Discussion on: Javascript Sock Merchant Challenge - Solution 2

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Replies for: I don't think you even need the last reducer. Just keep a running counter. function sockMerchant(n, ar) { const colors = {} let pairs = ...

@theodesp this was just me commenting on the code without changing the algorithm at all.
My own solution (which does use a counter) was posted separately:

Mihail Malostanidis

Apr 27

My own solution would be:

function stockAndCount( n, arr ) {
  let pairs = 0
  const unpaired = new Set()

  for (const color of arr)
    if (unpaired.has(color)) {
      pairs++
      unpaired.delete(color)
    } else unpaired.add(color)

  return pairs
}

This doesn't iterate again over the colors encountered, and uses the minimal memory required for the task: The running total of pairs and the unpaired colors, encountering which again would be the completion of a pair.