Searching · stretch

The first one big enough

items is sorted, possibly with repeats. Write first_at_least(items, x) returning the index of the first item >= x, or len(items) if there is none.

It is binary search that keeps going after a match: a match might not be the first one.

right answers
arguments left as they should be
fast enough at scale
without bisect, bisect_left, bisect_right, index

def first_at_least(items, x):
    for i in range(len(items)):
        if items[i] >= x:
            return i
    return len(items)

Run adds print(first_at_least([1, 3, 3, 3, 8], 3)) after your code, to try it.

Stuck on the idea rather than the code? The Binary search, the useful kind lesson walks through it.