Next Greater Element II
<p>Given a circular integer array <code>nums</code> (i.e., the next element of <code>nums[nums.length - 1]</code> is <code>nums[0]</code>), return <em>the <strong>next greater number</strong> for every element in</em> <code>nums</code>.</p>

<p>The <strong>next greater number</strong> of a number <code>x</code> is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn&#39;t exist, return <code>-1</code> for this number.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1]
<strong>Output:</strong> [2,-1,2]
Explanation: The first 1&#39;s next greater number is 2; 
The number 2 can&#39;t find next greater number. 
The second 1&#39;s next greater number needs to search circularly, which is also 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,3]
<strong>Output:</strong> [2,3,4,-1,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Stack; Array; Monotonic Stack
class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [-1] * n
        stk = []
        for i in range(n * 2 - 1, -1, -1):
            i %= n
            while stk and stk[-1] <= nums[i]:
                stk.pop()
            if stk:
                ans[i] = stk[-1]
            stk.append(nums[i])
        return ans
