ans to 0 to store the count of valid subarrays.st to store indices of elements.nums using index i.
a. While the stack is not empty and nums[i] is less than nums[st.top()], pop elements from the stack.
b. For each popped element, add i - st.top() to ans.
c. Push the current index i into the stack.nums.size() - st.top() to ans.ans.