JavaScript Algorithm: Array.diff

Implementing a difference function that subtracts one list from another

Photo by Crissy Jarvis on Unsplash

We are going to write a function called arrayDiff that will accept two arrays, a and b, as arguments.

You are given two arrays. The goal of the function is to remove all values in array a that is present in array b. The function should return the array after all the common values have been removed.

Here are some examples:

arrayDiff([], [4,5]) //output: []
arrayDiff([3,4], [3]) // output: [4]
arrayDiff([1,8,2], []) // output: [1, 8, 2]

In the first example, there are no values in a that are in b. The array is empty by default so the function will return an empty array.

In the second example, only 3 from a is found in b so that number is removed and the function returns [4].

In the last example, all values in a are not found in b so the function returns [1,8,2].

To begin, we will create a variable called arr1 that will contain all the values that are present in a but not in b. This is the array the function will return. We assign it an empty array.

let arr1 = [];

If you look at the first and third examples above, we notice that if b is empty, the function will return the a array as-is. This also happens when the a array is empty.

Because of that, we will write an if statement that checks if the array length of either a or b is zero. If it is, return a as-is.

if(b.length === 0 || a.length === 0){    
    return a;  
}

Along with the if-statement, we will also add in an else statement. Inside the else block we will iterate through the a array using the filter() method and filter out all values that exist b.

We assign the resulting array to arr1 and return it.

arr1 = a.filter(num => !(b.includes(num)));

Here is the full if-statement:

if(b.length === 0 || a.length === 0){    
    return a;  
}else{    
    arr1 = a.filter(num => !(b.includes(num)));
    return arr1; 
}