<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> both of the same length. The <strong>advantage</strong> of <code>nums1</code> with respect to <code>nums2</code> is the number of indices <code>i</code> for which <code>nums1[i] &gt; nums2[i]</code>.</p>

<p>Return <em>any permutation of </em><code>nums1</code><em> that maximizes its <strong>advantage</strong> with respect to </em><code>nums2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums1 = [2,7,11,15], nums2 = [1,10,4,11]
<strong>Output:</strong> [2,11,7,15]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums1 = [12,24,8,32], nums2 = [13,25,32,11]
<strong>Output:</strong> [24,32,8,12]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums2.length == nums1.length</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
        # 1. Node(value, source:nums1 or nums2)
        # 先nums1, nums2合并生成node, 按node value排序, 然后剔除nums2?

        # 2. 对于nums2, 找next greater element in nums1?
        if max(nums1) < min(nums2) or min(nums1) > max(nums2):
            return nums1
        nums1 = deque(sorted(nums1))
        res = [0] * len(nums1)
        # pq = nums1
        # heapq.heapify(pq)
        # nums2 from big to small
        leftindex = 0
        for b, i in sorted([-b, i] for i, b in enumerate(nums2)):
            candidateFromNums1 = nums1[-1]
            if -b < candidateFromNums1:
                res[i] = nums1.pop()
                # nums2[i] = nums1.pop()
            else:
                res[i] = nums1[leftindex]
                leftindex += 1
                # nums2[i] = nums1.popleft()
        return res
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