Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/imperative-programming/3/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Imperative programming / 3 / e / Solution by Codex 0 2026-10-31
An already-zero entry needs no new linked list node:
def zeroElement(vector: Node, index: Int): Unit = {
  require(vector != null)
  require(0 <= index && index < vector.index)
  var p = vector.next
  while ((p ne vector) && p.index < index) p = p.next
  if ((p ne vector) && p.index == index) {
    val left = p.prev
    val right = p.next
    left.next = right
    right.prev = left
  }
}
The full input representation invariant is a precondition. The search has the same loop invariant and loop variant as in part (d). At exit, either p is the only stored entry with that index or none exists. In the first case, bypassing it in both directions removes exactly that coordinate and preserves the order of all remaining entries. This also works when its two neighbors are the same list header, leaving a self-linked empty structure. The header itself is never removed. The requested coordinate is zero and all others are unchanged, in time complexity.
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