<p>You are given an integer <code>n</code> and an array of <strong>unique</strong> integers <code>blacklist</code>. Design an algorithm to pick a random integer in the range <code>[0, n - 1]</code> that is <strong>not</strong> in <code>blacklist</code>. Any integer that is in the mentioned range and not in <code>blacklist</code> should be <strong>equally likely</strong> to be returned.</p>

<p>Optimize your algorithm such that it minimizes the number of calls to the <strong>built-in</strong> random function of your language.</p>

<p>Implement the <code>Solution</code> class:</p>

<ul>
	<li><code>Solution(int n, int[] blacklist)</code> Initializes the object with the integer <code>n</code> and the blacklisted integers <code>blacklist</code>.</li>
	<li><code>int pick()</code> Returns a random integer in the range <code>[0, n - 1]</code> and not in <code>blacklist</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;Solution&quot;, &quot;pick&quot;, &quot;pick&quot;, &quot;pick&quot;, &quot;pick&quot;, &quot;pick&quot;, &quot;pick&quot;, &quot;pick&quot;]
[[7, [2, 3, 5]], [], [], [], [], [], [], []]
<strong>Output</strong>
[null, 0, 4, 1, 6, 1, 0, 4]

<strong>Explanation</strong>
Solution solution = new Solution(7, [2, 3, 5]);
solution.pick(); // return 0, any integer from [0,1,4,6] should be ok. Note that for every call of pick,
                 // 0, 1, 4, and 6 must be equally likely to be returned (i.e., with probability 1/4).
solution.pick(); // return 4
solution.pick(); // return 1
solution.pick(); // return 6
solution.pick(); // return 1
solution.pick(); // return 0
solution.pick(); // return 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= blacklist.length &lt;= min(10<sup>5</sup>, n - 1)</code></li>
	<li><code>0 &lt;= blacklist[i] &lt; n</code></li>
	<li>All the values of <code>blacklist</code> are <strong>unique</strong>.</li>
	<li>At most <code>2 * 10<sup>4</sup></code> calls will be made to <code>pick</code>.</li>
</ul>

class Solution {
public:

    vector<vector<int>>arr;
    int flag=1;
    int ttt;
    Solution(int n, vector<int>& tmp) 
    {     ttt=n;
          sort(tmp.begin(),tmp.end());
       
        if(tmp.size()){
        if(tmp[0]==0)
        {
            if(tmp.size()>1){
                if(tmp[1]>=2)
                arr.push_back({0,tmp[1]});
            }
            else{
                arr.push_back({0,n});
            }
        }
        else{

            arr.push_back({-1,tmp[0]});
        }

        if(tmp[tmp.size()-1]==n-1)
        {
               if(tmp.size()>1){
               if( tmp[tmp.size()-1]<=n-3)
                arr.push_back({tmp[tmp.size()-1],n-1});
            }
            else{
                arr.push_back({-1,n-1});
            } 
        }
        else
        {   
            if(tmp[tmp.size()-1]<=n-2)
            arr.push_back({tmp[tmp.size()-1],n});
        }

        for(int i=0;i<tmp.size()-1;i++){
            if(tmp[i+1]>=tmp[i]+2)
            arr.push_back({tmp[i],tmp[i+1]});
        }}
        else{
             flag=0;
        }
           
        sort(arr.begin(),arr.end());
    }
    
    int pick() {
        
           if(flag){
           int ind=rand()%arr.size();
           int a=arr[ind][0],b=arr[ind][1];
           int ans=rand()%(b-a-1);
           return a+ans+1;}
           else{
                 return rand()%ttt;
           }
        
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(n, blacklist);
 * int param_1 = obj->pick();
 */
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