Finding every second element in a repeating pattern

Question

Data with repeated 'i's followed by 'i's and/or 't's.

data = ['i','t','t','i','i','t','t','t']

Trying to retrieve the index of the last 't' in the pattern ['i','t','t']:

[2,6] // ['i','t','t','i','i','t','t','t'] # position of the returned 't's
      //   _______ ^       _______ ^

I'm looking for a non-recursive solution using (pure) functions only, using ramdajs for example.

Tried to use reduce and transduce, but unsuccessful sofar.

Answer 1

One approach would be to use R.aperture to iterate over a 3-element sliding window of the data list, then tracking the position of any sub-list that equals the pattern ['i', 't', 't'].

const data = ['i','t','t','i','i','t','t','t']

const isPattern = R.equals(['i', 't', 't'])

const reduceWithIdx = R.addIndex(R.reduce)

const positions = reduceWithIdx((idxs, next, idx) =>
  isPattern(next) ? R.append(idx + 2, idxs) : idxs
, [], R.aperture(3, data))

console.log(positions)

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A point-free version of this approach could look something like the following, though whether this is preferable comes down to a preference of style/readability.

const data = ['i','t','t','i','i','t','t','t']

const isPattern = R.equals(['i', 't', 't'])

const run = R.pipe(
  // create sliding window of 3 elements
  R.aperture(3),

  // zip sliding window with index
  R.chain(R.zip, R.compose(R.range(0), R.length)),

  // filter matching pattern
  R.filter(R.compose(isPattern, R.nth(1))),

  // extract index
  R.map(R.compose(R.add(2), R.head))
)

console.log(run(data))

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