<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]:
        nums1_sorted=sorted(nums1)
        l=[]
        for i in range(len(nums1)):
            l.append((i,nums2[i]))

        nums2_sorted=sorted(l,key=lambda x:x[1] ,reverse=True)
        start=0
        end=len(nums1)-1
        ans=[0]*len(nums1)
        while (start<=end):
            if (nums2_sorted and nums1_sorted[end]>nums2_sorted[0][1]):
                ans[nums2_sorted[0][0]]=nums1_sorted[end]
                end-=1
                nums2_sorted.pop(0)
            else:
                ans[nums2_sorted[0][0]]=nums1_sorted[start]
                start+=1
                nums2_sorted.pop(0)
        return ans

        
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