freeCodeCamp Challenge Guide: Diff Two Arrays

Guide
42

.filter(el => !arr2.includes(el))

what does ! before arr mean?

43

Hereā€™s My solution using indexOf() . Seems simple to understand. Well, hint helped me:sweat_smile:[spoiler]This text will be blurred

function diffArray(arr1, arr2) {
  var newArr = [];
  // Same, same; but different.
  
  var x = arr1.concat(arr2);
  
  for(var i = 0 ; i<x.length; i++){
    if(arr1.indexOf(x[i]) === -1 || arr2.indexOf(x[i]) === -1 )
      newArr.push(x[i]);
  }
    
  
  return newArr;
}

diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);

[/spoiler]

44

Hereā€™s My solution using indexOf() . Seems simple to understand. Well, hint helped me :sweat_smile:[spoiler]

function diffArray(arr1, arr2) {
  var newArr = [];
  // Same, same; but different.
  
  var x = arr1.concat(arr2);
  
  for(var i = 0 ; i<x.length; i++){
    if(arr1.indexOf(x[i]) === -1 || arr2.indexOf(x[i]) === -1 )
      newArr.push(x[i]);
  }
    
  
  return newArr;
}

diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);

[/spoiler]

45 
function diffArray(arr1, arr2) {
  return [...arr1, ...arr2].filter(item => !arr1.includes(item) || !arr2.includes(item));
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
1 Like
46

Had to read the hints to figure this out. Solution ended up being more or less Intermediate solution but not as optimized

function diffArray(arr1, arr2) {
  var arr = [];
  var newArr=[];
  // Same, same; but different.
  arr = arr1.concat(arr2);
  
  newArr = arr.filter(function(x){
    if((arr1.indexOf(x)===-1) || (arr2.indexOf(x)===-1)){
      return x;
    }
  });
 return newArr;
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
1 Like
47

I was originally writing something basically like this,

  1. merge and sort it out
  2. Delete duplicates
  3. Push to new array if conditions are met

but I couldnā€™t get the IF statement line to work quite right, I had most of it down ended up scraping it once i read the hints

e.g.

[1,1,2,2,3,3,4,5,5]

If the value [4] has no adjacent [4] next to it it must be unique

48

function diffArray(arr1, arr2) {

var newArr = [];

function resultDiffArray(a1,a2){
for(var i=0;i<a1.length;i++){
if(a2.indexOf(a1[i])==-1){
newArr.push(a1[i]);
}
}
}

resultDiffArray(arr1,arr2);
resultDiffArray(arr2,arr1);

return newArr;
}

49

So Iā€™m really new to this, and to be honest I wasnā€™t expecting this code to work but it didā€¦Iā€™m not really sure if I fully understand why it worked.

function diffArray(arr1, arr2) {
var newArr = [];
var otherArr= [];
newArr = arr1.concat(arr2);
for (var i = 0; i < newArr.length; i++) {
otherArr = newArr.filter(function(i){
return arr1.indexOf(i) < 0 || arr2.indexOf(i) < 0;
});
}
return otherArr;
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);

50

Simplified, with help from rmdawson71. This works!

function diffArray(arr1, arr2) {
 return arr1.concat(arr2).filter(item=>!arr1.includes(item) || !arr2.includes(item)
  );
}
51

my code

function diffArray(arr1, arr2) {
var newArr,temp,temp1;
// Same, same; but different.

temp=arr1.filter(function(el) 
          {
            return arr2.indexOf(el) == -1; 
          
          });

temp1=arr2.filter(function(el) 
          {
            return arr1.indexOf(el) == -1; 
          
          });

newArr=temp.concat(temp1);

return newArr;

}

diffArray([1, 2, 12, 3, 5], [1, 2, 3, 4, 5]);

53

Since my understanding of filter function is really poor I came up with this:

function diffArray(arr1, arr2) {
  var newArr = [];
  var result = [];
  // Same, same; but different.
  
  newArr = arr1.concat(arr2);
  
      for (var i=0; i<newArr.length; i++) {
        
        if (arr1.includes(newArr[i]) && !arr2.includes(newArr[i]) || !arr1.includes(newArr[i]) && arr2.includes(newArr[i])) {
          
          result.push(newArr[i]);
          console.log(result);
          
        } 
        
      }
  
  return result;
}

diffArray(['a', 'b', 'c', 'e'], ['a', 'b', 'c', 'd', 'e']);
54

my solution is a bit different I guess, using concat and splice

function diffArray(arr1, arr2) {
  var newArr = [];
  newArr = arr1.concat(arr2).sort(); //combine the two arrays in one and sort them
  for(i=0;i<newArr.length;i++){ //loop through the full new array
    if(newArr[i] == newArr[i+1]){ //if two consecutive elements are similar, delete them both
      newArr.splice(i,2);
      i = i-2;//after deleting for example element 0 and 1 while the loop is at count 0, next when it goes to element 1 it will be the old element 3 because 2 elements are gone, so if we remove two we count two steps back
    }
  }
  return newArr;
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
1 Like
55 
function diffArray(arr1, arr2) {
  // Same, same; but different.
  
  // Setup temp array to hold arr1 since filter() will change contents of arr1
  var tempArr1 = arr1; 
  return arr1.filter(item => arr2.indexOf(item) === -1).concat(arr2.filter(item => tempArr1.indexOf(item) === -1));
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);

This was my solution. Thoughts?

1 Like
58

Hello,

Could anyone explain me the solution below? I am not sure if I get what exactly those 3 dots do. I think they are ā€œspread syntaxā€ but still donā€™t understand how it works.

function diffArray(arr1, arr2) {
  return [
    ...diff(arr1, arr2),
    ...diff(arr2, arr1)
  ]
  
  function diff(a, b) {
    return a.filter(item => b.indexOf(item) === -1);
  }
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
1 Like