                }
            }
            break;
        default:
            while (j < a.length) {
                if (a[j] === b[k]) {
                    result.push(a[j]);
                    j++;
                    k++;
                } elseif (a[j] < b[k]) {
                    while (a[j] < b[k]) j++;
                } else {
                    while (b[k] < a[j]) k += ratio;
                    index = binarySearch(b, a[j], k - ratio + 1, k + 1 < b.length ? k + 1 : b.length - 1);
                    if (index > -1) {
                        result.push(a[j]);
                        j++;
                        k = index + 1;
                    } else {
                        j++;
                        k = k - ratio + 1;
                    }
                    if (k >= b.length) break;
                }
            }
    }
    return result;
}

functionintersectOrderedSets() {
    var shortest = 0;
    for (var i = 1; i < arguments.length; i++)
        if (arguments[i].length < arguments[shortest].length) shortest = i;
    var result = arguments[shortest];
    for (var i = 0, a, b, j, k, ratio, index; i < arguments.length; i++) {
        if (result.length === 0) return result;
        if (i === shortest) continue;
        a = result;
        b = arguments[i];
        result = intersect2OrderedSets(a, b);
    }
    return result;
}

How to use:

intersectOrderedSets(a,b,c);