Leftmost Big Enough — Solution & Editorial

Back to the Problem

A max segment tree answers this WITHOUT binary search: descend from the root, always preferring the left child when its max is ≥ x — O(log N) per query. (Binary search over prefix maxima costs an extra log.)

Complexity: O((N + Q) log N)

Watch out: The tree descent — not a scan — is the intended trick; a per-query scan is O(NQ).