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

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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"]
```

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]) );```