<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 {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        vector<int> ans;
        if (nums1.empty())
            return ans;
        sort(nums1.begin(), nums1.end());
        vector<bool> vis(nums1.size(), false);
        int notVisitedHead = 0;
        for (int i=0; i<nums2.size(); i++) {
            bool found = false;
            int start = notVisitedHead;
            int end = nums1.size() - 1;
            while (start < end) {
                int mid = (start + end)/2;
                if (nums1[mid] <= nums2[i]) {
                    start = mid + 1;
                }
                else {
                    end = mid;
                }
            }
            for (int j=start; j<nums1.size(); j++) {
                if (vis[j] == true)
                    continue;
                if (nums1[j] > nums2[i]) {
                    vis[j] = true;
                    found = true;
                    ans.push_back(nums1[j]);
                    if (j == notVisitedHead) {
                        while(notVisitedHead < nums1.size() && vis[notVisitedHead] == true)
                        notVisitedHead++;
                    }
                    break;
                }
            }
            if (found == true)
                continue;
            ans.push_back(nums1[notVisitedHead]);
            vis[notVisitedHead] = true;
            while(notVisitedHead < nums1.size() && vis[notVisitedHead] == true)
                        notVisitedHead++;
        }
        return ans;
    }
};
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