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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/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 / 2021 / Imperative programming / 2 / b / Solution by

Codex  0 2026-10-25

Compare every unordered pair of distinct starting positions:

def longestRepeated(): (Int, Int, Int) = {
  require(N >= 2)
  var besti = 0
  var bestj = 1
  var bestk = 0
  var i = 0
  while (i < N - 1) {
    var j = i + 1
    while (j < N) {
      val k = longestFrom(i, j)
      if (k > bestk) {
        besti = i; bestj = j; bestk = k
      }
      j += 1
    }
    i += 1
  }
  (besti, bestj, bestk)
}

After each comparison, the stored triple attains the largest common-prefix length among pairs processed so far; its positions remain distinct. The nested while loops enumerate exactly

N (N - 1) /2

pairs. Reversing the two positions does not change their common-prefix length, so no omitted ordered pair could improve the answer. Initialization also supplies a valid length-zero answer if no character repeats.

Write

K

for the final maximal length. Each comparison takes

O (K + 1)

time, including a possible first mismatch. Hence time is $O (N^{2} (K + 1))$ , at worst $O (N^{3})$ , and auxiliary space is

O (1)

. Writing merely

O (N^{2} K)

would miss the

K = 0

case. The explicit precondition excludes arrays with fewer than two valid starting positions; the requested distinct-position triple has no answer there under this domain convention.

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