I'm working through the Free Code Camp syllabus and I'm on to Intermediate Javascript Algorithims. This Pairwise problem was the last challenge in that section. The section came just after Object Oriented Javascript. So I figured they were looking for an OO solution, but the instructions included a link to MDN's array.reduce(). My solution doesn't use array.reduce() and I'd really appreciate some feedback on what I could have done better to make my code more compact and efficient. It feels a little clunky but passes all the tests.

The instructions

Return the sum of all indices of elements of 'arr' that can be paired with one other element to form a sum that equals the value in the second argument 'arg'. If multiple sums are possible, return the smallest sum. Once an element has been used, it cannot be reused to pair with another.

For example, pairwise([1, 4, 2, 3, 0, 5], 7) should return 11 because 4, 2, 3 and 5 can be paired with each other to equal 7.

pairwise([1, 3, 2, 4], 4) would only equal 1, because only the first two elements can be paired to equal 4, and the first element has an index of 0!

Remember to use RSAP if you get stuck. Try to pair program. Write your own code.

Here are some helpful links:

Array.reduce()

My Solution

function pairwise(arr, arg) {
    this.objects = [];
    var total = 0;

    function Element(value, index) {
        this.value = value;
        this.index = index;
        this.used = 0;
    }

    for (var i = 0; i < arr.length; i++) {
        this.objects.push(new Element(arr[i], i));
    }

    for (var j = 0; j < objects.length; j++) {
        if (objects[j].used === 0) {
            for (var k = 0; k < objects.length; k++) {
                if (objects[k].used === 0 && objects[k].index != objects[j].index) {
                    if (arg - objects[j].value == objects[k].value) {
                        total = total + objects[j].index + objects[k].index;
                        objects[j].used = 1;
                        objects[k].used = 1;
                        break;
                    }
                }
            }
        }
    }
    
    return total;
}

pairwise([1,1,1], 2);