Compare two arrays and find items that are missing in second array

Question

I have 2 arrays that they are identical at first but the user may remove (not able to add but just remove) items from the second array. I want to find items that are in the first array but not in the second one.

I can think of several ways to do this, but as these arrays can be very large, I'm curious to find out if anyone can offer a more efficient way:

$.grep( firstArray, function( n, i ){
  return $.inArray(n, secondArray) == -1;
});

Answer 1

You could try to make use of filter and indexOf array methods as below:

var firstArray = [1,2,3,4,5,6];
var secondArray = [3,4,6];

var result = firstArray.filter(item => secondArray.indexOf(item) == -1);

console.log(result);

Answer 2

Assuming the arrays have the same order, then you could filter with an index as closure.

var array1 = [1, 2, 3, 4, 5, 6, 7, 1, 2, 3],
    array2 = [2, 4, 6, 7, 2],
    missing = array1.filter((i => a => a !== array2[i] || !++i)(0));
    
console.log(missing);

Answer 3

Do a .filter on the array and check if the array 2 does not have the element using .includes

var a1 = [1,2,3,4,5,6];
var a2 = [1,3,5];

var absent = a1.filter(e=>!a2.includes(e));

console.log(absent);

Answer 4

function arr_diff (a1, a2) {
var a = [], diff = [];
for (var i = 0; i < a1.length; i++) {
    a[a1[i]] = true;
}
for (var i = 0; i < a2.length; i++) {
    if (a[a2[i]]) {
        delete a[a2[i]];
    } else {
        a[a2[i]] = true;
    }
}
for (var k in a) {
    diff.push(k);
}
return diff;

}

Use this function to get difference between two sets.
A:Set 1
B:Set 2
A-B:Elements that are present in A but, not in B
Basic set theory

Answer 5

As user can delete items. So you can add those deleted items in a different array.

var deltedItems = [];
var position = 0;

function onDeleteItem(value){
   deltedItems[position] = value;
   position++;
}

Here, you can find all your deleted items in deltedItems variable. By using this logic you can eliminate search cost of your program.