<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]:
        #sort nums2 arr and keep the original index
        num2_sort = []
        for i in range(len(nums2)):
            num2_sort.append((nums2[i], i))
        num2_sort.sort()
        #sort nums 1 also
        nums1.sort()

        num1_final = [-1] * len(nums1) #the permutation that maxmimize advantage
        remain_num = [] #leftover num that cannot fit
        i, j = 0, 0
        #for each element in nums2, greedily find nearest larger num in nums1 
        #and put it in nums2 element's original index for advantage
        while (i < len(nums1)):
            if nums1[i] > num2_sort[j][0]:
                idx = num2_sort[j][1]
                num1_final[idx] = nums1[i]
                j += 1
            else:
                remain_num.append(nums1[i])
            i += 1
        #we throw remaining num in nums1 to any position in nums2 since
        #it won't give us any advantage
        i = 0
        while remain_num:
            while num1_final[i] != -1:
                i += 1
            num1_final[i] = remain_num.pop()
        return num1_final


        
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