Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Imperative programming / 3 / e / Solution

The following method belongs inside the source class and uses the helper from part (d). It preserves the existing outgoing count, and a newly encountered user starts with exactly one incoming and zero outgoing payments.

def recordIncoming(user: Int): Unit = {
  require(1 <= user && user <= 99999999, "invalid user identity")
  val bucket = hash(user)
  val p = find(user, table(bucket))
  if (p == null) {
    table(bucket) = new PaymentsHash.Node(user, 1, 0, table(bucket))
  } else {
    require(p.countIn < Int.MaxValue, "incoming count is full")
    p.countIn += 1
  }
}

On a hit, exactly one field is incremented after the overflow check. On a miss, inserting at the bucket head preserves all existing records and gives precisely the required change from implicit zero to one. The given hash function must return an index in $[0,N)$ consistently for a given user. Its actual formula is not needed by this method. The nested private node class is accessible to its companion class in Scala, so retaining the source's PaymentsHash.Node visibility is valid.^[ref]