Alternating chars N times

Question

Write a function that returns a string length N of alternating strings: "A" and "B", starting with the 1st string. How could this be improved for be more performance or efficiency?

Eg: num = 5, should return "ABABA" & given num = 2, should return "AB".

function solution(num) {
	const str1 = 'A', str2 = 'B';
	let res = [];

	for(let i = 1; res.length < num; i++ ) { 
		res.push(i % 2 ? str1 : str2);
	}

  console.log(res.join(''))
}

solution(5) // return "ABABA"

Answer 1

A short code review;

• Your code does not work, it should return something instead it calls console.log

• Your for loop is icky, either update i and check i

  for(let i = 0; i < num; i++ ))

  or check res.length

  while(res.length < num)

• solution is too generic for this function name

• console.log() is missing a semi-colon

• Performance;

  • Checking whether i is odd every time seems like much, you can avoid that
  • RememberString.repeat(), you could repeat by calling "AB".repeat(num/2) and add an "A" if num is odd.

Obligatory rewrite

    //repeatAB(5) => "ABABA"
    function repeatAB(n) {
        return "AB".repeat(n/2) + (n%2 ? "A" : "");
    }

    console.log(repeatAB(5)) // returns "ABABA"

I ran this approach against the other approaches for the heck of it, this approach destroys the other approaches (at least visually):

http://jsbench.github.io/#11a2b8507a7f9290c6ef6cda596f8179

Answer 2

Why not use doubling to get a logarithmic number of steps?

    const str1 = 'A', str2 = 'B';
    let res = '';

    if (num >= 1)
        res = str1;              // 'A'
    if (num >= 2)
        res += str2;             // 'AB'

    while (2 * res.length < num) { 
        res += res;              // 'ABAB', 'ABABABAB', ...
    }
    if (res.length < num) {
        res += res.substring(0, num - res.length)
    }
    console.log(res)

EDIT

In the above code I try do create the resulting string of exactly the length required, which costs an additional conditional. We can also get rid of that at the cost of creating the string too long temporarily. The central part would then look like:

    while (res.length < num) { 
        res += res;              // 'ABAB', 'ABABABAB', ...
    }
    res = res.substring(0, num)

Answer 3

No need to use abbreviation when you can use meaningful name. For example, use result instead of res. Also if you can find a better name for your function, it will improve the readability. I will use alternateLetters but I think we can do better :)

To achieve your problem, you can use a for loop or a generate a range then use map to assign the letter.

function alternateLetters(num) {
    const a = 'A', b = 'B';
    const result = [...Array(num)].map((_, i) => i % 2 ? b : a);
    console.log(result.join(''))
}

alternateLetters(5) // return "ABABA"

if you prefer, you can also generate result with keys():

const result = [...Array(num).keys()].map(i => i % 2 ? b : a);

By the way I would make it more generic by passing 'A' and 'B' as parameter:

function alternateLetters(num, a, b) {
    const result = [...Array(num)].map((_, i) => i % 2 ? b : a);
    return result.join('');
}

console.log(alternateLetters(5, 'A', 'B')) // return "ABABA"

If we have more context about this problem, we may even do better. For example, are "A" and "B" always constant?

Answer 4

Optimization tips with benchmark demo tests:

Initial approach "issues":

• for(let i = 1; res.length < num; i++ ). Instead of calculation the resulting array length res.length on each loop iteration, you can just replace it with i <= num

• let res = []; ... res.push() ... res.join(). While the final result should eventually be a string we can declare the result holder as string and perform simple concatenation:

  let res = '';
  ...
  for (let i = 1; i <= num; i++ ) { 
      res += (i % 2 ? str1 : str2);
  }
  

---

The below approach is a bit more performant, in my opinion (as well as benchmarks shown that), using the mentioned improvements:

function cycle(num) {
    const str1 = 'A', str2 = 'B';
    let res = '';

    for (let i=1; i <= num; i++) { 
        res += (i % 2 ? str1 : str2);
    }
    console.log(res)
}

---

In prepared benchmarks I've named

• the initial function (for loop + res array) as cycle1
• my approach - as cycle2
• and the function from previous answer alternateLetters - named accordingly

Shared link on http://jsbench.github.io: http://jsbench.github.io/#be710bc0a4b288e024d5bd67286b1c8c

Another benchmark were run on https://jsbench.me/ and the results are as below:

Answer 5

Try using this,

if n == 1:
    print "A"
elif n%2 == 0:
    temp = (n/2);
    print "AB"*temp
else:
    temp = (n-1)/2
    print "AB"*temp + "A"

In above pseudo code, you can see if n is even then print "AB" * (n/2) times and if odd reduce one number and print "AB" divided by 2 times and add "A"