FreeCodeCampBeta -Iterate Through All an Array's Items Using For Loops - Question

JavaScript
1

Hello, is there a better way to do this challenge than mine:
Link: https://beta.freecodecamp.org/en/challenges/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops


function filteredArray(arr, elem) {
	let newArr = []; 
	let arrL = arr.length; //4
		
	for( let k = 0; k < arrL; k++ ) {
		for( let i = 0; i < arr[k].length; i++ ) {
			if( arr[k][i] == elem ) {
				arr[k].splice(0, arr[k].length);
			}
		}	
	}
	
	for( let d = 0; d < arr.length; d++) {
		if( arr[d].length > 0) {
			newArr.push( arr[d] );
		}
	}
	return newArr;

};

console.log(filteredArray([[10, 8, 3], [14, 6, 23], [3, 18, 6]], 18));

And by better, i’m sure this part can be a lot better:

	for( let d = 0; d < arr.length; d++) {
		if( arr[d].length > 0) {
			newArr.push( arr[d] );
		}
	}
	return newArr;

I’m looking for some thing to delete empty Array in Array.

5

Hi ;
This is my code . It is a shorter way with indexOf()

function filteredArray(arr, elem) {
let newArr = [];
// change code below this line
for( var i = 0 ; i<arr.length ; i++){
if (arr[i].indexOf(elem) === -1){
newArr.push(arr[i]);
}
}
// change code above this line
return newArr;
}

// change code here to test different cases:
console.log(filteredArray([[3, 2, 3], [1, 6, 3], [3, 13, 26], [19, 3, 9]], 3));

14 Likes
6 
function filteredArray(arr, elem) {
  let newArr = [...arr];
  // change code below this line
  for (let i = 0; i < newArr.length; i++) {
    for (let j = 0; j < newArr[i].length; j++) {
      if (newArr[i][j] == elem) {
        newArr.splice(i, 1);
        i--;   // <---
        break;
      }
    }
  }
  // change code above this line
  return newArr;
}
// change code here to test different cases:
console.log(filteredArray([["trumpets", 2], ["flutes", 4], ["saxophones", 2]], 2));
// return [["flutes", 4]]
1 Like
7

Great solution (even better considering that the indexOf method was a previous challenge).

8 
function filteredArray(arr, elem) {
  let newArr = [];
  // change code below this line
  let contador = 0;
  let longitud = 0;
    for (let i = 0; i<arr.length; i++){
      for (var j=0; j<arr[i].length;j++){
        longitud++;
        if (arr[i][j]==elem){
          contador++;
        }
          if (contador== 0 && longitud == arr[i].length ){
          newArr.push(arr[i]);
          } 
      }
      longitud = 0;
      contador= 0;
    }
  // change code above this line
  return newArr;
}

// change code here to test different cases:
console.log(filteredArray([ ["trumpets", 2], ["flutes", 4], ["saxophones", 2] ], 2));

Really messy solution, but it works

2 Likes
9

hey , can you please explain what the code do ?
i cant quite grasp how u r removing the array …

1 Like
10

Hi,
They’re testing each subarray to see if it contains elem using .indexOf() array method.
.indexOf(elem) returns -1 if the array does not contain elem
If the test comes back negative (does not contain elem), then that subarray gets pushed onto newArr.


let arr = [10,50,75,"one"];
console.log(arr.indexOf(75);  // logs 2, for index 2
console.log(arr.indexOf("cow"); // logs -1, there is no index for cow

function filteredArray(arr, elem) {
  let newArr = [];

  // loop through each subarray
  for (var i = 0; i < arr.length; i++) {
    // if subarray does not  contain elem, then push subarray onto newArr
    if (arr[i].indexOf(elem) === -1) { // -1 indicates elem not found in arr
      newArr.push(arr[i]);
    }
  }
  return newArr;
}
3 Likes
11

Thanks for taking the time to reply!
i understand the logic now .
Many Thanks ^^