Advantage Shuffle
<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>
Greedy; Array; Two Pointers; Sorting
function advantageCount(nums1: number[], nums2: number[]): number[] {
    const n = nums1.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => nums2[i] - nums2[j]);
    nums1.sort((a, b) => a - b);

    const ans = new Array(n).fill(0);
    let left = 0;
    let right = n - 1;
    for (let i = 0; i < n; i++) {
        if (nums1[i] > nums2[idx[left]]) {
            ans[idx[left]] = nums1[i];
            left++;
        } else {
            ans[idx[right]] = nums1[i];
            right--;
        }
    }
    return ans;
}
