Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/imperative-programming/4/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Imperative programming / 4 / c / Solution by Codex 0 2026-10-31
Use trie search with a letter multiset. Counts avoid generating the same result repeatedly when the input contains equal letters. The backtracking explores a letter value once, irrespective of how many indistinguishable copies remain.
def allWords(ls: String): List[String] = {
  val left = new Array[Int](26)
  var i = 0
  while (i < ls.length) {
    left(toIndex(ls.charAt(i))) += 1
    i += 1
  }
  var out = List.empty[String]

  def visit(t: Trie, prefix: String): Unit = {
    if (t.complete) out = prefix :: out
    var k = 0
    while (k < 26) {
      val child = t.children(k)
      if (left(k) > 0 && child != null) {
        left(k) -= 1
        visit(child, prefix + toChar(k))
        left(k) += 1
      }
      k += 1
    }
  }

  visit(this, "")
  out.reverse
}
At entry to visit, t is reached by prefix, and left(k) is the input multiplicity of letter minus its multiplicity in prefix. All these counts are nonnegative. Each recursive call consumes one available letter, and restoration makes sibling calls see the same counts as before that child. The recursion therefore terminates at depth at most , even when the dictionary contains longer words.
A completed prefix is emitted once, and the traversal continues so that longer words with the same prefix are also found. Every emitted word is in the dictionary and uses at most the available multiplicities. Conversely, every dictionary word meeting those bounds has an existing trie path whose next letter has positive remaining count at every step; the traversal reaches its terminal node. A trie has a unique root path for each word, and each child is visited at most once per reached node. Therefore
(1)
The boxed expression specifies the set of entries in the returned list, rather than its ordering. If the dictionary includes the empty string, the method includes it once. Counting costs ; visiting reachable nodes costs apart from string copying and output construction. The presented immutable-prefix construction additionally costs the sum of the prefix lengths it constructs. The recursion depth is at most , and the letter-count array has fixed size.
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