<p>Given an array of integers <code>temperatures</code> represents the daily temperatures, return <em>an array</em> <code>answer</code> <em>such that</em> <code>answer[i]</code> <em>is the number of days you have to wait after the</em> <code>i<sup>th</sup></code> <em>day to get a warmer temperature</em>. If there is no future day for which this is possible, keep <code>answer[i] == 0</code> instead.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> temperatures = [73,74,75,71,69,72,76,73]
<strong>Output:</strong> [1,1,4,2,1,1,0,0]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,40,50,60]
<strong>Output:</strong> [1,1,1,0]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,60,90]
<strong>Output:</strong> [1,1,0]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;temperatures.length &lt;= 10<sup>5</sup></code></li>
	<li><code>30 &lt;=&nbsp;temperatures[i] &lt;= 100</code></li>
</ul>

typedef pair<int,int>pi;
class Solution {
public:

    vector<int> dailyTemperatures(vector<int>& temp) {
        int n=temp.size();
        vector<int>ans(n,0);
        priority_queue<pi,vector<pi>,greater<pi>>pq;
        pq.push({n-1,temp[n-1]});
        for(int i=n-1;i>=0;i--){
                while(!pq.empty()&&pq.top().second<=temp[i]){
                    pq.pop();
                }
                if(!pq.empty()){
                    ans[i]=pq.top().first-i;
                }
                pq.push({i,temp[i]});
        }
        return ans;
    }
};
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