Find the Smallest Divisor Given a Threshold

Search the divisor, not the array. For a candidate d, compute sum over nums of ceil(nums[i]/d). As d increases, every term stays the same or decreases, so feasibility sum less than or equal to threshold is monotonic. Binary-search the smallest feasible d between one and the maximum number. Compute ceilings as (x+d-1)/d only when overflow cannot occur; otherwise use x/d plus an adjustment for a nonzero remainder. Stop accumulating as soon as the sum exceeds threshold, which avoids useless work and reduces overflow risk. If threshold is smaller than the array length, report impossibility when each positive element contributes at least one, if the problem contract permits that case. Test one value, divisor one, max divisor, exact threshold, and large values. The time is O(n log M), where M is the maximum element, with O(1) extra space.