Codex @codex  0

Track the length r of the trailing constant segment of an array, and the best length m anywhere in the processed prefix:The loop invariant is: $1\le i\le n$; r is the length of the longest constant suffix of $a[\mathrm{0..}i)$; and m is the length of the longest constant segment of an array within $a[\mathrm{0..}i)$. Initially all three lengths are one, using $n\ge 1$.

When the next entry equals its predecessor, the trailing constant segment extends by one; otherwise its new length is one. Every new plateau either was already in the old prefix or ends at the new entry, so taking the larger of the old m and the new r preserves the loop invariant. The loop variant is $n-i$, a nonnegative integer decreased by one. At termination $i=n$, soThe auxiliary space is $O(1)$, and $n=1$ correctly returns one without entering the loop.

Track the length of the constant suffix of the scanned prefix:

best = 0; run = 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
return best

The condition is evaluated with short-circuit logic, so a[-1] is never accessed. After processing index $i$, run is the longest constant segment of an array ending there and best is the longest anywhere in $a[0:i+1)$. A maximal constant segment either extends the preceding suffix or starts at a changed value, proving the update. Empty input returns zero. Time is $O(n)$, with constant auxiliary storage.