 Solution

Use binary search on an interval with unequal endpoint values. The array need not be sorted:The loop invariant isIt holds initially by the given endpoint inequality, which also implies $n\ge 1$. When $r-l>1$, $l<m<r$. If $b[l]\ne b[m]$, replacing $r$ maintains unequal endpoints. Otherwise $b[m]=b[l]\ne b[r]$, so replacing $l$ maintains them. This does not require monotonicity or even just two possible values.

The positive integer width $r-l$ is a loop variant, reduced to at most its old half rounded up. On exit $r=l+1$, givingThere are at most $\lceil {\log }_{2}n\rceil$ iterations. The additive constant handles $n=1$.