Partitioning input strings by even and odd indexes

Question

I have been trying to learn javascript for a year now, and have just embarked on HackerRank's 30 days of code. After a couple of days I finally solved this problem. Here is the task (if you click the link, press x to the signup page):

Task

Given a string S of length N that is indexed from 0 to N-1, print its even-indexed and odd-indexed characters as space-separated strings on a single line (see the Sample below for more detail).

Sample Input

2
Hacker
Rank

Sample Output

Hce akr
Rn ak

Here is my solution. Any performance optimizations I could make here?:

function processData(input) { //Enter your code here //split the string into an array let words = input.split("\n").slice(1), odd = "", even = "", newStr = ""; //turn each string inside the array into its own array let wordsArr = words.map((val) => { return val.split(""); }) //outer loop for the outer array for (let i = 0, len = wordsArr.length; i < len; i++){ //inner loop for the inner arrays for (let j = 0, len2 = wordsArr[i].length; j < len2; j++){ //if the index is even or odd, add to the respective strings if (j % 2 === 0){ even += wordsArr[i][j] } else if (j % 2 !== 0) { odd += wordsArr[i][j]; } //if the end of the array is detected, print the string //and set odd/even strings to be blank for the next iteration if (j == len2 - 1){ newStr = `${even} ${odd}`; even = ""; odd = ""; console.log(newStr); } } } } 

Answer 1

One possible solution that achieves the described result is:

`${s.replace(/./g, (s, pos) => [s, ''][pos & 1])} ${s.replace(/./g, (s, pos) => ['', s][pos & 1])}` 

...But in order to answer to your question:

// 0 let wordsArr = words.map((val) => { return val.split(""); }) // can be rewritten as let wordsArr = words.map(val => val.split('')) // 1 for (let i = 0, len = wordsArr.length; i < len; i++){ // personally I use i += 1 // following to Douglas Crockford // this is what actually happens under the hood // 2 if (j % 2 === 0){ even += wordsArr[i][j] } else if (j % 2 !== 0) { odd += wordsArr[i][j]; } // can be symplified to [even, odd][j & 1] += wordsArr[i][j]; 

Happy coding!

Answer 2

You can make a separate function, which takes a string and returns the two new strings. This simplifies the code by getting rid of nested loops, 2d arrays, and resetting temporary variables.

function processData(input) { let words = input.split('\n').slice(1).map(splitString); //This just prints the result in the specified format words.forEach(e => console.log(e.join(' '))); } 

The new splitString function can also be simplified. You can just use else instead of if else, since it's always true when the first condition is false. There is also no reason to check for the end of the array, since the loop will end when this happens. You don't even have to split the string into an array, since characters in a string for be accessed like array indexes and the both have the .length property.

function splitString(str) { let even = ''; let odd = ''; for(let i = 0; i < str.length; i++) { if(i%2 === 0) { even += str[i]; } else { odd += str[i]; } } return [even, odd]; } 

You can make this function even shorter if you use some of the built-in array functions, like filter or reduce.