Find missing element by comparing 2 arrays in Javascript

Question

For some reason I\'m having some serious difficulty wrapping my mind around this problem. I need this JS function that accepts 2 arrays, compares the 2, and then returns a s

Answer 1

Problem statement:

Find the element that is missing in the currentArray that was there in the previous array.

previousArray.filter(function(x) {  // return elements in previousArray matching...
    return !currentArray.includes(x);  // "this element doesn't exist in currentArray"
})

(This is as bad as writing two nested for-loops, i.e. O(N²) time. This can be made more efficient if necessary, by creating a temporary object out of currentArray, and using it as a hashtable for O(1) queries. For example:)

var inCurrent={}; currentArray.forEach(function(x){ inCurrent[x]=true });

So then we have a temporary lookup table, e.g.

previousArray = [1,2,3]
currentArray = [2,3];
inCurrent == {2:true, 3:true};

Then the function doesn't need to repeatedly search the currentArray every time which would be an O(N) substep; it can instantly check whether it's in currentArray in O(1) time. Since .filter is called N times, this results in an O(N) rather than O(N²) total time:

previousArray.filter(function(x) {
    return !inCurrent[x]
})

Alternatively, here it is for-loop style:

var inCurrent = {};
var removedElements = []
for(let x of currentArray)
    inCurrent[x] = true;
for(let x of previousArray)
    if(!inCurrent[x])
        removedElements.push(x)
        //break; // alternatively just break if exactly one missing element
console.log(`the missing elements are ${removedElements}`)

Or just use modern data structures, which make the code much more obvious:

var currentSet = new Set(currentArray);
return previousArray.filter(x => !currentSet.has(x))