Return even and odd numbers between dynamic ranges in javascript

Question

I am trying to write a code(function) that takes in two parameters a and b, and returns all the even number between a and b if a is greater than b else it should return the odd numbers between a and b.

this is my code.

function number_ranges (a, b) {

    let numbers = [];

    if (a > b){
        for (let i = b; i < a; i++){
            if (i > b){
                numbers.push(i);
            }
        }
    }else{
        for (let i = a; i < b; i++){
            if (i > a){
                numbers.push(i);
            }
        }
    }

    const result = numbers.filter(function(num){
        return a > b ? num % 2 === 0: num % 2 === 1;
    });
    return result;
}

I would like to see a different approach because i cant seem to be able to pass all the test cases

Why are you pushing all of the number in and than filtering? There is no need.... — epascarello
– epascarello, Commented Sep 27, 2018 at 13:59
Do you have an example of a test case where it's not working? — Explosion Pills
– Explosion Pills, Commented Sep 27, 2018 at 14:02
if(i > a) so why not just start it at i=a+1 and not have to do that check? — epascarello
– epascarello, Commented Sep 27, 2018 at 14:02
no the order doesnt matter but naturally it would be sorted since i would start iterating from the lowest.....correct me if i am wrong though i am a newbie — Oparah Dimkpa
– Oparah Dimkpa, Commented Sep 27, 2018 at 14:36

Nina Scholz · Accepted Answer · 2018-09-27 14:15:47Z

1

You could take a sningle loop and a check if the value is even or or in combination with a check for the order.

function getEvenOdd(a, b) {
    var inc = +(a < b) || -1,
        i,
        result = [];
     for (i = a; i !== b + inc; i += inc) {
         if (+(inc === 1) !== i % 2) {
             result.push(i);
         }
     }
     return result;
}

// even
console.log(getEvenOdd(3, 10).join(' ')); // [4, 6, 8, 10]
console.log(getEvenOdd(4, 10).join(' ')); // [4, 6, 8, 10]
console.log(getEvenOdd(3, 9).join(' '));  // [4, 6, 8]
console.log(getEvenOdd(4, 9).join(' '));  // [4, 6, 8]

// odd
console.log(getEvenOdd(10, 3).join(' ')); // [9, 7, 5, 3]
console.log(getEvenOdd(10, 4).join(' ')); // [9, 7, 5]
console.log(getEvenOdd(9, 3).join(' '));  // [9, 7, 5, 3]
console.log(getEvenOdd(9, 4).join(' '));  // [9, 7, 5]

answered Sep 27, 2018 at 14:15

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Luis felipe De jesus Munoz · Accepted Answer · 2018-09-27 17:26:53Z

1

I used this function f=(a,b)=>[...ArrayMath.max(a,b))].map((q,i)=>i).filter(z=>(a<b?!(z%2):z%2)&&z>=Math.min(a,b))

I first create a range from 1 to b as follow [...Array(Math.max(a,b))].map((q,i)=>i)

Then I filter with a<b?!(z%2):z%2. If a greater than b then a keep the even numbers, else the odds.

Also I keep only those number greater or equal to the first parameter using &&z>=Math.min(a,b)

f=(a,b)=>[...Array(Math.max(a,b))].map((q,i)=>i).filter(z=>(a<b?!(z%2):z%2)&&z>=Math.min(a,b))
console.log(f(10, 5))
console.log(f(5, 10))

edited Sep 27, 2018 at 17:26

answered Sep 27, 2018 at 14:13

Luis felipe De jesus Munoz

8,0133 gold badges38 silver badges54 bronze badges

3 Comments

Joe25 Over a year ago

Your function returns evens when a < b, in the opposite case it returns an empty array, please check the code.

Luis felipe De jesus Munoz Over a year ago

@Joe25 Thanks for noticing, I got it inverted (not too good at english :c).

Oparah Dimkpa Over a year ago

yeah it works pretty nice except that it returns the last number for the even number filter ....it's supposed to return only numbers in between ignoring the start and end point

Collectives™ on Stack Overflow

Return even and odd numbers between dynamic ranges in javascript

2 Answers 2

Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related