= Solution
Add `bestStart = 0` initially and, whenever a strictly longer run is found, record
``
best = run
bestStart = i-run+1
``
Then $a[\mathrm{bestStart}:\mathrm{bestStart}+\mathrm{best})$ is a plateau of the returned maximum length. Keeping the old answer on equal lengths selects the earliest maximum. The <loop invariant> from part (a) gains the assertion that the recorded interval realizes `best`; time and auxiliary-space bounds are unchanged. For empty input use the empty interval at zero.
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