<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>

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {

        stack<int>st;
        
        int n=temperatures.size();
        int output[n];
        
        if(temperatures.size()){
            output[n-1]=0;
            st.push(n-1);
        }

        for(int i=temperatures.size()-2;i>=0;i--){            
            while(!st.empty() && (temperatures[i]>=temperatures[st.top()])){
                    st.pop();
            }
            if(st.empty()){
             output[i]=0;
            }else if(temperatures[i]<temperatures[st.top()]){
                int index =st.top();
                int len=index-i;
                output[i]=len;
            }
            st.push(i);
        }   
        vector<int>finals;
        for(int i=0;i<n;i++){
            finals.push_back(output[i]);
        }
        return finals;
    }
};
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