= Solution
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 \b[$O(n)$ time and $O(1)$ auxiliary space]. Empty input yields length zero because no nonempty increasing segment exists.
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