Here is a cleaner and better way to solve the problem.

```lang-Python
# Return the smallest index where the element is bigger than `A[start_index]`.
# If `len(A)` is returned, no element is bigger than `A[start_index]`.
def next_bigger_element(start_index, A):
    lo, hi = start_index, len(A)
    while lo + 1 < hi:
        mid = (lo + hi) // 2
        if A[mid] == A[start_index]:
            lo = mid
        else:
            hi = mid
    return hi


def distinct_elements_at_least(k, A):
    if len(A) == 0:
        return k <= 0
    cur_index = 0
    count = 1
    # keep finding the next bigger element as long as less than `k` distinct
    # elements has been found and the end of the array has not been reached.
    while count < k and A[cur_index] != A[-1]:
        cur_index = next_bigger_element(cur_index, A)
        count += 1
    return count >= k
```

To find whether `A` contains at least 4 distinct elements, just call `distinct_elements_at_least(4, A)`.

This program works correctly for any given number `k`. For example, it can be used to check whether `A` has 0 element <!-- alway true --> or whether `A` has 7 distinct elements. For any fixed `k`, it works in $O(\log n)$ as desired.

------

If you do not mind `import bisect`, you may prefer the following shorter code, since method `next_bigger_element` is no longer needed.

```lang-Python
from bisect import bisect_right

def distinct_elements_at_least(k, A):
    if len(A) == 0:
        return k <= 0
    cur = 0
    count = 0
    while cur < len(A) and count < k:
        count += 1
        cur = bisect_right(A, A[cur], cur + 1)
    return count >= k
```