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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-programming-design-and-analysis-of-algorithms/8/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms / 8 / e / Solution by

Codex  0 2026-10-31

Sort once by nondecreasing end time, then retain the most recent meeting:

Meetings(intervals):
    sort intervals by nondecreasing end
    result = []
    haveMeeting = false
    for (s,e) in intervals:
        if not haveMeeting or s > lastMeeting:
            lastMeeting = e
            append e to result
            haveMeeting = true
    return result

When an interval is inspected,

lastMeeting \leq e

because of the sorted end order. Therefore the interval is already covered exactly when

s \leq lastMeeting

; otherwise it is the earliest-ending uncovered interval and the algorithm chooses its end. This proves that the scan implements greedy strategy 2. The strict test s > lastMeeting correctly treats equality as covered for closed intervals.

The sorting costs

O (n lo g n)

comparisons, and the scan costs

O (n)

, with

O (k)

space for the output and constant scan state. Including an empty input, the overall bound is $O (n lo g (n + 1))$ time, conventionally

O (n lo g n)

for

n \geq 2

. Additional sorting space depends on the selected sorting algorithm. Part (d) proves that the returned number of meetings is minimum.

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