left
to 0 and prod
to 1.ans
to 0 to store the count of valid subarrays.right
pointer.prod
by the current element nums[right]
.prod
is greater than or equal to k
, divide prod
by nums[left]
and increment left
.right - left + 1
to ans
to include all valid subarrays ending at right
.ans
as the final result.