How Would You Find the First Sorted Position Meeting a Lower-Bound Condition?

Invariant first: Define the answer is in [lo, hi). Everything left of lo is below target; everything from hi on is at least target. When lo equals hi, that is the first position at least target.

mid = (lo+hi)//2 has no overflow issue in Python, but be precise about updates: if a[mid] is at least target, hi = mid; else lo = mid + 1. Infinite loops almost always come from flipped boundary updates.

Specify the variant before coding: first at least, first above, last at most - they differ only in the comparator and which end you return. Interviewers rotate through them.

Close by testing: empty array, all smaller, all larger, duplicates - walk through them verbally.