 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/imperative-programming/2/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming / 2 / b / Solution by

Codex  0 2026-10-25

Set aside the selected pivot at the end, array partitioning the other entries, then move the pivot to the separation index:

def partition(l: Int, r: Int): Int = {
  require(0 <= l && l < r && r <= a.length)
  def swap(i: Int, j: Int): Unit = {
    val t = a(i); a(i) = a(j); a(j) = t
  }
  val chosen = bestIndex(l, r)
  val x = a(chosen)
  swap(chosen, r - 1)
  var p = l
  var j = l
  while (j < r - 1) {
    if (a(j) < x) {
      swap(p, j)
      p += 1
    }
    j += 1
  }
  swap(p, r - 1)
  p
}

The loop invariant is

l \leq p \leq j \leq r - 1

a [l, p) < x

a [p, j) \geq x

, and

a (r - 1) = x

. The segment's multiset is unchanged and all entries outside it are unchanged. A low entry is swapped into the low segment; a high entry simply extends the high segment. The loop variant is

r - 1 - j

. At exit, exchanging the reserved pivot with a(p) gives $a [l, p) < x$ , $a (p) = x$ , and $a [p + 1, r) \geq x$ . The displaced value at the old pivot position was already at least

x

. Length one is handled by self-swaps and an empty loop. Duplicate pivot values belong to the high side. This costs

Θ (r - l)

time and

O (1)

auxiliary space.

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