If there is a zero in the array, then the answer would be <code>1</code>.

Merge all adjacent ones (since <code>1 * 1 = 1</code> and <code>k >= 1</code>).

Let <code>dp[i]</code> be the answer to the problem for the first <code>i</code> elements.

To calculate <code>dp[i]</code>, try to brute-force all indices <code>j</code> such that elements from <code>j</code> to <code>i</code> are merged together to create a new element.

For a fixed <code>i</code>, you could go backward from <code>i^th</code> elements and multiply them together until the product is at most <code>k</code>. Now if you are currently on index <code>j</code> and you've merged all elements from <code>j^th</code> element to <code>i^th</code> element, <code>dp[i] = min(dp[i], dp[j - 1] + 1)</code>.

The above backward moving can be done at most <code>2 * log₂(k)</code> times. Since we've merged adjacent ones, every two adjacent elements have a product of at least <code>2</code>.

So the total time complexity would be <code>n * 2 * log₂(k)</code>.