<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) {
        int n = temperatures.size();
        vector<int> ans(n);

        ans[0] = 0;
        reverse(temperatures.begin(), temperatures.end());
        stack<int> s;
        s.push(temperatures[0]);
        vector<int> v(1001);

        for (int i = 1; i < n; i++) {
            int k = 0;
            while (!s.empty() && s.top() <= temperatures[i]) {
                k++;

                s.pop();
            }

            if (!s.empty())
                ans[i] = i - v[s.top()];
            else
                ans[i] = 0;
            s.push(temperatures[i]);
            v[temperatures[i]] = i;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    
}
};

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