Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Imperative programming (resit) / 3 / c

Increasing here means nondecreasing, so equal neighbors remain in the same run. Change only the suffix-reset test:

best = 0; run = 0; bestStart = 0
for i = 0,...,n-1:
    if i == 0 or a[i] < a[i-1]: run = 1
    else: run = run+1
    if run > best:
        best = run; bestStart = i-run+1
return (bestStart,best)

After index $i$, run is the longest nondecreasing suffix of the prefix and the recorded interval is a longest nondecreasing segment seen so far. The adjacent-comparison condition implies the all-pairs condition by transitivity. The scan uses $O(n)$ time and $O(1)$ auxiliary space. Empty input yields length zero because no nonempty increasing segment exists.