How to get the difference between two arrays in JavaScript with repeating values

Question

How can I check to see if one array is contained in another and return the missing values? I have found ways to do this in this post but none account for repeating values in the arrays. For example, I am trying to do something like this:

getDiff([1, 2, 3], [1, 2, 3, 4]) --> []

getDiff([1, 2, 2, 3], [1, 2, 3, 4]) --> [2]

getDiff(["A", "B", "C"], ["B", "B", "A", "C"]) --> []

getDiff(["B", "B", "B"], [3, 2, 1]) --> ["B", "B", "B"]

If you post what you have tried, someone can more readily help you fix/finish it. — Scott Hunter, Jun 30 '21 at 19:53
If the order of the result doesn't matter, this would be simplified by sorting the two arrays. — Scott Hunter, Jun 30 '21 at 19:54

raina77ow · Accepted Answer · 2021-06-30T20:34:38.863

One possible approach:

function getRepeatingDiff(source, diffed) {
  const diffedCounter = diffed.reduce((acc, el) => {
    acc[el] = (acc[el] || 0) + 1
    return acc;
  }, {});
  
  const diff = source.reduce((acc, el) => {
    return diffedCounter[el]-- > 0 ? acc : [ ...acc, el ];
  }, []);
  
  return diff;
}

console.log( getRepeatingDiff([1, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff([1, 3, 2, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff(["A", "B", "C"], ["B", "B", "A", "C"]) );
console.log( getRepeatingDiff(["B", "B", "B"], [3, 2, 1]) );

Essentially, it's a two-step process: calculate the number of items in diffed array, then go through source array one more time - and add a new item into diff resulting array for each copy missing in diffedCounter. This particular implementation has one deficiency though: as it uses Object to collect the counts, it doesn't differentiate between 3 as number and '3' as a string when counting elements.

This might be fixed in two different ways: either switch to Map (but that'll make code more complicated, as there's no such things as decrementing for map values) - or just use type prefixes when creating a counter keys. Here's the approach based on former:

function getRepeatingDiff(source, diffed) {
  const diffedCounter = diffed.reduce((map, el) => {
    const prev = map.get(el);
    return map.set(el, (prev || 0 ) + 1);
  }, new Map());
  
  const diff = source.reduce((acc, el) => {
    const remainingCount = diffedCounter.get(el);
    if (remainingCount) diffedCounter.set(el, remainingCount - 1);
    return remainingCount ? acc : [ ...acc, el ];
  }, []);
  
  return diff;
}

console.log( getRepeatingDiff([1, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff([1, 3, 2, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff(["A", "B", "C"], ["B", "B", "A", "C"]) );
console.log( getRepeatingDiff(["B", "B", "B"], [3, 2, 1]) );

I realized it's way too inefficient to go through the source array twice, so updated my answer. It still doesn't differentiate between numbers and strings though. — raina77ow, Jun 30 '21 at 20:16

How to get the difference between two arrays in JavaScript with repeating values

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