Check if every element in one array is in a second array

Solution 1:

Do you have to support crummy browsers? If not, the every function should make this easy.

If arr1 is a superset of arr2, then each member in arr2 must be present in arr1

var isSuperset = arr2.every(function(val) { return arr1.indexOf(val) >= 0; });

Here's a fiddle

EDIT

So you're defining superset such that for each element in arr2, it occurs in arr1 the same number of times? I think filter will help you do that (grab the shim from the preceding MDN link to support older browsers):

var isSuperset = arr2.every(function (val) { 
    var numIn1 = arr1.filter(function(el) { return el === val;  }).length;
    var numIn2 = arr2.filter(function(el) { return el === val;  }).length;
    return numIn1 === numIn2;   
});

Updated Fiddle

END EDIT

If you do want to support older browsers, the MDN link above has a shim you can add, which I reproduce here for your convenience:

if (!Array.prototype.every)  
{  
  Array.prototype.every = function(fun /*, thisp */)  
  {  
    "use strict";  

    if (this == null)  
      throw new TypeError();  

    var t = Object(this);  
    var len = t.length >>> 0;  
    if (typeof fun != "function")  
      throw new TypeError();  

    var thisp = arguments[1];  
    for (var i = 0; i < len; i++)  
    {  
      if (i in t && !fun.call(thisp, t[i], i, t))  
        return false;  
    }  

    return true;  
  };  
}  

EDIT

Note that this will be an O(N²) algorithm, so avoid running it on large arrays.

Solution 2:

One option is to sort the two arrays, then traverse both, comparing elements. If an element in the sub-bag candidate is not found in the super-bag, the former is not a sub-bag. Sorting is generally O(n*log(n)) and the comparison is O(max(s,t)), where s and t are the array sizes, for a total time complexity of O(m*log(m)), where m=max(s,t).

function superbag(sup, sub) {
    sup.sort();
    sub.sort();
    var i, j;
    for (i=0,j=0; i<sup.length && j<sub.length;) {
        if (sup[i] < sub[j]) {
            ++i;
        } else if (sup[i] == sub[j]) {
            ++i; ++j;
        } else {
            // sub[j] not in sup, so sub not subbag
            return false;
        }
    }
    // make sure there are no elements left in sub
    return j == sub.length;
}

If the elements in the actual code are integers, you can use a special-purpose integer sorting algorithm (such as radix sort) for an overall O(max(s,t)) time complexity, though if the bags are small, the built-in Array.sort will likely run faster than a custom integer sort.

A solution with potentially lesser time-complexity is to create a bag type. Integer bags are particularly easy. Flip the existing arrays for the bags: create an object or an array with the integers as keys and a repeat count for values. Using an array won't waste space by creating as arrays are sparse in Javascript. You can use bag operations for sub-bag or super-bag checks. For example, subtract the super from the sub candidate and test if the result non-empty. Alternatively, the contains operation should be O(1) (or possibly O(log(n))), so looping over the sub-bag candidate and testing if the super-bag containment exceeds the sub-bag's containment for each sub-bag element should be O(n) or O(n*log(n)).

The following is untested. Implementation of isInt left as an exercise.

function IntBag(from) {
    if (from instanceof IntBag) {
        return from.clone();
    } else if (from instanceof Array) {
        for (var i=0; i < from.length) {
            this.add(from[i]);
        }
    } else if (from) {
        for (p in from) {
            /* don't test from.hasOwnProperty(p); all that matters
               is that p and from[p] are ints
             */
            if (isInt(p) && isInt(from[p])) {
                this.add(p, from[p]);
            }
        }
    }
}
IntBag.prototype=[];
IntBag.prototype.size=0;
IntBag.prototype.clone = function() {
    var clone = new IntBag();
    this.each(function(i, count) {
        clone.add(i, count);
    });
    return clone;
};
IntBag.prototype.contains = function(i) {
    if (i in this) {
        return this[i];
    }
    return 0;
};
IntBag.prototype.add = function(i, count) {
    if (!count) {
        count = 1;
    }
    if (i in this) {
        this[i] += count;
    } else {
        this[i] = count;
    }
    this.size += count;
};
IntBag.prototype.remove = function(i, count) {
    if (! i in this) {
        return;
    }
    if (!count) {
        count = 1;
    }
    this[i] -= count;
    if (this[i] > 0) {
        // element is still in bag
        this.size -= count;
    } else {
        // remove element entirely
        this.size -= count + this[i];
        delete this[i];
    }
};
IntBag.prototype.each = function(f) {
    var i;
    foreach (i in this) {
        f(i, this[i]);
    }
};
IntBag.prototype.find = function(p) {
    var result = [];
    var i;
    foreach (i in this.elements) {
        if (p(i, this[i])) {
            return i;
        }
    }
    return null;
};
IntBag.prototype.sub = function(other) {
    other.each(function(i, count) {
        this.remove(i, count);
    });
    return this;
};
IntBag.prototype.union = function(other) {
    var union = this.clone();
    other.each(function(i, count) {
        if (union.contains(i) < count) {
            union.add(i, count - union.contains(i));
        }
    });
    return union;
};
IntBag.prototype.intersect = function(other) {
    var intersection = new IntBag();
    this.each(function (i, count) {
        if (other.contains(i)) {
            intersection.add(i, Math.min(count, other.contains(i)));
        }
    });
    return intersection;
};
IntBag.prototype.diff = function(other) {
    var mine = this.clone();
    mine.sub(other);
    var others = other.clone();
    others.sub(this);
    mine.union(others);
    return mine;
};
IntBag.prototype.subbag = function(super) {
    return this.size <= super.size
       && null !== this.find(
           function (i, count) {
               return super.contains(i) < this.contains(i);
           }));
};

See also "comparing javascript arrays" for an example implementation of a set of objects, should you ever wish to disallow repetition of elements.

Solution 3:

No one has posted a recursive function yet and those are always fun. Call it like arr1.containsArray( arr2 ).

Demo: http://jsfiddle.net/ThinkingStiff/X9jed/

Array.prototype.containsArray = function ( array /*, index, last*/ ) {

    if( arguments[1] ) {
        var index = arguments[1], last = arguments[2];
    } else {
        var index = 0, last = 0; this.sort(); array.sort();
    };

    return index == array.length
        || ( last = this.indexOf( array[index], last ) ) > -1
        && this.containsArray( array, ++index, ++last );

};

Solution 4:

Found this on github lodash library. This function use built in functions to solve the problem. .includes() , .indexOf() and .every()

var array1 = ['A', 'B', 'C', 'D', 'E'];
var array2 = ['B', 'C', 'E'];
var array3 = ['B', 'C', 'Z'];
var array4 = [];

function arrayContainsArray (superset, subset) {
  if (0 === subset.length) {
    return false;
  }
  return subset.every(function (value) {
    return (superset.includes(value));
  });
}

 function arrayContainsArray1 (superset, subset) {
   if (0 === subset.length) {
     return false;
   }
   return subset.every(function (value) {
     return (superset.indexOf(value) >= 0);
   });
}

console.log(arrayContainsArray(array1,array2)); //true
console.log(arrayContainsArray(array1,array3)); //false
console.log(arrayContainsArray(array1,array4)); //false

console.log(arrayContainsArray1(array1,array2)); //true
console.log(arrayContainsArray1(array1,array3)); //false
console.log(arrayContainsArray1(array1,array4)); //false