<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 {
    const bool victor = [](){
        iostream::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        return true;
    }();
public:
    vector<int>&& dailyTemperatures(vector<int>& temps) {
        vector<int> stack;
        const int N = temps.size();

        for (int i = 0; i < N; ++i) {
            while (stack.size() && temps[i] > temps[stack.back()]) {
                temps[stack.back()] = i - stack.back();
                stack.pop_back();
            }
            stack.emplace_back(i);
        }

        while (stack.size()) {
            temps[stack.back()] = 0;
            stack.pop_back();
        }
        return move(temps);
    }
};
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