A function that returns an array which is the intersection of two other arrays

Question

function arraysCommon(array1, array2) {
  return array1.filter(x => array2.includes(x));
}

This function does not work the way I want it to. For instance given array1 = [1,2,3,2,1] and array2 = [5,4,3,2,1] it returns [1,2,3,2,1], since the elements 1,2,3 are seen in both arrays. But I want it to return [1,2,3] in that order since 1,2,3 are seen only once in array2 and are treated as seperate entities.

So pretty much the functionality should be that

• Each element in the first array can map to at most one element in the second array.
• Duplicated elements in each array are treated as separate entities.
• the first array determines the order

I have attempted to loop through the arrays and check and compare the number of duplicates in each array but I can't seem to get the logic working correctly. Is there a different way to approach this?

I've attached an image of two Venn diagrams that might clarify the difference

Answer 1

Unfortunately, it gets more complicated because you need to know what numbers you have already added. In this case you need a temporary array to hold the result. We also need to track if a number exists in the array two times.

Try this:

function arraysCommon(array1, array2) {
  //Copy array2 by duplicating and spreading the elements into another array.
  var copyArray2 = [...array2];
  //Temperary Array
  var temp = [];
  for (let x of array1) {
      //Check if the element is in the first array and not already added to the temp array
      if (copyArray2.includes(x)) {
          temp.push(x);
          //Remove item from copy array2 so it cannot be used anymore
          copyArray2.splice(copyArray2.indexOf(x), 1);
      }
  } 
  //Return the temp array
  return temp;
}

console.log(arraysCommon([1,2,3,2,1], [5,4,3,2,1]))
console.log(arraysCommon([1,2,3,2,1], [2,2,3,3,4]))