<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) {
        sort(nums1.begin(), nums1.end());
        int n = nums1.size();
//[2,7,11,15] => t t t t 
//[1,10,4,11]  
//2 11 7 15
        for(auto i: nums1) cout<<i<<" ";
        cout<<endl;
        int i = 0, j = 0;
        vector<bool> flags(n, false); 
        vector<int> ans;
        while( j < n) {

            int idx = upper_bound(nums1.begin(), nums1.end(), nums2[j]) - nums1.begin();
            cout<<idx<<endl;
            while(idx < n && flags[idx] == true) idx++;
           
            if(idx == n) {
                j++;
                ans.push_back(-1);
                continue;
            }
            ans.push_back(nums1[idx]);
            flags[idx] = true;
             j++;
        }
         i = 0;
        for(int j =0; j < ans.size(); j++) {
            
            if(ans[j] == -1) {
                cout<<"hello";
                while(i < n && flags[i]) i++;
                ans[j] = nums1[i];
                flags[i] = true;
            }
        }

        while(i<n) {
            while(i < n && flags[i]) i++;
            if(i < n)
{            ans.push_back(nums1[i]);
            flags[i] = true;}
        }
        return ans;
    }
};


//6475 13574 14234 15448 19893  => f t f t f 
//[14234,6475,19893,15448,13574]

//15448 13574 
63400