CODEWITHSUNDEEP
javascriptarrays
Assaf
Assaf

Reputation: 1124

JavaScript - better approach to a problem of spliting an array

The problem I have to solve is the following -

splitEven(a) – The function accepts an array of integers and would return an array that contains the values that are located in the even indices of the original array, sorted in ascending order. For example, if the function gets the array [3, 1, 6, 7, 4] it would return the array [3, 4, 6]

This is my solution - 

    function splitEven(a){
    var b = [];
    var even = function(element){
        return element % 2 === 0;
    }
    for (var i = 0; i < a.length; i++) {
        var c = even(a[i]);
        if (c) {
            b.push(a[i])
        }
    }
    return b;
}

However I feel it is not best practice to do what I have done. I am familier with Java, and I think I tend to solve problems not in the right way.

Can you think of a better approach for this problem, to improve my practice?

Upvotes: 1

Views: 64

Answers (3)

Geshode
Geshode

Reputation: 3764

You could just increase by 2 instead of 1 and in the end use sort().

function splitEven(array){
    var b = [];
    for (i = 0; i < array.length; i += 2){
      b.push(array[i]);
    }

    b.sort((a,b) => {return a-b;});

    return b;
}

Upvotes: 1

jorgeRamirez
jorgeRamirez

Reputation: 31

Another approach to simplify your code is to use an IF statement nested in your FOR loop.

function splitEven(array) {
  var even = [];

  for (let i = 0; i < array.length; i++) {
    if (array[i] % 2 == 0) {
      even.push(array[i]);
    }
  }
  return even;
}

Upvotes: 1

CertainPerformance
CertainPerformance

Reputation: 370819

You can use filter to create an array that contains only the even-numbered indicies in the original array, and then sort() that array:

const splitEven = arr => arr
  .filter((_, i) => i % 2 === 0)
  .sort((a, b) => a - b);

console.log(
  splitEven([3, 1, 6, 7, 4])
);
console.log(
  splitEven([3, 1, 6, 7, 4, 11, 12])
);

Upvotes: 6

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