Htydjtk

I have the following question:

Write a function that returns true if all integers in an array are factors of a number, and false otherwise.

I tried the code below:

function checkFactors(factors, num) {



  for (let i=0; i<factors.length; i++){

    let element = factors[i];

      console.log(element)



    if (num % element !== 0){

      return false 

    }

    else {

      return true

    }

  }

}



console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false

My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.

Just place return true out of for loop,

If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number

Let's understand by 1st example

• On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.


• So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true

function checkFactors(factors, num) {



  for (let i=0; i<factors.length; i++){

    let element = factors[i];

    if (num % element !== 0){

      return false 

    }

  }

  return true

}







console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false

console.log(checkFactors([1, 2], 2))

In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like

1. keep the failing case return value inside for loop


2. keep the passing case return value at the end of function

JS have a inbuilt method Array.every for such cases

function checkFactors(factors, num) {

   return factors.every(element => num % element === 0);

}

console.log(checkFactors([1, 2, 3, 8], 12)); 

console.log(checkFactors([1, 2], 2));

You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:

Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹

You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.

You realize that your boss actually wanted you to do:

Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²

Or in code:

 // ¹

  function checkChocolates(chocolates) {

    for(const chocolate of chocolates) {

       if(isTasty(chocolate)) {

         return true;

       } else {

         return false;

       }

    }

 }



 // ²

 function checkChocolates(chocolates) {

   for(const chocolate of chocolates) {

     if(isTasty(chocolate)) {

       continue; // this could be omitted, as a loop keeps looping nevertheless

     } else {

       return false;

     }

   }

   return true;

 }

As this is a very common task in programming, there is already a shorter way to express this:

 if(chocolates.every(isTasty)) {

   alert("all chocolates are fine");

 } else {

    alert("Oh, that doesnt taste good");

 }

whereas isTasty is a function taking a chocolate and returning either true or false.

If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)

Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.

The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.

Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.

function checkFactors(factors, num) {

  let flag = true;

  for (let i=0; i<factors.length; i++){

    let element = factors[i];

    if (num % element !== 0){

      flag = false;

      break;

    }

  }

  return flag;

}



console.log(checkFactors([1, 2, 3, 8], 12)); 

console.log(checkFactors([1, 2], 2));

console.log(checkFactors([2, 4, 3, 6, 9], 12));

console.log(checkFactors([3, 5, 2, 6, 9], 15));

console.log(checkFactors([4, 2, 8, 1], 16));

You can also use Array.every to check the same in a concise way:

function checkFactors(factors, num) {

   return factors.every(element => num % element === 0);

}

console.log(checkFactors([1, 2, 3, 8], 12)); 

console.log(checkFactors([1, 2], 2));

console.log(checkFactors([2, 4, 3, 6, 9], 12));

console.log(checkFactors([3, 5, 2, 6, 9], 15));

console.log(checkFactors([4, 2, 8, 1], 16));

Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.

function checkFactors(factors, num) {



  for (let i=0; i<factors.length; i++){

    let element = factors[i];

      console.log(element)



    if (num % element !== 0){

      return false 

    }

  }

  return true;

}