Bean Counting WITHOUT using for loop and regex

Question

I tried to solve the following problem without using the for loop:

"Write a function countBs that takes a string as its only argument and returns a number that indicates how many uppercase “B” characters there are in the string.
console.log(countBs("BBC")); // → 2

Thank you for your help.

Here is the code I wrote so far which doesn't work:

function countBs(word) {
  let count = 0
  if (word.length == 0) {
    return count;
  } else {
    if (word[word.length - 1] == 'B') {
      count++
    }
    return countBs(word.slice(0, word.length - 1))
  }
}
console.log(countBs("BBC"))

Are you supposed to use recursion for this, or why this approach? — CBroe, Jul 07 '20 at 11:39
Does this answer your question? [How to count string occurrence in string?](https://stackoverflow.com/questions/4009756/how-to-count-string-occurrence-in-string) — Argee, Jul 07 '20 at 11:39
@CBroe I was just trying to understand if recursion can be used for this without using for loop or regex. — Lucas Erlacher, Jul 07 '20 at 11:48
@JohnnyMopp thank you for your reply. I tried to implement but it still returns undefined. — Lucas Erlacher, Jul 07 '20 at 11:49
Sure recursion _could_ be used here - but it has little advantage over a simple loop. If anything, it has disadvantages, due to the call stack size recursive calls will cause. — CBroe, Jul 07 '20 at 11:50
@JohnnyMopp you're right. I placed it in the wrong bracket. Thanks a lot for your solution. — Lucas Erlacher, Jul 07 '20 at 12:08
`function countBs(word, count = 0) { if (word.length == 0) { return count; } else { if (word[word.length - 1] == 'B') { count++ } return countBs(word.slice(0, word.length - 1), count) } } console.log(countBs("BBC"))` would be how you can modify your code... — Argee, Jul 07 '20 at 12:25
Because a string can be destructured into an array, we can write a fairly simply recursive version: `const countBs = ([c, ...cs]) => c == undefined ? 0 : (c === 'B' ? 1 : 0) + countBs(cs)`. We could fairly easily convert it to tail-recursive, if TCO ever becomes a JS reality. — Scott Sauyet, Jul 07 '20 at 12:37

score 0 · Answer 1 · answered Jul 07 '20 at 11:41

0

You can easily use regex for this case:

function countBs(word) {
  const matches = word.match(/B/g);

  return matches && matches.length || 0;
}

Basically it globally searches for occurrences of 'B'. If it find it, it returns the length of it, if the matches are null it will return 0.

Simpler less expressive:

function countBs(word) {
  return (word.match(/B/g) || []).length;
}

answered Jul 07 '20 at 11:41

Poul Kruijt

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Thank you Poul for your answer. I would like to understand why my version is not working without using regex. – Lucas Erlacher Jul 07 '20 at 11:47
@LucasErlacher because you initialize the count to 0 at every iteration... if you give the current count to your function, it works well... see my comment above for reference – Argee Jul 07 '20 at 12:27
Thank you @Argee – Lucas Erlacher Jul 08 '20 at 16:16

Bean Counting WITHOUT using for loop and regex

1 Answers1